Week 4: Preparation

Reading Material

We recommend that you read the textbook. Watching YouTube videos on the week’s topics can be useful, but it should not replace proper preparation for the week’s program and is not recommended as a standalone approach.

Read and study the following:

• Long Day: The rest of Chapter 2

• Short Day: Thema Exercise 2 Theme 2: Data Matrices and Dimensional Reduction

• Python demo04

Key Concepts

After reading, you should be able to explain the following key concepts:

• Symmetric and Hermitian matrices

• The spectral theorem

• Diagonalization

• Orthogonal diagonalization

This week, we will explore these key concepts in great detail. We expect you to have familiarized yourself with these topics before lectures.

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Preparatory Exercises

I: Identifying Type of Matrix

Consider a \(2\times 2\) matrix given by:

\[\begin{equation*} A = \begin{bmatrix} 2 & 1-i \\ 1+i & 3 \end{bmatrix}. \end{equation*}\]

Answer the following questions:

1. Is \(A\) symmetric? (A matrix is symmetric if \(A^T = A\).)

2. Is \(A\) Hermitian? (A matrix is Hermitian if \(A^* = A\), where \(A^*\) is the conjugated transpose.)

3. Is \(A\) normal? (A matrix \(A\) is normal, if \(AA^* = A^*A\).)

Hint

First, compare \(A^T\) with \(A\) (without finding the complex conjugate) and then \(A^*\) with \(A\).

Answer

1. \(A^T = \begin{bmatrix} 2 & 1+i \\ 1-i & 3 \end{bmatrix} \neq A\), Since \((A^T)_{1,2} \neq A_{1,2}\). Hence \(A\) is not symmetric.

2. \(A^* = \begin{bmatrix} 2 & 1-i \\ 1+i & 3 \end{bmatrix} = A\). Hence \(A\) is Hermitian.

3. Since \(A\) is Hermitian, it is also automatically normal.

II: Diagonalization of a Symmetric \(2\times 2\) Matrix

Consider the real symmetric matrix

\[\begin{equation*} B = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}. \end{equation*}\]

1. Find the eigenvalues: Find the eigenvalues of \(B\) by finding the roots of the characteristic polynomial \(\det(B - \lambda I) = 0\).

2. Find eigenvectors: Find for each eigenvalue an associated eigenvector.

3. Normalize the eigenvectors: Choose eigenvectors with a norm of \(1\).

4. Orthogonal diagonalization: Show, that \(B\) is orthogonally diagonalizable by finding an orthogonal matrix \(Q\) and a diagonal matrix \(\Lambda\) with \(B = Q \Lambda Q^T\).

Answer 1

Eigenvalues:

\[\begin{equation*} \lambda^2 - 4\lambda + 3 = 0 \end{equation*}\]

has the roots \(\lambda_1 = 1, \lambda_2 = 3\)

Answer 2

Eigenvectors:

• \(\lambda = 1\): \(\pmb{v}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\).

• \(\lambda = 3\): \(\pmb{v}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\).

Answer 3

Unit eigenvectors:

• \(\lambda = 1\): \(\pmb{q}_1 = \begin{bmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \end{bmatrix}\).

• \(\lambda = 3\): \(\pmb{q}_2 = \begin{bmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{bmatrix}\).

Answer 4

Orthogonal matrix:

\[\begin{equation*} Q = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}, \quad \Lambda = \begin{bmatrix} 1 & 0 \\ 0 & 3 \end{bmatrix}. \end{equation*}\]

Hence \(B = Q \Lambda Q^T\).

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