Question

For each of the matrices in Exercises 1 through 6, find an orthonormal eigenbasis. Do not use technology.

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

Short answer

The orthonormal eigenbasis for the given matrix are $\frac{1}{3}\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right],\frac{1}{3}\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right],\frac{1}{3}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]$.

Step-by-step solution

Define symmetric matrix

• In linear algebra, a symmetric matrix is a square matrix that does not change when its transpose is calculated.
• A symmetric matrix is defined as one whose transpose is identical to the matrix itself.
• A square matrix of size n x nis symmetric if B^T= B.

Find the eigenvalues of the given matrix

Given,

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

According to theorem 7.2.1,

$\det \left(A-\lambda I_n\right)=0$

The formula for finding eigenspace is

$E_{\lambda }=\ker \left(A-\lambda I_n\right)=\left\{\overrightarrow{v}\text{in}{\mathrm{ℝ}}^n:A\overline{v}=\lambda \overline{v}\right\}$

Now we need to find an orthonormal eigenbasis for the given matrix A.

$$\begin{gathered} \Rightarrow det\left(\left[\begin{array}{ccc}0& 2& 2\\ 2& 1& 0\\ 2& 0& -1\end{array}\right]-\left[\begin{array}{ccc}\lambda & 0& 0\\ 0& \lambda & 0\\ 0& 0& \lambda \end{array}\right]\right)=0 \\ =det\left(\left[\begin{array}{ccc}0-\lambda & 2-0& 2-0\\ 2-0& 1-\lambda & 0-0\\ 2-0& 0-0& -1-\lambda \end{array}\right]\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow -\lambda (1-\lambda )(-1-\lambda )-2\left(2\right(-1-\lambda )-0)+2(0-2(1-\lambda \left)\right)=0 \\ \Rightarrow \left(-\lambda +{\lambda }^2\right)(-1-\lambda )-2(-2-2\lambda )-4(1-\lambda )=0 \\ \Rightarrow \lambda +{\lambda }^2-{\lambda }^2-{\lambda }^3+4+4\lambda -4+4\lambda =0 \\ \Rightarrow -{\lambda }^3+9\lambda =0 \\ \Rightarrow -\lambda (\lambda -3)(\lambda +3)=0 \\ \Rightarrow \lambda =0,-3,3 \end{gathered}$$

The eigenvalues of the given matrix A are 0,-3,3

Find the eigenspaces for  λ=0

Eigenspace for$\lambda =0$

$$\begin{gathered} E_0=\ker \left(A-0I_3\right) \\ \Rightarrow E_0=ker\left[\begin{array}{ccc}0& 2& 2\\ 2& 1& 0\\ 2& 0& -1\end{array}\right] \end{gathered}$$

Now find the kernel of the matrix.

$\Rightarrow \left[\begin{array}{ccc}0& 2& 2\\ 2& 1& 0\\ 2& 0& -1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

From the above equation we get the solution as

$$\begin{gathered} 2x_2+2x_3=0\Rightarrow 2x_2=-2x_3\Rightarrow x_2=-x_3 \\ 2x_1+x_2=0\Rightarrow 2x_1=-x_2\Rightarrow x_1=-\frac{1}{2}{x}_2 \\ 2x_1-x_3=0\Rightarrow 2x_1=x_3\Rightarrow x_1=\frac{1}{2}{x}_3 \end{gathered}$$

Therefore, the eigenvector of the given matrix can be represented as:

$\Rightarrow \left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}\frac{1}{2}{x}_3\\ -x_3\\ x_3\end{array}\right]=\frac{1}{2}{x}_3\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right],x_3\in r$

Hence the eigenspace for $\lambda =0$is $E_0=span\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right]$.

Find the eigenspaces for λ=-3

Eigenspace for$\lambda =-3$

$$\begin{gathered} E_{-3}=\ker \left(A+3I_3\right) \\ =ker\left(\left[\begin{array}{ccc}0& 2& 2\\ 2& 1& 0\\ 2& 0& -1\end{array}\right]+\left[\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]\right) \\ =ker\left(\left[\begin{array}{ccc}0+3& 2+0& 2+0\\ 2+0& 1+3& 0+0\\ 2+0& 0+0& -1+3\end{array}\right]\right) \\ =ker\left(\left[\begin{array}{ccc}3& 2& 2\\ 2& 4& 0\\ 2& 0& 2\end{array}\right]\right) \end{gathered}$$

Find the kernel of the matrix.

$\Rightarrow \left(\left[\begin{array}{ccc}3& 2& 2\\ 2& 4& 0\\ 2& 0& 2\end{array}\right]\right)\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

Multiply first row with 1/3 to get new first row.

$\left(\frac{1}{3}\times R_1\to R_1\right)\Rightarrow \left(\left[\begin{array}{ccc}1& \frac{2}{3}& \frac{2}{3}\\ 2& 4& 0\\ 2& 0& 2\end{array}\right]\right)\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new second row:

$\left(R_2-2R_1\to R_2\right)\Rightarrow \left(\left[\begin{array}{ccc}1& \frac{2}{3}& \frac{2}{3}\\ 0& \frac{8}{3}& -\frac{4}{3}\\ 2& 0& 2\end{array}\right]\right)\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new third row:

$\left(R_3-2R_1\to R_3\right)\Rightarrow \left(\left[\begin{array}{ccc}1& \frac{2}{3}& \frac{2}{3}\\ 0& \frac{8}{3}& -\frac{4}{3}\\ 0& -\frac{4}{3}& \frac{2}{3}\end{array}\right]\right)\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

New second row:

