MT2175: Complex Matrices and Vector Spaces

A complex number \(z\) can be expressed in three forms:

1. Cartesian Form: \(z = a + ib\), where \(a = \text{Re}(z)\) is the real part and \(b = \text{Im}(z)\) is the imaginary part.
2. Polar Form: \(z = r(\cos\theta + i\sin\theta)\), where \(r = |z|\) is the modulus (or absolute value) and \(\theta = \arg(z)\) is the argument (or angle).
3. Exponential Form: \(z = re^{i\theta}\), which is derived from the polar form using Euler's formula, \(e^{i\theta} = \cos\theta + i\sin\theta\).

For any real number \(\theta\) and any integer \(n\), De Moivre's Formula states: \[ (\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta) \] In exponential form, this is equivalent to \((e^{i\theta})^n = e^{in\theta}\). This formula is extremely useful for finding powers and roots of complex numbers.

A complex number \(z\) has \(n\) distinct \(n\)-th roots. If \(z = r(\cos\theta + i\sin\theta)\), its \(n\)-th roots are given by: \[ w_k = \sqrt[n]{r} \left( \cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right) \right) \] for \(k = 0, 1, 2, \dots, n-1\). Geometrically, the roots lie on a circle of radius \(\sqrt[n]{r}\) and are equally spaced by an angle of \(2\pi/n\).

A complex vector space is a vector space where the scalars are complex numbers (from \(\mathbb{C}\)) instead of real numbers. It must satisfy the same ten vector space axioms as a real vector space, including closure under addition and scalar multiplication, associativity, commutativity, existence of a zero vector, additive inverses, and distributive properties. The space \(\mathbb{C}^n\) is a key example of an \(n\)-dimensional complex vector space.

A complex matrix is simply a matrix whose entries are complex numbers. All the standard matrix operations like addition, scalar multiplication (with complex scalars), and matrix multiplication apply to complex matrices in the same way they do to real matrices.

For two vectors \(\mathbf{x} = (x_1, \dots, x_n)\) and \(\mathbf{y} = (y_1, \dots, y_n)\) in \(\mathbb{C}^n\), the standard complex inner product is defined as: \[ \langle \mathbf{x}, \mathbf{y} \rangle = \sum_{i=1}^n x_i \bar{y}_i = x_1\bar{y}_1 + x_2\bar{y}_2 + \dots + x_n\bar{y}_n \] This can also be written in matrix form as \(\mathbf{y}^*\mathbf{x}\), where \(\mathbf{y}^*\) is the conjugate transpose (Hermitian conjugate) of \(\mathbf{y}\).

A complex inner product \(\langle \cdot, \cdot \rangle\) on a vector space \(V\) must satisfy the following properties for all \(\mathbf{x}, \mathbf{y}, \mathbf{z} \in V\) and \(\alpha \in \mathbb{C}\):

1. Conjugate Symmetry: \(\langle \mathbf{x}, \mathbf{y} \rangle = \overline{\langle \mathbf{y}, \mathbf{x} \rangle}\)
2. Linearity in the first argument: \(\langle \alpha\mathbf{x} + \mathbf{y}, \mathbf{z} \rangle = \alpha\langle \mathbf{x}, \mathbf{z} \rangle + \langle \mathbf{y}, \mathbf{z} \rangle\)
3. Positive-definiteness: \(\langle \mathbf{x}, \mathbf{x} \rangle \ge 0\) and \(\langle \mathbf{x}, \mathbf{x} \rangle = 0 \iff \mathbf{x} = \mathbf{0}\).

Note that linearity is only in the first argument. For the second argument, it is conjugate-linear: \(\langle \mathbf{x}, \alpha\mathbf{y} \rangle = \bar{\alpha}\langle \mathbf{x}, \mathbf{y} \rangle\).

The Hermitian conjugate (or conjugate transpose) of a complex matrix \(A\), denoted \(A^*\), is obtained by taking the transpose of the matrix and then taking the complex conjugate of each entry. \[ A^* = \bar{A}^T \] For example, if \(A = \begin{pmatrix} 1+i & 2 \\ 3i & 4-i \end{pmatrix}\), then \(A^* = \begin{pmatrix} 1-i & -3i \\ 2 & 4+i \end{pmatrix}\).

Let \(A\) be a Hermitian matrix (\(A = A^*\)) with eigenvalue \(\lambda\) and corresponding eigenvector \(\mathbf{v} \neq \mathbf{0}\). So, \(A\mathbf{v} = \lambda\mathbf{v}\).
Consider the inner product \(\langle A\mathbf{v}, \mathbf{v} \rangle\). \[ \langle A\mathbf{v}, \mathbf{v} \rangle = \langle \lambda\mathbf{v}, \mathbf{v} \rangle = \lambda \langle \mathbf{v}, \mathbf{v} \rangle = \lambda ||\mathbf{v}||^2 \] Also, using the property \(\langle A\mathbf{x}, \mathbf{y} \rangle = \langle \mathbf{x}, A^*\mathbf{y} \rangle\), we have: \[ \langle A\mathbf{v}, \mathbf{v} \rangle = \langle \mathbf{v}, A^*\mathbf{v} \rangle = \langle \mathbf{v}, A\mathbf{v} \rangle = \langle \mathbf{v}, \lambda\mathbf{v} \rangle = \bar{\lambda} \langle \mathbf{v}, \mathbf{v} \rangle = \bar{\lambda} ||\mathbf{v}||^2 \] Equating the two expressions gives \(\lambda ||\mathbf{v}||^2 = \bar{\lambda} ||\mathbf{v}||^2\). Since \(\mathbf{v} \neq \mathbf{0}\), \(||\mathbf{v}||^2 \neq 0\), so we can divide by it to get \(\lambda = \bar{\lambda}\). This implies that \(\lambda\) must be a real number.

Let \(A\) be a Hermitian matrix. Let \(\lambda_1\) and \(\lambda_2\) be two distinct eigenvalues (\(\lambda_1 \neq \lambda_2\)) with corresponding eigenvectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\). We know \(A\mathbf{v}_1 = \lambda_1\mathbf{v}_1\) and \(A\mathbf{v}_2 = \lambda_2\mathbf{v}_2\). Consider \(\langle A\mathbf{v}_1, \mathbf{v}_2 \rangle\): \[ \langle A\mathbf{v}_1, \mathbf{v}_2 \rangle = \langle \lambda_1\mathbf{v}_1, \mathbf{v}_2 \rangle = \lambda_1 \langle \mathbf{v}_1, \mathbf{v}_2 \rangle \] Also, \(\langle A\mathbf{v}_1, \mathbf{v}_2 \rangle = \langle \mathbf{v}_1, A^*\mathbf{v}_2 \rangle = \langle \mathbf{v}_1, A\mathbf{v}_2 \rangle = \langle \mathbf{v}_1, \lambda_2\mathbf{v}_2 \rangle = \bar{\lambda}_2 \langle \mathbf{v}_1, \mathbf{v}_2 \rangle\). Since eigenvalues of a Hermitian matrix are real, \(\bar{\lambda}_2 = \lambda_2\). So, \(\lambda_1 \langle \mathbf{v}_1, \mathbf{v}_2 \rangle = \lambda_2 \langle \mathbf{v}_1, \mathbf{v}_2 \rangle\). \[ (\lambda_1 - \lambda_2) \langle \mathbf{v}_1, \mathbf{v}_2 \rangle = 0 \] Since \(\lambda_1 \neq \lambda_2\), we must have \(\langle \mathbf{v}_1, \mathbf{v}_2 \rangle = 0\). Thus, the eigenvectors are orthogonal.

Let \(U\) be an \(n \times n\) matrix with columns \(\mathbf{u}_1, \mathbf{u}_2, \dots, \mathbf{u}_n\). The matrix \(U^*\) has rows \(\mathbf{u}_1^*, \mathbf{u}_2^*, \dots, \mathbf{u}_n^*\). The product \(U^*U\) is an \(n \times n\) matrix where the entry in the \(i\)-th row and \(j\)-th column is \((\mathbf{u}_i^* \mathbf{u}_j) = \langle \mathbf{u}_j, \mathbf{u}_i \rangle\). \[ (U^*U)_{ij} = \langle \mathbf{u}_j, \mathbf{u}_i \rangle \] The matrix \(U\) is unitary if and only if \(U^*U = I\). This is equivalent to the condition that the entries of the product matrix are \((U^*U)_{ij} = \delta_{ij}\) (the Kronecker delta). So, \(\langle \mathbf{u}_j, \mathbf{u}_i \rangle = 1\) if \(i=j\) and \(\langle \mathbf{u}_j, \mathbf{u}_i \rangle = 0\) if \(i \neq j\). This is precisely the definition of the set of vectors \(\{\mathbf{u}_1, \dots, \mathbf{u}_n\}\) being an orthonormal set. Since there are \(n\) such vectors in \(\mathbb{C}^n\), they form an orthonormal basis.

