Question

Verify that each of the following matrices is Hermitian. Find its eigenvalues and eigenvectors, write a unitary matrix U which diagonalizes \(\mathrm{H}\) by a similarity transformation, and show that \(U^{-1} H U\) is the diagonal matrix of eigenvalues. $$\left(\begin{array}{cc} 3 & 1-i \\ 1+i & 2 \end{array}\right)$$

Short answer

The matrix is Hermitian. The eigenvalues are 4 and 1. The unitary matrix that diagonalizes it is \( U = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1-i}{2} \ \frac{1+i}{\sqrt{3}} & -\frac{1}{2} \end{pmatrix} \).

Step-by-step solution

- Verify Hermitian Property

A matrix is Hermitian if it is equal to its conjugate transpose. For the given matrix: \[ H = \begin{pmatrix} 3 & 1-i \ 1+i & 2 \end{pmatrix} \]Calculate the conjugate transpose, \[ H^* = \begin{pmatrix} 3 & 1+i \ 1-i & 2 \end{pmatrix} \]Since \[H = H^*\], the matrix is Hermitian.

- Find Eigenvalues

To find the eigenvalues of \(H\), solve the characteristic equation:\[ \det(H - \lambda I) = 0 \]Where \(I\) is the identity matrix and \(\lambda\) represents the eigenvalues:\[ \det\begin{pmatrix} 3-\lambda & 1-i \ 1+i & 2-\lambda \end{pmatrix} = 0 \]Expanding the determinant,\[ (3-\lambda)(2-\lambda) - (1-i)(1+i) = 0 \]\[ \lambda^2 - 5\lambda + 4 = 0 \]Solving for \(\lambda\), the eigenvalues are \(\lambda_1 = 4\) and \(\lambda_2 = 1\).

- Find Eigenvectors

For each eigenvalue \(\lambda\), solve \((H - \lambda I)\mathbf{v} = 0\). For \(\lambda = 4\):\[ \begin{pmatrix} -1 & 1-i \ 1+i & -2 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = 0 \]Solving, the eigenvector is \(\mathbf{v}_1 = \begin{pmatrix} 1 \ 1+i \end{pmatrix}\).For \(\lambda = 1\):\[ \begin{pmatrix} 2 & 1-i \ 1+i & 1 \end{pmatrix} \begin{pmatrix} v_1 \ v_2 \end{pmatrix} = 0 \]Solving, the eigenvector is \(\mathbf{v}_2 = \begin{pmatrix} 1-i \ -1 \end{pmatrix}\).

- Form the Unitary Matrix U

Normalize the eigenvectors to form the columns of the unitary matrix \(U\). For \(\mathbf{v}_1\):\[ \|\mathbf{v}_1\| = \sqrt{1^2 + (1+i)(1-i)} = \sqrt{3} \]\(\mathbf{u}_1 = \begin{pmatrix} \frac{1}{\sqrt{3}} \ \frac{1+i}{\sqrt{3}} \end{pmatrix}\)For \(\mathbf{v}_2\):\[ \|\mathbf{v}_2\| = \sqrt{(1-i)(1+i) + (-1)^2} = 2 \]\(\mathbf{u}_2 = \begin{pmatrix} \frac{1-i}{2} \ -\frac{1}{2} \end{pmatrix}\)Thus,\[ U = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1-i}{2} \ \frac{1+i}{\sqrt{3}} & -\frac{1}{2} \end{pmatrix} \]

- Diagonalize the Matrix

Compute \(U^{-1} H U\). First, find \(U^{-1}\) which is the conjugate transpose of \(U\) because \(U\) is unitary:\[ U^{-1} = U^* = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1+i}{\sqrt{3}} \ \frac{1-i}{2} & -\frac{1}{2} \end{pmatrix} \]Then,\[ U^{-1} H U = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1+i}{\sqrt{3}} \ \frac{1-i}{2} & -\frac{1}{2} \end{pmatrix} \begin{pmatrix} 3 & 1-i \ 1+i & 2 \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1-i}{2} \ \frac{1+i}{\sqrt{3}} & -\frac{1}{2} \end{pmatrix} \]This results in the diagonal matrix \(\text{diag}(4, 1)\).

Key concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental concepts in linear algebra. They come from solving the equation \[ A \mathbf{v} = \lambda \mathbf{v} \]where \( A \) is a square matrix, \( \mathbf{v} \) is a nonzero vector called an eigenvector, and \( \lambda \) is the eigenvalue associated with that eigenvector.
Let's break this down:

• Eigenvalues (\(\lambda\)) tell us how much the eigenvector is stretched or shrunk during the transformation.
• Eigenvectors (\(\mathbf{v}\)) point in the direction that remains unchanged when a transformation is applied, except for a scaling factor.

In the exercise, we found the eigenvalues for the Hermitian matrix by solving the characteristic polynomial, and determined the eigenvectors by substituting back into the equations.

Diagonalization

Diagonalization is the process of converting a matrix into a diagonal matrix. This simplification makes many calculations easier, like raising a matrix to a power.
A matrix is diagonalizable if it has a complete set of linearly independent eigenvectors. In simpler terms:

• We can express the original matrix as \[ A = U D U^{-1} \] where \(U \) is a matrix of eigenvectors, \( D \) is a diagonal matrix of eigenvalues, and \(U^{-1} \) is the inverse of \( U \).

In the exercise, we demonstrated diagonalization by computing \( U^{-1} H U \), resulting in the diagonal matrix of eigenvalues.

Unitary Matrices

Unitary matrices are special kinds of matrices where the conjugate transpose is equal to its inverse. Mathematically, a matrix \( U \) is unitary if \[ U^* U = UU^* = I \], where \( U^* \) is the conjugate transpose and \( I \) is the identity matrix.

• Unitary matrices preserve length and angles, making them essential in quantum mechanics and other fields.

In our exercise, we created a unitary matrix using normalized eigenvectors. We used this matrix to diagonalize \( H \).

Linear Algebra

Linear algebra is a branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. The core concepts include vectors, matrices, determinants, eigenvalues, and eigenvectors.
It is crucial for fields like physics, computer science, and engineering.

• Linear transformations, which map vectors to vectors, are often represented by matrices.
• Understanding properties like eigenvalues and eigenvectors helps simplify complex transformations.

The exercise provided a practical application of these concepts in verifying and manipulating a Hermitian matrix.

Matrix Conjugate Transpose

The conjugate transpose of a matrix, also known as the Hermitian transpose, is obtained by taking the transpose and then taking the complex conjugate of each entry. Mathematically, for a matrix \( A \), its conjugate transpose \( A^* \) is:

• First, transpose the matrix (swap rows and columns).
• Then, take the complex conjugate of each element.

In our exercise, we verified that the matrix \( H \) is Hermitian by showing that \[ H = H^* \], meaning it is equal to its own conjugate transpose. Hermitian matrices have real eigenvalues and orthogonal eigenvectors, making them particularly useful in various applications.