$\left(\frac{3}{8}\times R_2\to R_2\right)\Rightarrow \left(\left[\begin{array}{ccc}1& \frac{2}{3}& \frac{2}{3}\\ 0& 1& -\frac{1}{2}\\ 0& -\frac{4}{3}& \frac{2}{3}\end{array}\right]\right)\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

New third row:

$\left(R_3+\frac{4}{3}\times R_2\to R_3\right)\Rightarrow \left(\left[\begin{array}{ccc}1& \frac{2}{3}& \frac{2}{3}\\ 0& 1& -\frac{1}{2}\\ 0& 0& 0\end{array}\right]\right)\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

New first row:

$B^T=B\left(R_1-\frac{2}{3}\times R_2\to R_1\right)\Rightarrow \left(\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& -\frac{1}{2}\\ 0& 0& 0\end{array}\right]\right)\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\left(1\right)$

From equation (1) we get,

$$\begin{gathered} x_1+x_3=0\Rightarrow x_1=-x_3 \\ {x}_2-\frac{1}{2}{x}_3=0\Rightarrow x_2=\frac{1}{2}{x}_3 \end{gathered}$$

Therefore the eigenvector is given as

$\Rightarrow \left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}{x}_3\\ \frac{1}{2}{x}_3\\ x_3\end{array}\right]=\frac{1}{2}{x}_3\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right],x_3\in R$

Hence the eigenspace for $\lambda =-3$is $\Rightarrow E_{-3}=span\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]$.

Find the eigenspaces for λ=3

Eigenspace for$\lambda =3$

$$\begin{gathered} E_3=\ker \left(A-3I_3\right) \\ =ker\left(\left[\begin{array}{ccc}0& 2& 2\\ 2& 1& 0\\ 2& 0& -1\end{array}\right]-\left[\begin{array}{ccc}3& 0& 0\\ 0& 3& 0\\ 0& 0& 3\end{array}\right]\right) \\ =ker\left[\begin{array}{ccc}-3& 2& 2\\ 2& -2& 0\\ 2& 0& -4\end{array}\right] \end{gathered}$$

Finding the kernel:

$\Rightarrow \left[\begin{array}{ccc}-3& 2& 2\\ 2& -2& 0\\ 2& 0& -4\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new first row:

$\left(-\frac{1}{3}\times R_1\to R_1\right)\Rightarrow \left[\begin{array}{ccc}1& -\frac{2}{3}& -\frac{2}{3}\\ 2& -2& 0\\ 2& 0& -4\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new second row:

$\left(R_2-2R_1\to R_2\right)\Rightarrow \left[\begin{array}{ccc}1& -\frac{2}{3}& -\frac{2}{3}\\ 0& -\frac{2}{3}& \frac{4}{3}\\ 2& 0& -4\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$$\left(-\frac{1}{3}\times R_1\to R_1\right)\Rightarrow \left[\begin{array}{ccc}1& -\frac{2}{3}& -\frac{2}{3}\\ 2& -2& 0\\ 2& 0& -4\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new third row:

$\left(R_3-2R_1\to R_3\right)\Rightarrow \left[\begin{array}{ccc}1& -\frac{2}{3}& -\frac{2}{3}\\ 0& -\frac{2}{3}& \frac{4}{3}\\ 0& \frac{4}{3}& -\frac{8}{3}\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new second row:

$\left(-\frac{3}{2}\times R_2\to R_2\right)\Rightarrow \left[\begin{array}{ccc}1& -\frac{2}{3}& -\frac{2}{3}\\ 0& 1& -2\\ 0& \frac{4}{3}& -\frac{8}{3}\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new third row:

$\left(R_3-\frac{4}{3}\times R_2\to R_3\right)\Rightarrow \left[\begin{array}{ccc}1& -\frac{2}{3}& -\frac{2}{3}\\ 0& 1& -2\\ 0& 0& 0\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$

To get new first row:

$\left(R_1+\frac{2}{3}\times R_2\to R_1\right)\Rightarrow \left[\begin{array}{ccc}1& 0& -2\\ 0& 1& -2\\ 0& 0& 0\end{array}\right]=\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]\left(2\right)$

From the above equation we get,

$$\begin{gathered} x_1-2x_3=0\Rightarrow x_1=2x_3 \\ {x}_2-2x_3=0\Rightarrow x_2=2x_3 \end{gathered}$$

The eigenvectors are:

$\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}2x_3\\ 2x_3\\ x_3\end{array}\right]=x_3\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right],x_3\in R$$\Rightarrow \left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}2x_3\\ 2x_3\\ x_3\end{array}\right]=x_3\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right],x_3\in R$

Hence the eigenspace for$\lambda =3$is$E_3=span\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]$

Find the orthonormal eigenbasis

The eigenspaces are perpendicular to each other. The orthonormal eigenbasis can be calculated as follows:

$$\begin{gathered} \frac{1}{\sqrt{{\left(1\right)}^2+{\left(-2\right)}^2+2^2}}\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right]\Rightarrow \frac{1}{3}\left[\begin{array}{c}1\\ 2\\ 2\end{array}\right] \\ \frac{1}{\sqrt{{(-2)}^2+{\left(1\right)}^2+2^2}}\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right]\Rightarrow \frac{1}{3}\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right] \\ \frac{1}{\sqrt{2^2+2^2+1^2}}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]\Rightarrow \frac{1}{3}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right] \end{gathered}$$

Hence the orthonormal eigenbasis are $\frac{1}{3}\left[\begin{array}{c}1\\ -2\\ 2\end{array}\right],\frac{1}{3}\left[\begin{array}{c}-2\\ 1\\ 2\end{array}\right],\frac{1}{3}\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right]$.