A square complex matrix \(A\) is said to be unitarily diagonalisable if there exists a unitary matrix \(U\) such that \(U^*AU = D\), where \(D\) is a diagonal matrix.
The columns of the unitary matrix \(U\) are an orthonormal basis of eigenvectors for \(A\), and the diagonal entries of \(D\) are the corresponding eigenvalues.

An \(n \times n\) complex matrix \(A\) is called normal if it commutes with its Hermitian conjugate. That is, \[ AA^* = A^*A \] This class of matrices is important because a matrix is unitarily diagonalisable if and only if it is normal.

Let \(A\) be a Hermitian matrix. By definition, \(A = A^*\). We need to show that \(AA^* = A^*A\).
Starting with the left side: \[ AA^* = A(A) = A^2 \] Starting with the right side: \[ A^*A = (A)A = A^2 \] Since both sides are equal to \(A^2\), we have \(AA^* = A^*A\). Therefore, any Hermitian matrix is normal.

Let \(U\) be a unitary matrix. By definition, \(UU^* = U^*U = I\). We need to show that \(UU^* = U^*U\).
This is true by the very definition of a unitary matrix. Both \(UU^*\) and \(U^*U\) are equal to the identity matrix \(I\), so they are equal to each other. Therefore, any unitary matrix is normal.

A square complex matrix \(A\) is unitarily diagonalisable if and only if it is a normal matrix.
This is a fundamental result in linear algebra. It means that the set of matrices that can be diagonalised using a unitary transformation (which preserves lengths and angles) is precisely the set of normal matrices.

If \(A\) is a normal matrix, it is unitarily diagonalisable. Let \(\lambda_1, \dots, \lambda_n\) be its eigenvalues and \(\mathbf{u}_1, \dots, \mathbf{u}_n\) be a corresponding orthonormal basis of eigenvectors. The spectral decomposition of \(A\) is the expression: \[ A = \lambda_1 \mathbf{u}_1\mathbf{u}_1^* + \lambda_2 \mathbf{u}_2\mathbf{u}_2^* + \dots + \lambda_n \mathbf{u}_n\mathbf{u}_n^* \] This can be written as \(A = \sum_{i=1}^n \lambda_i E_i\), where \(E_i = \mathbf{u}_i\mathbf{u}_i^*\) is the matrix that represents the orthogonal projection onto the subspace spanned by the eigenvector \(\mathbf{u}_i\).

The matrices \(E_i = \mathbf{u}_i\mathbf{u}_i^*\) in the spectral decomposition of a normal matrix have the following properties:

1. They are Hermitian: \(E_i^* = (\mathbf{u}_i\mathbf{u}_i^*)^* = (\mathbf{u}_i^*)^*\mathbf{u}_i^* = \mathbf{u}_i\mathbf{u}_i^* = E_i\).
2. They are idempotent (they are projections): \(E_i^2 = E_i\).
3. They are mutually orthogonal: \(E_i E_j = 0\) for \(i \neq j\).
4. They sum to the identity: \(\sum_{i=1}^n E_i = I\).

If \(A = \sum \lambda_i E_i\) is the spectral decomposition of \(A\), then a square root \(B\) can be found by taking the square root of the eigenvalues: \[ B = \sum \sqrt{\lambda_i} E_i \] where \(\sqrt{\lambda_i}\) can be any of the two complex square roots of \(\lambda_i\). This gives \(2^n\) possible square roots in general. Then, \(B^2 = (\sum \sqrt{\lambda_i} E_i)^2 = \sum (\sqrt{\lambda_i})^2 E_i^2 = \sum \lambda_i E_i = A\), using the properties \(E_i E_j = 0\) for \(i \neq j\) and \(E_i^2 = E_i\).

The Gram-Schmidt process transforms a linearly independent set \(\{\mathbf{v}_i\}\) into an orthonormal set \(\{\mathbf{u}_i\}\) that spans the same subspace.

1. Set \(\mathbf{w}_1 = \mathbf{v}_1\). Normalize to get \(\mathbf{u}_1 = \frac{\mathbf{w}_1}{||\mathbf{w}_1||}\).
2. Project \(\mathbf{v}_2\) onto \(\mathbf{u}_1\) and subtract it: \(\mathbf{w}_2 = \mathbf{v}_2 - \langle \mathbf{v}_2, \mathbf{u}_1 \rangle \mathbf{u}_1\). Normalize to get \(\mathbf{u}_2 = \frac{\mathbf{w}_2}{||\mathbf{w}_2||}\).
3. For \(\mathbf{v}_k\), subtract its projections onto all previous orthonormal vectors: \(\mathbf{w}_k = \mathbf{v}_k - \sum_{j=1}^{k-1} \langle \mathbf{v}_k, \mathbf{u}_j \rangle \mathbf{u}_j\). Normalize to get \(\mathbf{u}_k = \frac{\mathbf{w}_k}{||\mathbf{w}_k||}\).

The determinant of a square matrix \(A\) is the product of its eigenvalues (counting multiplicities). \[ \det(A) = \lambda_1 \lambda_2 \cdots \lambda_n \] This is true for both real and complex matrices. It can be seen by considering the characteristic polynomial \(p(\lambda) = \det(A - \lambda I)\). Setting \(\lambda=0\) gives \(p(0) = \det(A)\). Also, from the factored form \(p(\lambda) = (-1)^n(\lambda - \lambda_1)\cdots(\lambda - \lambda_n)\), setting \(\lambda=0\) gives \(p(0) = (-1)^n(-\lambda_1)\cdots(-\lambda_n) = \lambda_1\cdots\lambda_n\).

The trace of a square matrix \(A\), denoted \(\text{tr}(A)\), is the sum of its diagonal entries. It is also equal to the sum of its eigenvalues (counting multiplicities). \[ \text{tr}(A) = \sum_{i=1}^n a_{ii} = \sum_{i=1}^n \lambda_i \] This can be shown by examining the coefficient of the \(\lambda^{n-1}\) term in the characteristic polynomial.

We want to prove that \(U\) is unitary \(\iff \langle U\mathbf{x}, U\mathbf{y} \rangle = \langle \mathbf{x}, \mathbf{y} \rangle\) for all \(\mathbf{x}, \mathbf{y} \in \mathbb{C}^n\).
(\(\Rightarrow\)) Assume \(U\) is unitary, so \(U^*U = I\). Then \[ \langle U\mathbf{x}, U\mathbf{y} \rangle = (U\mathbf{y})^*(U\mathbf{x}) = (\mathbf{y}^*U^*) (U\mathbf{x}) = \mathbf{y}^*(U^*U)\mathbf{x} = \mathbf{y}^*I\mathbf{x} = \mathbf{y}^*\mathbf{x} = \langle \mathbf{x}, \mathbf{y} \rangle \] (\(\Leftarrow\)) Assume \(\langle U\mathbf{x}, U\mathbf{y} \rangle = \langle \mathbf{x}, \mathbf{y} \rangle\) for all \(\mathbf{x}, \mathbf{y}\). This means \(\mathbf{y}^*U^*U\mathbf{x} = \mathbf{y}^*\mathbf{x}\). Let \(\mathbf{y} = \mathbf{e}_i\) and \(\mathbf{x} = \mathbf{e}_j\) (standard basis vectors). Then \(\mathbf{e}_i^* (U^*U) \mathbf{e}_j = (U^*U)_{ij}\) and \(\mathbf{e}_i^*\mathbf{e}_j = \delta_{ij}\). Therefore, \((U^*U)_{ij} = \delta_{ij}\), which means \(U^*U = I\). So \(U\) is unitary.

