Week 4: Preparation

Week 4: Preparation#

Reading Material#

  • Review: We recommend that you review the Mathematics 1a curriculum on diagonalization

  • Reading: In Chapter 1, read Examples 1.4.5 through 1.4.8. Read the rest of Chapter 2, so Sections 2.7 through 2.9

  • Python: Demo 4

Key Concepts#

Long Day will cover:

  • Quadratic equations of elliptic and hyperbolic geometry

  • Symmetric and Hermitian matrices

  • The spectral theorem

  • Diagonalization

  • Orthogonal diagonalization

  • Reduction of quadratic forms

Short Day is dedicated to Theme 2: Data Matrices and Dimensional Reduction.

Preparatory Exercises#

I: Identifying Type of Matrix#

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

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

  1. Is \(A\) symmetric?

  2. Is \(A\) Hermitian?

  3. Is \(A\) normal?

Hint

Remember that a matrix is symmetric if \(A^T = A\), Hermitian if \(A^* = A\) (where \(A^*\) is the conjugated transpose, also called the adjoint matrix) and 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 of \(B\).

  2. Find for each eigenvalue an associated eigenvector.

  3. Normalize the eigenvectors.

  4. Show that \(B\) is orthogonally diagonalizable.

Hint

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

Hint

To normalize the eigenvectors, you can either divide each of them by their own norm, or you can simply choose a unit eigenvector (with a norm of 1) to start with.

Hint

To show that \(B\) is orthogonally diagonalizable you must find an orthogonal matrix \(Q\) and a diagonal matrix \(\Lambda\) such that \(B = Q \Lambda Q^T\).

Answer 1

The characteristic equation

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

has the roots \(\lambda_1 = 1, \lambda_2 = 3\). These are the eigenvalues.

Answer 2

  • An eigenvector for \(\lambda = 1\) is \(\pmb{v}_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}\).

  • An eigenvector for \(\lambda = 3\) is \(\pmb{v}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\).

Note that there are many other correct choices.

Answer 3

  • A unit eigenvector for \(\lambda = 1\) is \(\pmb{q}_1 = \begin{bmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \end{bmatrix}\).

  • A unit eigenvector for \(\lambda = 3\) is \(\pmb{q}_2 = \begin{bmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{bmatrix}\).

Answer 4

Using the following orthogonal matrix and diagonal matrix:

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

we have an orthogonal diagonalization of\(B\), since \(B = Q \Lambda Q^T\).

Contents