Matrix Orthonormal Diagonalization

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Example: Apply the Gram-Schmidt process to find an orthonormal matrix $P$ that diagonalizes

$A = \begin{bmatrix}0&2&2 \\ 2&0&2\\ 2&2&0 \end{bmatrix}$

Solution

The characteristic polynomial of the matrix $A$ is given by

$P(\lambda) = \text{det}(\lambda I_3 - A) =$ $\text{det}\begin{pmatrix}\begin{bmatrix} \lambda & -2 & -2\\ -2 & \lambda & -2\\ -2 & -2 & -2 \end{bmatrix} \end{pmatrix}$

$= \lambda(\lambda^{2} - 4) - 4(\lambda + 2) - 4(\lambda + 2)$
$= \lambda(\lambda + 2)(\lambda -2) - 8(\lambda + 2)$
$= (\lambda + 2)(\lambda^{2} - 2\lambda - {\rm 8})$
$= (\lambda + 2)^{2}(\lambda -4) = 0 \nonumber \ (\text{characteristic equation})$

Solving for $\lambda$ , we obtain $\lambda_1 = \lambda_2 = -2$ , $\lambda_3 = 4.$

Note that $-2$ is the eigenvalue of multiplicity $2.$

Next, we find the eigenvectors associated with $\lambda = -2$ by solving the homogeneous system

$(-2I_3 - A)X = 0,$

$\begin{bmatrix} -2&-2&-2\\ -2&-2&-2\\ -2&-2&-2 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix}0\\ 0\\ 0 \end{bmatrix}$

Writing the system above in an augmented matrix and applying elementary row operations, we have

$\begin{bmatrix}1& 1& 1 & 0\\ 0&0&0&0\\ 0&0&0&0 \end{bmatrix}$

which implies $x_1 + x_2 + x_3 = 0.$

Choosing $x_3 = r, \ x_2 = t,$ we get

$\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} -t - r\\ t \\ r \end{bmatrix} = \begin{bmatrix} -1\\ 1\\ 0 \end{bmatrix} + r\begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}$

Thus, basis for the solution space of the linear system consists of the eigenvectors

$u_1 = \begin{bmatrix} -1\\ 1\\ 0, \end{bmatrix} \ u_2 = \begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix}$

Since $u_1$ and $u_2$ are not orthogonal, we use the Gram-Schmidt process to obtain an orthonormal basis.

Let $v_1 = u_1$

$v_2 = u_2 - \frac{u_2 . v_1}{v_1.v_1}v_1$

$= \begin{bmatrix} -1\\ 0\\ 1 \end{bmatrix} - \begin{bmatrix} -1\\ 1\\ 0 \end{bmatrix}$

$= \begin{bmatrix} -\frac{1}{2}\\ -\frac{1}{2}\\ 1 \end{bmatrix}$

The set $\begin{Bmatrix}v_1, \ v_2 \end{Bmatrix}$ is orthogonal.

Normalizing $\begin{Bmatrix}v_1, \ v_2 \end{Bmatrix}$ , we obtain

$w_1 = \frac{1}{||v_1||}v_1 = \frac{1}{\sqrt{2}} \begin{bmatrix} -1\\ 1\\ 0 \end{bmatrix} = \begin{bmatrix} -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\\ 0 \end{bmatrix},$

$w_2 = \frac{1}{||v_2||}v_2 = \frac{\sqrt{2}}{\sqrt{3}}\begin{bmatrix} -\frac{1}{2}\\ -\frac{1}{2}\\ 1 \end{bmatrix}$
$= \begin{bmatrix} -\frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{6}}\\ \sqrt{\frac{2}{3}}\end{bmatrix}.$

Now, we find the basis for the eigenvector associated with $\lambda_3 = 4$ by solving the homogeneous system

$(4I_3 - A)X = 0,$

$\begin{bmatrix} 4&-2&-2\\ -2&4&-2\\ -2&-2&4\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}.$

Writing the homogeneous system above in augmented matrix and applying elementary row operations

$\begin{bmatrix} 4&-2&-2&0\\ -2&4&-2&0\\ -2&-2&4&0 \end{bmatrix}$

$\sim \begin{bmatrix} 4&-2&-2&0\\ 0&3&-3&0\\ 0&6&-6&0 \end{bmatrix}$

$\sim \begin{bmatrix} 1&-\frac{1}{2}&-\frac{1}{2}&0\\ 0&1&-1&0\\ 0&0&0&0 \end{bmatrix}.$

If we choose $x_3 = r,$

$\begin{bmatrix}x_1\\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}r + \frac{1}{2}r\\ r\\ r \end{bmatrix} = r\begin{bmatrix}1\\ 1\\ 1 \end{bmatrix}.$

Hence a basis for the solution space consists of the eigenvector

$u_3 = \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$

Normalizing $u_3,$ we have

$w_3 = \frac{1}{\sqrt{3}}\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\end{bmatrix}.$

Therefore, the orthonormal matrix $P$ that diagonalizes $A$ is

$\begin{bmatrix} -\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\ 0&\frac{2}{\sqrt{6}}&\frac{1}{\sqrt{3}}\end{bmatrix}.$

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Matrix Orthonormal Diagonalization

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