Yes. If \(A\) is unitarily diagonalisable, there exists a unitary matrix \(U\) and a diagonal matrix \(D\) such that \(A = UDU^*\). We need to show \(AA^* = A^*A\). \[ A^* = (UDU^*)^* = (U^*)^*D^*U^* = UD^*U^* \] (Note: \(D\) is diagonal, so \(D^* = \bar{D}^T = \bar{D}\). \[ AA^* = (UDU^*)(UD^*U^*) = UDD^*U^* \] \[ A^*A = (UDU^*)^*(UDU^*) = (UD^*U^*)(UDU^*) = UD^*DU^* \] Since diagonal matrices commute (\\(DD^* = D^*D\\)), we have \(AA^* = A^*A\). Thus, \(A\) is normal.

First, write \(-8i\) in exponential form. The modulus is \(r=8\). The angle is \(\theta = -\pi/2\). So, \(-8i = 8e^{i(-\pi/2 + 2k\pi)}\). We want to solve \(z^3 = 8e^{i(-\pi/2 + 2k\pi)}\). The roots are: \[ z_k = \sqrt[3]{8} e^{i\frac{-\pi/2 + 2k\pi}{3}} = 2e^{i(\frac{-\pi}{6} + \frac{2k\pi}{3})} \] For \(k=0, 1, 2\):

• \(k=0: z_0 = 2e^{-i\pi/6} = 2(\cos(-\pi/6) + i\sin(-\pi/6)) = 2(\frac{\sqrt{3}}{2} - \frac{1}{2}i) = \sqrt{3} - i\)
• \(k=1: z_1 = 2e^{i(-\pi/6 + 2\pi/3)} = 2e^{i\pi/2} = 2(0+i) = 2i\)
• \(k=2: z_2 = 2e^{i(-\pi/6 + 4\pi/3)} = 2e^{i7\pi/6} = 2(\cos(7\pi/6) + i\sin(7\pi/6)) = 2(-\frac{\sqrt{3}}{2} - \frac{1}{2}i) = -\sqrt{3} - i\)

1. Hermitian? We check if \(A=A^*\). \[ A^* = \begin{pmatrix} 1 & \overline{1-i} \\ \overline{1+i} & 2 \end{pmatrix} = \begin{pmatrix} 1 & 1+i \\ 1-i & 2 \end{pmatrix} = A \] Yes, \(A\) is Hermitian.
2. Normal? Since all Hermitian matrices are normal, \(A\) is normal.
3. Unitary? We check if \(AA^* = I\). Since \(A=A^*\), we check \(A^2=I\). \[ A^2 = \begin{pmatrix} 1 & 1+i \\ 1-i & 2 \end{pmatrix} \begin{pmatrix} 1 & 1+i \\ 1-i & 2 \end{pmatrix} = \begin{pmatrix} 1+(1+i)(1-i) & 1+i+2(1+i) \\ 1-i+2(1-i) & (1-i)(1+i)+4 \end{pmatrix} = \begin{pmatrix} 3 & \dots \\ \dots & \dots \end{pmatrix} \neq I \] No, \(A\) is not unitary.

The norm \(||\mathbf{v}||\) is calculated as \(\sqrt{\langle \mathbf{v}, \mathbf{v} \rangle}\). \[ \langle \mathbf{v}, \mathbf{v} \rangle = v_1\bar{v}_1 + v_2\bar{v}_2 + v_3\bar{v}_3 \] \[ = (1)(\overline{1}) + (i)(\overline{i}) + (1-i)(\overline{1-i}) \] \[ = (1)(1) + (i)(-i) + (1-i)(1+i) \] \[ = 1 - i^2 + (1 - i^2) = 1 + 1 + (1+1) = 4 \] Therefore, the norm is \(||\mathbf{v}|| = \sqrt{4} = 2\).

The subspace is a line. We just need to find a unit vector in the same direction as \(\mathbf{v}_1\). This is a 1-dimensional basis. First, find the norm of \(\mathbf{v}_1\): \[ ||\mathbf{v}_1|| = \sqrt{\langle \mathbf{v}_1, \mathbf{v}_1 \rangle} = \sqrt{1\cdot\bar{1} + i\cdot\bar{i}} = \sqrt{1(1) + i(-i)} = \sqrt{1+1} = \sqrt{2} \] The orthonormal basis is given by the single vector \(\mathbf{u}_1\): \[ \mathbf{u}_1 = \frac{\mathbf{v}_1}{||\mathbf{v}_1||} = \frac{1}{\sqrt{2}}(1, i) = \left(\frac{1}{\sqrt{2}}, \frac{i}{\sqrt{2}}\right) \] The basis is \(\left\{ \left(\frac{1}{\sqrt{2}}, \frac{i}{\sqrt{2}}\right) \right\}\).

The eigenvalues \(\lambda\) of a unitary matrix \(U\) all have a modulus of 1, i.e., \(|\lambda| = 1\). They lie on the unit circle in the complex plane. Proof: Let \(\mathbf{v}\) be an eigenvector for \(\lambda\). Then \(U\mathbf{v} = \lambda\mathbf{v}\). Since unitary transformations preserve the norm, we have \(||U\mathbf{v}|| = ||\mathbf{v}||\). So, \(||\lambda\mathbf{v}|| = |\lambda| \cdot ||\mathbf{v}|| = ||\mathbf{v}||\). Since \(\mathbf{v} \neq \mathbf{0}\), we can divide by \(||\mathbf{v}||\) to get \(|\lambda| = 1\).

1. Check if normal: \(A^* = \begin{pmatrix} 2 & i \\ -i & 2 \end{pmatrix} = A\). Since A is Hermitian, it is normal.
2. Find eigenvalues: \(\det(A - \lambda I) = (2-\lambda)^2 - (-i)(i) = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = (\lambda-3)(\lambda-1) = 0\). So \(\lambda_1=3, \lambda_2=1\).
3. Find eigenvectors: For \(\lambda_1=3\): \(A-3I = \begin{pmatrix} -1 & i \\ -i & -1 \end{pmatrix} \to \begin{pmatrix} 1 & -i \\ 0 & 0 \end{pmatrix}\). Eigenvector \(\mathbf{v}_1 = (i, 1)\). For \(\lambda_2=1\): \(A-I = \begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix} \to \begin{pmatrix} 1 & i \\ 0 & 0 \end{pmatrix}\). Eigenvector \(\mathbf{v}_2 = (-i, 1)\).
4. Orthonormalize: \(||\mathbf{v}_1|| = \sqrt{i(-i)+1(1)} = \sqrt{2}\). \(||\mathbf{v}_2|| = \sqrt{(-i)i+1(1)} = \sqrt{2}\). \(\mathbf{u}_1 = \frac{1}{\sqrt{2}}(i, 1)\), \(\mathbf{u}_2 = \frac{1}{\sqrt{2}}(-i, 1)\).
5. Form U and D: \(U = \frac{1}{\sqrt{2}}\begin{pmatrix} i & -i \\ 1 & 1 \end{pmatrix}\), \(D = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}\).

The process is almost identical, but the type of transformation matrix used is different.

• A real symmetric matrix \(A\) is diagonalised by a real orthogonal matrix \(P\) (where \(P^T = P^{-1}\)), such that \(P^TAP = D\).
• A complex Hermitian matrix \(A\) is diagonalised by a unitary matrix \(U\) (where \(U^* = U^{-1}\)), such that \(U^*AU = D\).

The concept of an orthogonal matrix is the real-valued special case of a unitary matrix. The process in both cases involves finding an orthonormal basis of eigenvectors.

If a normal matrix has real eigenvalues, then it must be a Hermitian matrix.
Proof Sketch: Since the matrix \(A\) is normal, it is unitarily diagonalisable, so \(A = UDU^*\) for some unitary \(U\) and diagonal \(D\). The diagonal entries of \(D\) are the eigenvalues of \(A\). If the eigenvalues are real, then \(D\) is a real matrix, which means \(D^* = D^T = D\). Now we check if \(A\) is Hermitian: \[ A^* = (UDU^*)^* = (U^*)^*D^*U^* = UDU^* = A \] Thus, \(A\) is Hermitian.

