MATH 2362 – Symmetric matrices and diagonalization (7.2)

Example (last time): Diagonalize the symmetric matrix

${“version”:”1.1″,”math”:”A=\left[\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right]”}$

Characteristic polynominal: $f(\lambda )=(\lambda -1{)}^2(\lambda -4)$

${“version”:”1.1″,”math”:”f(\lambda)=(\lambda-1)^{2}(\lambda-4)”}$

Eigenvectors: $\underset{―}{\lambda =1}:{\overrightarrow{v}}_1=\left[\begin{array}{c}-1\\ 1\\ 0\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]$ (and their non-zero linear combinations)

${“version”:”1.1″,”math”:”\underline{\lambda=4}: \vec{v}_{3}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]”}$ (and its no-zero multiples)

Then ${“version”:”1.1″,”math”:”P=\left[\begin{array}{lll} \vec{v}_{1} & \vec{v}_{2} & \vec{v}_{3} \end{array}\right]=\left[\begin{array}{ccc} -1 & -1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{array}\right]”}$ diagonalizes A, with ${“version”:”1.1″,”math”:”\small P^{-1} A P=D=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{array}\right]”}$

Theorem: If A is symmetric, any two eigenvectors associated with distinct eigenvalues are orthogonal.

Definition: A square matrix P is called orthogonal is P is invertible and ${“version”:”1.1″,”math”:”\small P^{-1}=P^{T} \text {. }”}$

Theorem: A matrix P is orthogonal if and only if the columns (and rows) of P form an orthonormal set.

MATH 2362 – Orthogonal Diagonalization (7.2)

Defintion: A square matrix A is called orthogonally diagonalizable if there is an orthogonal matrix diagonalizing A, i.e. there is an orthogonal P such that ${“version”:”1.1″,”math”:”\small P^{-1} A P=D”}$.

Theorem: An n x n matrix A is orthogonally diagonalizable if and only if A is symmetric.

Proof:

${“version”:”1.1″,”math”:”\small \Longleftarrow”}$

Idea of proof: if A is symmetric matrix, all its eigenspaces are full-dimensional (i.e. geometric multiplicity=algebraic multiplicity for every eigenvalue).

${“version”:”1.1″,”math”:”\small \Longrightarrow”}$

Corollary: Symmetric matrices are always diagonalizable.

Example: Orthogonally diagonalize the symmetric matrix ${“version”:”1.1″,”math”:”\small A=\left[\begin{array}{lll} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array}\right]”}$

Characteristic polynominal: ${“version”:”1.1″,”math”:”\small f(\lambda)=(\lambda-1)^{2}(\lambda-4)”}$

Eigenvectors:

${“version”:”1.1″,”math”:”\underline{\lambda=1}: \vec{v}_{1}=\left[\begin{array}{c} -1 \\ 1 \\ 0 \end{array}\right], \vec{v}_{2}=\left[\begin{array}{c} -1 \\ 0 \\ 1 \end{array}\right]”}$ (and their non-zero linear combinations)

${“version”:”1.1″,”math”:”\underline{\lambda=4}: \vec{v}_{3}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right]”}$ (and its no-zero multiples)

Then ${“version”:”1.1″,”math”:”P=\left[\begin{array}{lll} \vec{v}_{1} & \vec{v}_{2} & \vec{v}_{3} \end{array}\right]=\left[\begin{array}{ccc} -1 & -1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{array}\right]”}$ diagonalizes A, with  ${“version”:”1.1″,”math”:”\small P^{-1} A P=D=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{array}\right]”}$