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7.2 Orthogonal Diagonalization: Class Notes

Definition

If $A$ and $B$ are square matrices, then we say that $B$ is orthogonally similar to $A$ (or that $A$ and $B$ are orthogonally similar matrices) if there is an orthogonal matrix $P$ such that $B=P^{T}AP$.

If $A$ is orthogonally similar to some diagonal matrix, say $P^{T}AP=D$, then we say $A$ is orthogonally diagonalizable and that $P$ orthogonally diagonalizes $A$.

Overview:

$$A\to {\lambda }_i\to {𝐩}_i\to \text{Gram–Schmidt}\to P(\text{orthogonal})$$

$$P^{T}AP=D(P^{-1}=P^T)$$

Theorem 7.2.1

#2

Find the characteristic equation of the given symmetric matrix, and then by inspection determine the dimensions of the eigenspaces.

$$\left[\begin{array}{ccc}1& -4& 2\\ -4& 1& -2\\ 2& -2& -2\end{array}\right]$$

Solution

Note that this matrix is symmetric and so is orthogonally diagonalizable.

Set $det(\lambda I-A)=0$ and expand along the first row:

$$\left|\begin{array}{ccc}\lambda -1& 4& -2\\ 4& \lambda -1& 2\\ -2& 2& \lambda +2\end{array}\right|=0$$

$$(\lambda -1)\left|\begin{array}{cc}\lambda -1& 2\\ 2& \lambda +2\end{array}\right|-4\left|\begin{array}{cc}4& 2\\ -2& \lambda +2\end{array}\right|-2\left|\begin{array}{cc}4& \lambda -1\\ -2& 2\end{array}\right|=0$$

$$(\lambda -1)({\lambda }^2+\lambda -6)-4(4\lambda +8+4)-2(8+2\lambda -2)=0$$

Each piece has a factor of $(\lambda +3)$:

$$(\lambda -1)(\lambda +3)(\lambda -2)-4·4(\lambda +3)-2·2(\lambda +3)=0$$

$$\begin{array}{cc}\hfill {\displaystyle (\lambda +3)[(\lambda -1)(\lambda -2)-16-4]}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (\lambda +3)({\lambda }^2-3\lambda -18)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (\lambda +3{)}^2(\lambda -6)}& {\displaystyle =0}\hfill \end{array}$$

Since the matrix is symmetric, the dimension of each eigenspace is the multiplicity of its eigenvalue:

Theorem 7.2.2

To orthogonally diagonalize an $n\times n$ symmetric matrix:

• Step 1: Find a basis for each eigenspace of $A$.
• Step 2: Apply the Gram–Schmidt process to each basis to obtain an orthonormal basis for each eigenspace.
• Step 3: Form $P$ whose columns are the vectors from Step 2.

#11

Find a matrix $P$ that orthogonally diagonalizes $A$, and determine $P^{-1}AP$.

$$A=\left[\begin{array}{ccc}2& -1& -1\\ -1& 2& -1\\ -1& -1& 2\end{array}\right]$$

Solution

Since $P$ is orthogonal, $P^{-1}AP=P^{T}AP=D$. Set $det(\lambda I-A)=0$ and expand along the first row:

$$\left|\begin{array}{ccc}\lambda -2& 1& 1\\ 1& \lambda -2& 1\\ 1& 1& \lambda -2\end{array}\right|=0$$

$$(\lambda -2)\left|\begin{array}{cc}\lambda -2& 1\\ 1& \lambda -2\end{array}\right|-1\left|\begin{array}{cc}1& 1\\ 1& \lambda -2\end{array}\right|+1\left|\begin{array}{cc}1& \lambda -2\\ 1& 1\end{array}\right|=0$$

$$\begin{array}{cc}\hfill {\displaystyle (\lambda -2)[(\lambda -2{)}^2-1]-(\lambda -2-1)+(1-\lambda +2)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (\lambda -2)({\lambda }^2-4\lambda +3)-(\lambda -3)-(\lambda -3)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (\lambda -2)(\lambda -3)(\lambda -1)-2(\lambda -3)}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (\lambda -3)[(\lambda -2)(\lambda -1)-2]}& {\displaystyle =0}\hfill \\ \hfill {\displaystyle (\lambda -3)({\lambda }^2-3\lambda )}& {\displaystyle =0\Rightarrow \lambda (\lambda -3{)}^2=0}\hfill \end{array}$$

Factor out the common $(\lambda -3)$. The eigenvalues are $\lambda =0$ (multiplicity 1) and $\lambda =3$ (multiplicity 2).

$\lambda =0$: Row reduce $0I-A$:

$$\left[\begin{array}{ccc}-2& 1& 1\\ 1& -2& 1\\ 1& 1& -2\end{array}\right]\to \left[\begin{array}{ccc}-2& 1& 1\\ 0& 1& -1\\ 0& 0& 0\end{array}\right]\to \left[\begin{array}{ccc}-2& 0& 2\\ 0& 1& -1\\ 0& 0& 0\end{array}\right]$$

$R_2\to R_1+2R_2$ gives $(0,-3,3)$ and $R_3\to R_1+2R_3$ gives $(0,3,-3)$. Then $R_2\to -\frac{1}{3}{R}_2$ and $R_3\to R_3+3R_2$ give the middle matrix, and $R_1\to R_1-R_2$ gives the last one.

So $x_1=x_3$ and $x_2=x_3$:

$${\displaystyle 𝐱=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]t\Rightarrow {𝐩}_1=\frac{(1,1,1)}{\Vert (1,1,1)\Vert }=\left[\begin{array}{c}1/\sqrt{3}\\ 1/\sqrt{3}\\ 1/\sqrt{3}\end{array}\right]}$$

$\lambda =3$: Row reduce $3I-A$:

$$\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]\to \left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$$

So $x_1=-x_2-x_3$. Let $s=x_2$, $t=x_3$:

$$𝐱=\left[\begin{array}{c}-1\\ 1\\ 0\end{array}\right]s+\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]t$$

Any nonzero multiples of these work as basis vectors, so use

$${𝐮}_1=\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right],{𝐮}_2=\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right]$$

These are not orthogonal, so apply Gram–Schmidt:

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle {𝐯}_1}& {\displaystyle ={𝐮}_1}\hfill \\ \hfill {\displaystyle {𝐯}_2}& {\displaystyle ={𝐮}_2-\frac{⟨{𝐮}_2,{𝐯}_1⟩}{\Vert {𝐯}_1{\Vert }^2}{𝐯}_1=\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right]-\frac{1}{2}\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right]=\left[\begin{array}{c}1/2\\ 1/2\\ -1\end{array}\right]}\hfill \end{array}}$$

Use ${𝐯}_2\prime =\left[\begin{array}{c}1\\ 1\\ -2\end{array}\right]$ (twice ${𝐯}_2$) to avoid fractions. Then

$${\displaystyle {𝐩}_2=\frac{{𝐯}_1}{\Vert {𝐯}_1\Vert }=\left[\begin{array}{c}1/\sqrt{2}\\ -1/\sqrt{2}\\ 0\end{array}\right],{𝐩}_3=\frac{{𝐯}_2\prime }{\Vert {𝐯}_2\prime \Vert }=\left[\begin{array}{c}1/\sqrt{6}\\ 1/\sqrt{6}\\ -2/\sqrt{6}\end{array}\right]}$$