1. Check Normal: \(A^* = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\). \(AA^* = I\), \(A^*A = I\). So A is normal (it is unitary).
2. Eigen-problem: Eigenvalues are \(\lambda_1=i, \lambda_2=-i\). Corresponding eigenvectors are \(\mathbf{v}_1=(1, i), \mathbf{v}_2=(1, -i)\).
3. Orthonormalize: \(||\mathbf{v}_1|| = \sqrt{2}, ||\mathbf{v}_2|| = \sqrt{2}\). So \(\mathbf{u}_1 = \frac{1}{\sqrt{2}}(1, i), \mathbf{u}_2 = \frac{1}{\sqrt{2}}(1, -i)\).
4. Find Projection Matrices \(E_i = \mathbf{u}_i\mathbf{u}_i^*\): \[ E_1 = \frac{1}{2}\begin{pmatrix} 1 \\ i \end{pmatrix}\begin{pmatrix} 1 & -i \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix} \] \[ E_2 = \frac{1}{2}\begin{pmatrix} 1 \\ -i \end{pmatrix}\begin{pmatrix} 1 & i \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix} \] 5. Decomposition: \(A = \lambda_1 E_1 + \lambda_2 E_2 = i E_1 - i E_2\). \[ A = \frac{i}{2}\begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix} - \frac{i}{2}\begin{pmatrix} 1 & i \\ -i & 1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} i & 1 \\ -1 & i \end{pmatrix} - \frac{1}{2}\begin{pmatrix} i & -1 \\ 1 & i \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \]

Yes. A real matrix is diagonalisable if it has \(n\) linearly independent eigenvectors. It does not need to be symmetric.
However, a real matrix is orthogonally diagonalisable (i.e., with an orthogonal matrix \(P\) such that \(P^TAP=D\)) if and only if it is symmetric. This is a much stronger condition.

The key difference lies in the symmetry property.

• A real inner product is symmetric: \(\langle \mathbf{x}, \mathbf{y} \rangle = \langle \mathbf{y}, \mathbf{x} \rangle\).
• A complex inner product is conjugate symmetric (or Hermitian): \(\langle \mathbf{x}, \mathbf{y} \rangle = \overline{\langle \mathbf{y}, \mathbf{x} \rangle}\).

This change is necessary to ensure that the norm-squared, \(\langle \mathbf{x}, \mathbf{x} \rangle\), is a non-negative real number for complex vectors. If it were symmetric, \(\langle i\mathbf{v}, i\mathbf{v} \rangle = i^2\langle \mathbf{v}, \mathbf{v} \rangle = -||\mathbf{v}||^2\), which would violate the positive-definiteness property.

If \(A\) is a real matrix (\\(A = \bar{A}\\)) and \(A\mathbf{v} = \lambda\mathbf{v}\\), then by taking the complex conjugate of the entire equation, we get: \\[ \overline{A\mathbf{v}} = \overline{\lambda\mathbf{v}} \] \\[ \bar{A}\bar{\mathbf{v}} = \bar{\lambda}\bar{\mathbf{v}} \] Since \(A\) is real, \(\bar{A} = A\\), so: \\[ A\bar{\mathbf{v}} = \bar{\lambda}\bar{\mathbf{v}} \] This shows that \(\bar{\lambda}\) is also an eigenvalue of \(A\), with corresponding eigenvector \(\bar{\mathbf{v}}\). Complex eigenvalues of real matrices always come in conjugate pairs.

Multiplying a complex number \(z\) by \(i\) corresponds to rotating the vector representing \(z\) in the complex plane by \(90^\circ\) (or \(\pi/2\) radians) counter-clockwise.
This can be seen using the exponential form. Let \(z = re^{i\theta}\). Since \(i = e^{i\pi/2}\), the product is: \\[ iz = (e^{i\pi/2})(re^{i\theta}) = re^{i(\theta + \pi/2)} \] The new vector has the same modulus \(r\) but its argument is increased by \(\pi/2\).

No. The set of Hermitian matrices is not closed under multiplication by a general complex scalar.
Let \(A\) be a non-zero Hermitian matrix (\\(A=A^*\\)). Let \(\alpha = i\\). We check if \(iA\) is Hermitian: \\[ (iA)^* = \bar{i}A^* = (-i)A = -A \] For \(iA\) to be Hermitian, we would need \(iA = (iA)^* = -A\\), which implies \((i+1)A = 0\\). Since \(A\) is non-zero, this is not true.
However, the set of Hermitian matrices does form a real vector space, since it is closed under multiplication by real scalars.

A unitary matrix is the complex analogue of a real orthogonal matrix. A real matrix \(Q\) is orthogonal if \(Q^T Q = I\). A complex matrix \(U\) is unitary if \(U^* U = I\).
If a unitary matrix \(U\) happens to have only real entries, then its Hermitian conjugate \(U^*\) is the same as its transpose \(U^T\). In this case, the condition \(U^*U=I\) becomes \(U^TU=I\), which is the definition of an orthogonal matrix. Therefore, a real unitary matrix is an orthogonal matrix.

If \(E_i = \mathbf{u}_i\mathbf{u}_i^*\) is the projection matrix onto the eigenspace for \(\lambda_i\) (spanned by \(\mathbf{u}_i\)), and \(\mathbf{v}\\) is in that eigenspace, then \(\mathbf{v} = c\mathbf{u}_i\) for some scalar \(c\). Then: \\[ E_i\mathbf{v} = E_i(c\mathbf{u}_i) = c(E_i\mathbf{u}_i) = c(\mathbf{u}_i\mathbf{u}_i^*) \mathbf{u}_i \] Since \(\mathbf{u}_i^*\mathbf{u}_i = ||\mathbf{u}_i||^2 = 1\), this becomes: \\[ c(\mathbf{u}_i)(1) = c\mathbf{u}_i = \mathbf{v} \] So, \(E_i\mathbf{v} = \mathbf{v}\). The projection of a vector already in the subspace is the vector itself.

If \(\mathbf{v}\\) is in the eigenspace for \(E_j\), then \(\mathbf{v} = c\mathbf{u}_j\) for some scalar \(c\). Then: \\[ E_i\mathbf{v} = E_i(c\mathbf{u}_j) = c(E_i\mathbf{u}_j) = c(\mathbf{u}_i\mathbf{u}_i^*) \mathbf{u}_j \] Since the eigenvectors \(\{\mathbf{u}_k\}\) form an orthonormal set, \(\mathbf{u}_i^*\mathbf{u}_j = \langle \mathbf{u}_j, \mathbf{u}_i \rangle = 0\) for \(i \neq j\). Therefore: \\[ c(\mathbf{u}_i)(0) = \mathbf{0} \] So, \(E_i\mathbf{v} = \mathbf{0}\). The projection of a vector onto an orthogonal subspace is the zero vector.

Yes. A real matrix is diagonalisable if it has \(n\) linearly independent eigenvectors. It does not need to be symmetric.
However, a real matrix is orthogonally diagonalisable (i.e., with an orthogonal matrix \(P\) such that \(P^TAP=D\)) if and only if it is symmetric. This is a much stronger condition.

The characteristic equation is \(\det(A - \lambda I) = (2-\lambda)^2 = 0\). The only eigenvalue is \(\lambda=2\).

• Algebraic Multiplicity: The eigenvalue \(\lambda=2\) is a root of multiplicity 2 in the characteristic polynomial. So, the algebraic multiplicity is 2.
• Geometric Multiplicity: We find the dimension of the eigenspace \(N(A-2I)\). \\(A-2I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\). The null space is spanned by the vector \((1, 0)\). The dimension of the eigenspace is 1. So, the geometric multiplicity is 1.

Since the geometric multiplicity (1) is less than the algebraic multiplicity (2), the matrix is not diagonalisable.

Yes. The proof is very similar to the Hermitian case. Let \(A\mathbf{v}_1 = \lambda_1\mathbf{v}_1\) and \(A\mathbf{v}_2 = \lambda_2\mathbf{v}_2\) with \(\lambda_1 \neq \lambda_2\). Consider \(\langle A\mathbf{v}_1, \mathbf{v}_2 \rangle = \lambda_1 \langle \mathbf{v}_1, \mathbf{v}_2 \rangle\). Also, \(\langle A\mathbf{v}_1, \mathbf{v}_2 \rangle = \langle \mathbf{v}_1, A^*\mathbf{v}_2 \rangle\). We know \(A^*\mathbf{v}_2 = \bar{\lambda}_2\mathbf{v}_2\) (an eigenvector of \(A^*\)). So \(\langle \mathbf{v}_1, A^*\mathbf{v}_2 \rangle = \langle \mathbf{v}_1, \bar{\lambda}_2\mathbf{v}_2 \rangle = \lambda_2 \langle \mathbf{v}_1, \mathbf{v}_2 \rangle\). Thus, \(\lambda_1 \langle \mathbf{v}_1, \mathbf{v}_2 \rangle = \lambda_2 \langle \mathbf{v}_1, \mathbf{v}_2 \rangle\), which means \((\lambda_1 - \lambda_2) \langle \mathbf{v}_1, \mathbf{v}_2 \rangle = 0\). Since \(\lambda_1 \neq \lambda_2\), we must have \(\langle \mathbf{v}_1, \mathbf{v}_2 \rangle = 0\).

Cartesian: \(z = 1+i\).
1. Modulus: \(r = |z| = \sqrt{1^2 + 1^2} = \sqrt{2}\).
2. Argument: The point \((1,1)\) is in the first quadrant. \(\tan\theta = 1/1 = 1\), so \(\theta = \pi/4\).
Polar Form: \[ z = \sqrt{2}(\cos(\pi/4) + i\sin(\pi/4)) \] Exponential Form: \[ z = \sqrt{2}e^{i\pi/4} \]

First, convert \(1+i\) to exponential or polar form. From the previous card, \(1+i = \sqrt{2}e^{i\pi/4}\).
Now, apply De Moivre's formula: \\[ (1+i)^8 = (\sqrt{2}e^{i\pi/4})^8 = (\sqrt{2})^8 (e^{i\pi/4})^8 \] \\[ = 2^4 e^{i(8\pi/4)} = 16 e^{i2\pi} \] Since \(e^{i2\pi} = \cos(2\pi) + i\sin(2\pi) = 1 + 0i = 1\), the result is: \\[ (1+i)^8 = 16(1) = 16 \]

The process is almost identical, but the type of transformation matrix used is different.

• A real symmetric matrix \(A\) is diagonalised by a real orthogonal matrix \(P\) (where \(P^T = P^{-1}\)), such that \(P^TAP = D\).
• A complex Hermitian matrix \(A\) is diagonalised by a unitary matrix \(U\) (where \(U^* = U^{-1}\)), such that \(U^*AU = D\).

The concept of an orthogonal matrix is the real-valued special case of a unitary matrix. The process in both cases involves finding an orthonormal basis of eigenvectors.

Yes.
A matrix is Hermitian if \(A = A^*\). The Hermitian conjugate \(A^*\) is defined as \(\bar{A}^T\).
If \(A\) is a real matrix, then all its entries are real, so taking the complex conjugate has no effect: \(\bar{A} = A\). Therefore, for a real matrix, \(A^* = A^T\).
If \(A\) is also symmetric, then \(A = A^T\).
Combining these, if \(A\) is real and symmetric, then \(A = A^T = A^*\). Thus, \(A\) is Hermitian.

No, not always. Let \(A\) and \(B\) be Hermitian, so \(A=A^*\) and \(B=B^*\). We check the product \(AB\): \\[ (AB)^* = B^*A^* = BA \] For \(AB\) to be Hermitian, we would need \((AB)^* = AB\). This means we need \(BA = AB\). The product of two Hermitian matrices is Hermitian if and only if the matrices commute.

Yes. If \(A\) is normal, then \(AA^* = A^*A\). We need to check if \((A^2)(A^2)^* = (A^2)^*(A^2)\\).
LHS: \((A^2)(A^2)^* = A^2(A^*)^2 = A A A^* A^*\). Since \(A\) is normal, we can swap \(A\) and \(A^*\): \(A(AA^*)A^* = A(A^*A)A^* = A^*A A A^* = A^* (AA^*) A = A^*(A^*A)A = (A^*)^2 A^2 = (A^2)^*A^2\). So, \((A^2)(A^2)^* = (A^2)^*(A^2)\). The statement is true.

The matrix \(E_i\) is the orthogonal projection from \(\mathbb{C}^n\) onto the one-dimensional subspace spanned by the eigenvector \(\mathbf{u}_i\).
For any vector \(\mathbf{v} \in \mathbb{C}^n\), the vector \(E_i\mathbf{v}\) is the projection of \(\mathbf{v}\) onto the line defined by \(\mathbf{u}_i\). \\[ E_i\mathbf{v} = (\mathbf{u}_i\mathbf{u}_i^*)\mathbf{v} = \mathbf{u}_i(\mathbf{u}_i^*\mathbf{v}) = \mathbf{u}_i \langle \mathbf{v}, \mathbf{u}_i \rangle = \langle \mathbf{v}, \mathbf{u}_i \rangle \mathbf{u}_i \] This is the standard formula for orthogonal projection of \(\mathbf{v}\) onto the unit vector \(\mathbf{u}_i\).

Yes. A real matrix can have complex eigenvalues, and the eigenvectors corresponding to these complex eigenvalues will themselves be complex.
For example, the rotation matrix \(A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) is real, but its eigenvalues are \(\pm i\) and its eigenvectors are \((1, -i)\) and \((1, i)\), which are complex.

Yes. This happens when the matrix is not diagonalisable over \(\mathbb{R}\\) but is diagonalisable over \(\mathbb{C}\).
For a matrix to be diagonalisable over a field, it must have a full set of \(n\) linearly independent eigenvectors in the vector space over that field.
A real matrix like \(A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) has no real eigenvalues, so it has no eigenvectors in \(\mathbb{R}^2\) and cannot form a basis. However, it has two distinct complex eigenvalues \(\pm i\) and thus two linearly independent eigenvectors in \(\mathbb{C}^2\), which form a basis for \(\mathbb{C}^2\).

The sum of the projection matrices from a spectral decomposition is the identity matrix: \\[ \sum_{i=1}^n E_i = E_1 + E_2 + \dots + E_n = I \] This is because \(I = UU^*\). If \(U = [\mathbf{u}_1 | \dots | \mathbf{u}_n]\), then \(UU^* = \sum_{i=1}^n \mathbf{u}_i\mathbf{u}_i^* = \sum_{i=1}^n E_i\). This reflects the fact that projecting a vector onto every basis vector of an orthonormal basis and summing the results reconstructs the original vector.

Using the properties of the projection matrices \(E_i\), we have: \\[ A^k = (\sum \lambda_i E_i)^k = \sum \lambda_i^k E_i \] This works because when expanding the power, all cross terms \(E_i E_j\) for \(i \neq j\) are zero, and \(E_i^k = E_i\) for \(k \ge 1\).
This provides a very efficient way to compute powers of a unitarily diagonalisable matrix.

We check if \(AA^*=A^*A\). \\[ A^* = \begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix} \] \\[ AA^* = \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix} \begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix} = \begin{pmatrix} 1-i^2 & -i+i \\ i-i & -i^2+1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \] \\[ A^*A = \begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix} \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix} = \begin{pmatrix} 1-i^2 & i-i \\ -i+i & -i^2+1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \] Since \(AA^* = A^*A\), the matrix is normal.

We can view \(\mathbb{C}^n\) as a real vector space by restricting the scalars to be real numbers (\\(\mathbb{R}\)).
A vector \(\mathbf{v} \in \mathbb{C}^n\) can be written as \(\mathbf{v} = \mathbf{a} + i\mathbf{b}\) where \(\mathbf{a}, \mathbf{b} \in \mathbb{R}^n\). A basis for \(\mathbb{C}^n\) as a real vector space can be formed by taking the standard basis vectors of \(\mathbb{R}^n\) and their multiples by \(i\). For example, a basis for \(\mathbb{C}^2\) as a real vector space is \(\{(1,0), (i,0), (0,1), (0,i)\}\).
The dimension of \(\mathbb{C}^n\) as a real vector space is \(2n\).

They are equal: \((A^*)^{-1} = (A^{-1})^*\).
Proof: We know \(AA^{-1} = I\). Taking the Hermitian conjugate of both sides: \\[ (AA^{-1})^* = I^* \] \\[ (A^{-1})^* A^* = I \] This equation shows, by definition of an inverse, that \((A^{-1})^*\) is the inverse of \(A^*\). Therefore, \((A^*)^{-1} = (A^{-1})^*\).

We want to solve \(z^3 = 1\). In exponential form, \(1 = 1 \cdot e^{i(0 + 2k\pi)}\). The roots are given by: \\[ z_k = \sqrt[3]{1} e^{i\frac{2k\pi}{3}} \quad \text{for } k=0, 1, 2 \]

• \(k=0: z_0 = e^{i0} = 1\)
• \(k=1: z_1 = e^{i2\pi/3} = \cos(2\pi/3) + i\sin(2\pi/3) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\)
• \(k=2: z_2 = e^{i4\pi/3} = \cos(4\pi/3) + i\sin(4\pi/3) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\)

These are often denoted \(1, \omega, \omega^2\).

No. It fails on multiple conditions.

1. Zero Vector: The zero matrix is not unitary, since \(0^*0 = 0 \neq I\). A subspace must contain the zero vector.
2. Closure under Addition: If \(U_1\) and \(U_2\) are unitary, \(U_1+U_2\) is generally not unitary. For example, \(I\) is unitary, but \(I+I=2I\) is not, since \((2I)^*(2I) = 4I \neq I\).

Therefore, the set of unitary matrices is not a subspace.

We want to show \(||A\mathbf{x}||^2 = ||A^*\mathbf{x}||^2\). \\[ ||A\mathbf{x}||^2 = \langle A\mathbf{x}, A\mathbf{x} \rangle = \langle \mathbf{x}, A^*A\mathbf{x} \rangle \] \\[ ||A^*\mathbf{x}||^2 = \langle A^*\mathbf{x}, A^*\mathbf{x} \rangle = \langle \mathbf{x}, (A^*)^*A^*\mathbf{x} \rangle = \langle \mathbf{x}, AA^*\mathbf{x} \rangle \] Since \(A\) is normal, \(A^*A = AA^*\). Therefore, \\[ \langle \mathbf{x}, A^*A\mathbf{x} \rangle = \langle \mathbf{x}, AA^*\mathbf{x} \rangle \] This implies \(||A\mathbf{x}||^2 = ||A^*\mathbf{x}||^2\), and since norms are non-negative, \(||A\mathbf{x}|| = ||A^*\mathbf{x}||\).

We know \(A\mathbf{v} = \lambda\mathbf{v}\). We want to show \(A^*\mathbf{v} = \bar{\lambda}\mathbf{v}\). Consider the matrix \(B = A - \lambda I\). Since \(A\) is normal, \(B\) is also normal: \\[ B^*B = (A^* - \bar{\lambda}I)(A - \lambda I) = A^*A - \lambda A^* - \bar{\lambda}A + |\lambda|^2 I \] \\[ BB^* = (A - \lambda I)(A^* - \bar{\lambda}I) = AA^* - \bar{\lambda}A - \lambda A^* + |\lambda|^2 I \] Since \(AA^*=A^*A\), we have \(BB^*=B^*B\). From the previous card, we know \(||B\mathbf{v}|| = ||B^*\mathbf{v}||\). Since \(\mathbf{v}\\) is an eigenvector of \(A\) for \(\lambda\), \(B\mathbf{v} = (A-\lambda I)\mathbf{v} = \mathbf{0}\). So \(||B\mathbf{v}||=0\). This means \(||B^*\mathbf{v}||=0\), which implies \(B^*\mathbf{v}=\mathbf{0}\). \\[ (A^* - \bar{\lambda}I)\mathbf{v} = \mathbf{0} \implies A^*\mathbf{v} = \bar{\lambda}\mathbf{v} \] Thus, \(\mathbf{v}\\) is an eigenvector of \(A^*\) with eigenvalue \(\bar{\lambda}\).

A matrix \(A\) is skew-Hermitian if \(A^* = -A\).
The eigenvalues of a skew-Hermitian matrix are purely imaginary (or zero).
Proof: Let \(A\mathbf{v} = \lambda\mathbf{v}\). Then \(\langle A\mathbf{v}, \mathbf{v} \rangle = \lambda ||\mathbf{v}||^2\). Also, \(\langle A\mathbf{v}, \mathbf{v} \rangle = \langle \mathbf{v}, A^*\mathbf{v} \rangle = \langle \mathbf{v}, -A\mathbf{v} \rangle = \langle \mathbf{v}, -\lambda\mathbf{v} \rangle = -\bar{\lambda} \langle \mathbf{v}, \mathbf{v} \rangle = -\bar{\lambda} ||\mathbf{v}||^2\). So, \(\lambda ||\mathbf{v}||^2 = -\bar{\lambda} ||\mathbf{v}||^2\), which means \(\lambda = -\bar{\lambda}\). If \(\lambda = a+ib\), then \(a+ib = -(a-ib) = -a+ib\), which implies \(a = -a\), so \(a=0\). Thus \(\lambda = ib\) is purely imaginary.

Yes. We need to check if \((A+I)(A+I)^* = (A+I)^*(A+I)\). \\[ (A+I)^* = A^* + I^* = A^* + I \] LHS: \((A+I)(A^*+I) = AA^* + A + A^* + I\).
RHS: \((A^*+I)(A+I) = A^*A + A^* + A + I\).
Since \(A\) is normal, \(AA^* = A^*A\). Therefore, the LHS and RHS are equal, and \(A+I\) is normal.

1. Normalize \(\mathbf{v}_1\): \\(||\mathbf{v}_1||^2 = 1^2 + |i|^2 + 0^2 = 1+1=2 \implies ||\mathbf{v}_1|| = \sqrt{2}\). \\(\mathbf{u}_1 = \frac{1}{\sqrt{2}}(1, i, 0)\).
2. Find \(\mathbf{w}_2\): \\(\langle \mathbf{v}_2, \mathbf{u}_1 \rangle = \frac{1}{\sqrt{2}}(1\cdot\bar{1} + 2\cdot\bar{i} + (1+i)\cdot\bar{0}) = \frac{1-2i}{\sqrt{2}}\). \\(\mathbf{w}_2 = \mathbf{v}_2 - \langle \mathbf{v}_2, \mathbf{u}_1 \rangle \mathbf{u}_1 = (1,2,1+i) - \frac{1-2i}{\sqrt{2}} \frac{1}{\sqrt{2}}(1,i,0) = (1,2,1+i) - \frac{1-2i}{2}(1,i,0)\) \\(= (1,2,1+i) - (\frac{1-2i}{2}, \frac{i+2}{2}, 0) = (\frac{1+2i}{2}, \frac{2-i}{2}, 1+i)\).
3. Normalize \(\mathbf{w}_2\): \\(||\mathbf{w}_2||^2 = \frac{5}{4} + \frac{5}{4} + 2 = \frac{18}{4} = \frac{9}{2} \implies ||\mathbf{w}_2|| = \frac{3}{\sqrt{2}}\). \\(\mathbf{u}_2 = \frac{\sqrt{2}}{3} (\frac{1+2i}{2}, \frac{2-i}{2}, 1+i) = \frac{1}{3\sqrt{2}}(1+2i, 2-i, 2+2i)\). The basis is \(\{\mathbf{u}_1, \mathbf{u}_2\}\).

The determinant is the product of the eigenvalues and the trace is the sum of the eigenvalues.

• \\(\det(A) = (a+ib)(a-ib) = a^2 - (ib)^2 = a^2 + b^2\). The determinant is a real number.
• \\(\text{tr}(A) = (a+ib) + (a-ib) = 2a\). The trace is a real number.

This is consistent with the fact that for a real matrix, both the determinant and trace must be real numbers.

Let \(U\) and \(V\) be two \(n \times n\) unitary matrices. By definition, \(U^*U = I\) and \(V^*V = I\). We want to show that their product, \(UV\), is also unitary. We check if \((UV)^*(UV) = I\). \\[ (UV)^*(UV) = (V^*U^*)(UV) \] Using associativity of matrix multiplication: \\[ = V^*(U^*U)V = V^*(I)V = V^*V = I \] Thus, the product \(UV\) is unitary.

Yes. Let \(D\) be a diagonal matrix. Its entries are \(d_{ij} = 0\) for \(i \neq j\). The Hermitian conjugate \(D^*\) is the conjugate transpose. The transpose of a diagonal matrix is itself. So \(D^* = \bar{D}\), which is also a diagonal matrix with entries \(\bar{d}_{ii}\) on the diagonal. Since diagonal matrices commute, we have: \\[ DD^* = D\bar{D} = \bar{D}D = D^*D \] The entry \((DD^*)_{ii} = d_{ii}\bar{d}_{ii} = |d_{ii}|^2\). The entry \((D^*D)_{ii} = \bar{d}_{ii}d_{ii} = |d_{ii}|^2\). All off-diagonal entries are zero. Therefore, any diagonal matrix is normal.

Consider \(A = \begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix}\). We checked in a previous card that it is normal.

• It is not diagonal.
• It is not Hermitian, since \(A^* = \begin{pmatrix} 1 & -i \\ -i & 1 \end{pmatrix} \neq A\).
• It is not unitary, since \(AA^* = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \neq I\).

This matrix is normal but does not fall into the other categories.

We have already proven this. Let \(\mathbf{v}\\) be an eigenvector of a normal matrix \(A\) with eigenvalue \(\lambda\). We showed that \(||(A-\lambda I)\mathbf{v}|| = ||(A-\lambda I)^*\mathbf{v}||\). Since \((A-\lambda I)\mathbf{v} = \mathbf{0}\), it follows that \(||(A^*-\bar{\lambda}I)\mathbf{v}|| = 0\), which means \((A^*-\bar{\lambda}I)\mathbf{v} = \mathbf{0}\). This shows that \(A^*\mathbf{v} = \bar{\lambda}\mathbf{v}\). Therefore, \(\mathbf{v}\\) is an eigenvector of \(A^*\) (with eigenvalue \(\bar{\lambda}\)). The set of eigenvectors is the same.

The key difference is related to eigenvalues. By the Fundamental Theorem of Algebra, any \(n \times n\) complex matrix has \(n\) complex eigenvalues (counting multiplicities). A real matrix may not have any real eigenvalues.
This means a complex matrix is more 'likely' to be diagonalisable. A complex matrix is diagonalisable if and only if for every eigenvalue, the geometric multiplicity equals the algebraic multiplicity.
A real matrix can fail to be diagonalisable over \(\mathbb{R}\) simply by not having enough real eigenvalues, even if it would be diagonalisable over \(\mathbb{C}\).

Yes. Let \(B = U^*AU\). We check if \(BB^*=B^*B\). \\[ B^* = (U^*AU)^* = U^*A^*(U^*)^* = U^*A^*U \] LHS: \(BB^* = (U^*AU)(U^*A^*U) = U^*A(UU^*)A^*U = U^*AIA^*U = U^*AA^*U\).
RHS: \(B^*B = (U^*A^*U)(U^*AU) = U^*A^*(UU^*)AU = U^*A^*IAU = U^*A^*AU\).
Since \(A\) is normal, \(AA^*=A^*A\), so the LHS and RHS are equal. Thus, \(B\) is normal. This property is called "unitarily invariant".

Since \(A\) is diagonal, it is normal. Its spectral decomposition is simple. The eigenvalues are \(\lambda_1=8, \lambda_2=-i\). The projection matrices are \(E_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\) and \(E_2 = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}\). So \(A = 8E_1 - iE_2\). We want \(B = \lambda_1^{1/3}E_1 + \lambda_2^{1/3}E_2\). The cube roots of 8 are \(2, 2e^{i2\pi/3}, 2e^{i4\pi/3}\). Let's pick \(\sqrt[3]{8}=2\). The cube roots of \(-i = e^{-i\pi/2}\) are \(e^{-i\pi/6}, e^{i\pi/2}, e^{i7\pi/6}\). Let's pick \(\sqrt[3]{-i} = e^{i\pi/2} = i\). One possible matrix \(B\) is: \\[ B = 2\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + i\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & i \end{pmatrix} \] Checking: \(B^3 = \begin{pmatrix} 2^3 & 0 \\ 0 & i^3 \end{pmatrix} = \begin{pmatrix} 8 & 0 \\ 0 & -i \end{pmatrix} = A\).

The Spectral Theorem provides a complete classification of which matrices can be unitarily diagonalised. It states that this is possible if and only if the matrix is normal (\\(AA^*=A^*A\\)).
This is a powerful theoretical tool because it tells us exactly which class of matrices has the 'nice' property of possessing an orthonormal basis of eigenvectors. This includes Hermitian, skew-Hermitian, and unitary matrices as special cases.

We use Euler's formula: \(e^{i\theta} = \cos\theta + i\sin\theta\). \\[ z = 3(\cos(\pi/3) + i\sin(\pi/3)) \] We know that \(\cos(\pi/3) = 1/2\) and \(\sin(\pi/3) = \sqrt{3}/2\). \\[ z = 3\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = \frac{3}{2} + i\frac{3\sqrt{3}}{2} \] So, the Cartesian form is \(\frac{3}{2} + i\frac{3\sqrt{3}}{2}\).

Yes. For a real matrix \(A\), the normal condition \(AA^*=A^*A\) becomes \(AA^T=A^TA\).
A symmetric matrix (\\(A=A^T\\)) is always normal.
However, a real orthogonal matrix (\\(A^T=A^{-1}\\)) is also normal: \(AA^T = A A^{-1} = I\) and \(A^TA = A^{-1}A = I\).
The rotation matrix \(A = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\) for \(\theta \neq k\pi\) is orthogonal but not symmetric, and it is normal.

No, it is not unique.

1. The order of eigenvalues in the diagonal matrix \(D\) can be changed.
2. Changing the order of eigenvalues in \(D\) corresponds to changing the order of the corresponding eigenvector columns in the transformation matrix \(P\) (or \(U\)).
3. Eigenvectors are not unique; any non-zero scalar multiple of an eigenvector is also an eigenvector. While this scalar is often chosen to normalize the vector in unitary diagonalisation, one could still multiply by a scalar of modulus 1 (like \(i\) or \(-1\)) to get a different orthonormal eigenvector.

The main challenge is determining if there are enough linearly independent eigenvectors.
For a matrix to be diagonalisable, the geometric multiplicity (the dimension of the eigenspace) of each eigenvalue must be equal to its algebraic multiplicity (the multiplicity of the root in the characteristic polynomial).
If an eigenvalue \(\lambda\) is repeated \(k\) times, you must be able to find \(k\) linearly independent eigenvectors for that \(\lambda\). If you cannot, the matrix is not diagonalisable. For symmetric/Hermitian/normal matrices, this is guaranteed to be possible.

The determinant is the product of the eigenvalues and the trace is the sum of the eigenvalues.

• \\(\det(A) = 1 \cdot 1 \cdot 5 = 5\)
• \\(\text{tr}(A) = 1 + 1 + 5 = 7\)

This holds regardless of whether the matrix is diagonalisable.

We expand \(||\mathbf{u}+\mathbf{v}||^2\) using the definition of the norm: \\[ ||\mathbf{u}+\mathbf{v}||^2 = \langle \mathbf{u}+\mathbf{v}, \mathbf{u}+\mathbf{v} \rangle \] Using the properties of the inner product: \\[ = \langle \mathbf{u}, \mathbf{u} \rangle + \langle \mathbf{u}, \mathbf{v} \rangle + \langle \mathbf{v}, \mathbf{u} \rangle + \langle \mathbf{v}, \mathbf{v} \rangle \] \\[ = ||\mathbf{u}||^2 + \langle \mathbf{u}, \mathbf{v} \rangle + \overline{\langle \mathbf{u}, \mathbf{v} \rangle} + ||\mathbf{v}||^2 \] Since \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal, \(\langle \mathbf{u}, \mathbf{v} \rangle = 0\). Therefore, the middle two terms are zero. \\[ ||\mathbf{u}+\mathbf{v}||^2 = ||\mathbf{u}||^2 + ||\mathbf{v}||^2 \]

The procedure is the same as for real eigenvalues and eigenvectors, but the arithmetic is done in \(\mathbb{C}\).

1. Find Eigenvalues: Solve the characteristic equation \(\det(A - \lambda I) = 0\) for \(\lambda\). By the Fundamental Theorem of Algebra, there will be \(n\) roots in \(\mathbb{C}\) (counting multiplicities).
2. Find Eigenvectors: For each eigenvalue \(\lambda_i\), find the null space of the matrix \((A - \lambda_i I)\). Any non-zero vector in \(N(A - \lambda_i I)\) is an eigenvector corresponding to \(\lambda_i\). This involves solving a system of linear equations with complex coefficients.

The process is structurally identical, but the inner product used is different.
In \(\mathbb{R}^n\), the projection formula uses the real dot product: \(\text{proj}_{\mathbf{u}}(\mathbf{v}) = \frac{\mathbf{v} \cdot \mathbf{u}}{||\mathbf{u}||^2}\mathbf{u}\).
In \(\mathbb{C}^n\), the projection formula must use the complex inner product: \(\text{proj}_{\mathbf{u}}(\mathbf{v}) = \frac{\langle \mathbf{v}, \mathbf{u} \rangle}{||\mathbf{u}||^2}\mathbf{u}\).
Because the complex inner product is conjugate symmetric, the order matters: \(\langle \mathbf{v}, \mathbf{u} \rangle\) is not the same as \(\langle \mathbf{u}, \mathbf{v} \rangle\). The formula \(\mathbf{w}_k = \mathbf{v}_k - \sum_{j=1}^{k-1} \langle \mathbf{v}_k, \mathbf{u}_j \rangle \mathbf{u}_j\) must be used with care regarding the order of the vectors in the inner product.

If \(A\) is Hermitian, its eigenvalues are real.
If \(A\) is unitary, its eigenvalues have a modulus of 1.
The only real numbers with a modulus of 1 are \(1\) and \(-1\).
Therefore, if a matrix is both unitary and Hermitian, its eigenvalues must be either 1 or -1.

No. Let \(A\) and \(B\) be normal matrices. We check if \(A+B\) is normal. \\((A+B)(A+B)^* = (A+B)(A^*+B^*) = AA^* + AB^* + BA^* + BB^*\) \\((A+B)^*(A+B) = (A^*+B^*)(A+B) = A^*A + A^*B + B^*A + B^*B\) Since \(AA^*=A^*A\) and \(BB^*=B^*B\), for the sum to be normal we need \(AB^* + BA^* = A^*B + B^*A\). This is not true in general.

The matrix that performs this projection is the \(i\)-th projection matrix \(E_i\) from the spectral decomposition. It is given by the outer product: \\[ E_i = \mathbf{u}_i \mathbf{u}_i^* \] where \(\mathbf{u}_i\) is the normalized eigenvector (i.e., \(||\mathbf{u}_i||=1\)).
If you have an orthogonal eigenvector \(\mathbf{v}_i\) that is not normalized, the projection matrix is \(\frac{1}{||\mathbf{v}_i||^2} \mathbf{v}_i \mathbf{v}_i^*\).

It must be a normal matrix.
The Spectral Theorem states that a matrix is unitarily diagonalisable if and only if it is normal. A matrix is unitarily diagonalisable if and only if it has an orthonormal basis of eigenvectors. Therefore, having a complete set of orthonormal eigenvectors is equivalent to being a normal matrix.

The powers of \(i\) cycle with a period of 4: \(i^1=i, i^2=-1, i^3=-i, i^4=1\). To find \(i^{2025}\), we can find the remainder of 2025 when divided by 4. \\[ 2025 = 4 \times 506 + 1 \] So, \(2025 \equiv 1 \pmod{4}\). Therefore, \\[ i^{2025} = i^1 = i \]

Yes. If \(A\) is normal, \(AA^*=A^*A\). We want to check if \(A^{-1}(A^{-1})^* = (A^{-1})^*(A^{-1})\). We know \((A^{-1})^* = (A^*)^{-1}\). Start with \(AA^*=A^*A\). Multiply on the left by \(A^{-1}\) and on the right by \((A^*)^{-1}\): \\[ A^{-1}(AA^*) (A^*)^{-1} = A^{-1}(A^*A)(A^*)^{-1} \] \\[ (A^{-1}A)A^*(A^*)^{-1} = A^{-1}A^*A(A^*)^{-1} \] \\[ I = A^{-1}A^*A(A^*)^{-1} \] Now multiply on the left by \((A^*)^{-1}\) and on the right by \(A^{-1}\): \\[ (A^*)^{-1}A^{-1} = (A^{-1})^*A^{-1} = A^{-1}(A^*)^{-1} \] This shows \((A^{-1})^*A^{-1} = A^{-1}(A^{-1})^*\), so \(A^{-1}\) is normal.

The main difference is the treatment of eigenvectors within each eigenspace.

• For a general diagonalisation, you just need to find a basis of eigenvectors for each eigenspace. The collection of all these basis vectors must form a set of \(n\) linearly independent vectors.
• For a unitary diagonalisation, you must find an orthonormal basis for each eigenspace. This often requires applying the Gram-Schmidt process to the basis vectors found for any eigenspace with dimension greater than 1.

Eigenvectors from different eigenspaces of a normal matrix are automatically orthogonal, so you only need to apply Gram-Schmidt within each eigenspace.

A real skew-symmetric matrix is also a skew-Hermitian matrix, because if \(A\) is real, \(A^* = \bar{A}^T = A^T\). So \(A^* = -A\).
The eigenvalues of any skew-Hermitian matrix are purely imaginary or zero. Therefore, the eigenvalues of a real skew-symmetric matrix are purely imaginary or zero.

The complex number is \(z = 2-2i\). This corresponds to the point \((2, -2)\) in the complex plane.

• Modulus: \(|z| = \sqrt{2^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}\).
• Argument: The point is in the fourth quadrant. The angle with the positive real axis is \(\tan\theta = -2/2 = -1\). The principal argument in \((-\pi, \pi]\) is \(\theta = -\pi/4\).

Yes. Let \(B = A - cI\). We check if \(BB^* = B^*B\). \\[ B^* = (A - cI)^* = A^* - \bar{c}I^* = A^* - \bar{c}I \] LHS: \(BB^* = (A-cI)(A^*-\bar{c}I) = AA^* - cA^* - \bar{c}A + c\bar{c}I\).
RHS: \(B^*B = (A^*-\bar{c}I)(A-cI) = A^*A - cA^* - \bar{c}A + \bar{c}cI\).
Since \(A\) is normal (\\(AA^*=A^*A\)) and \(c\bar{c}=\bar{c}c=|c|^2\), the LHS and RHS are equal. Thus, \(A-cI\) is normal.

Not necessarily. A matrix is invertible if and only if 0 is not one of its eigenvalues.
If \(A\) is unitarily diagonalisable, it means \(A = UDU^*\) where \(D\) contains the eigenvalues of \(A\). The determinant of \(A\) is the product of its eigenvalues. \\[ \det(A) = \det(UDU^*) = \det(U)\det(D)\det(U^*) = \det(D) = \lambda_1\lambda_2\cdots\lambda_n \] If one of the eigenvalues is 0, then \(\det(A)=0\) and \(A\) is not invertible. For example, the zero matrix is normal and diagonal, but not invertible.

They are orthogonal complements of each other in \(\mathbb{C}^m\) (where \(A\) is \(m \times n\)). \\[ R(A)^{\perp} = N(A^*) \] This is a fundamental theorem of linear algebra. It means that every vector in the column space of \(A\) is orthogonal to every vector in the null space of its Hermitian conjugate.

If \(\lambda\) is an eigenvalue of \(A\) with eigenvector \(\mathbf{v}\), then \(A\mathbf{v} = \lambda\mathbf{v}\). Then \(A^2\mathbf{v} = A(A\mathbf{v}) = A(\lambda\mathbf{v}) = \lambda(A\mathbf{v}) = \lambda(\lambda\mathbf{v}) = \lambda^2\mathbf{v}\). So, if \(\lambda\) is an eigenvalue of \(A\), then \(\lambda^2\) is an eigenvalue of \(A^2\).
The eigenvalues of \(A^2\) are:

• \(1^2 = 1\)
• \(i^2 = -1\)
• \((-i)^2 = -1\)

The eigenvalues of \(A^2\) are 1, -1, -1.

No. If \(A\) is Hermitian, \(A^*=A\). Let's check the conjugate transpose of \(iA\): \\[ (iA)^* = \bar{i}A^* = (-i)A = -A \] For \(iA\) to be Hermitian, we would need \(iA = -A\), which implies \((i+1)A=0\). This is not true for any non-zero matrix \(A\).
In fact, if \(A\) is Hermitian, then \(iA\) is skew-Hermitian.

Multiplication and division become much simpler.

• Multiplication: To multiply two complex numbers, you multiply their moduli and add their arguments. \\[ (r_1e^{i\theta_1})(r_2e^{i\theta_2}) = r_1r_2e^{i(\theta_1+\theta_2)} \]
• Division: To divide two complex numbers, you divide their moduli and subtract their arguments. \\[ \frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}} = \frac{r_1}{r_2}e^{i(\theta_1-\theta_2)} \]

This is much faster than multiplying or dividing in Cartesian form, which requires FOIL and multiplying by the conjugate.

The concept of an orthogonal matrix in \(\mathbb{R}^n\) (where \(Q^TQ=I\)) is generalized to a unitary matrix in \(\mathbb{C}^n\).
The transpose operation is replaced by the Hermitian conjugate (conjugate transpose), denoted by \(*\).
A real matrix \(Q\) is orthogonal if \(Q^T = Q^{-1}\).
A complex matrix \(U\) is unitary if \(U^* = U^{-1}\).
Just as the columns of an orthogonal matrix form an orthonormal basis for \(\mathbb{R}^n\) with the real dot product, the columns of a unitary matrix form an orthonormal basis for \(\mathbb{C}^n\) with the complex inner product.