Question

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

4.$A=\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 1\\ 1& 1& 1\end{array}\right]$

Short answer

The orthonormal Eigenbasis of symmetric matrix is $\frac{1}{\sqrt{3}}\left[\begin{array}{c}-1\\ -1\\ 1\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}-1\\ 1\\ 0\end{array}\right],\frac{1}{\sqrt{6}}\left[\begin{array}{c}1\\ 1\\ 2\end{array}\right]$.

Step-by-step solution

 Step 1: Definition and Theorem of Eigenvalues.

Theorem 7.2.1

Consider an n x n matrix A and a scalar $\lambda$. Then $\lambda$is an eigenvalue of A if (and only if)

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

This is called the characteristic equation (or the secular equation) of matrix A.

Definition 7.3.1

Consider an eigenvalue $\lambda$of an n x n matrix A. Then the kernel of the matrix $A-\lambda I_n$is called the Eigenspace associated with $\lambda$, denoted by E_$\lambda$ :

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

Find the value of eigenspace

Now consider the given matrix as represented below. We have to find an orthonormal eigenbasis of the matrix A.

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

Now by Theorem 7.2.1, solving the characteristic equation of the given matrix A as represented below.

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

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

Therefore the eigenvalues of the given matrix A are -1,0 and 2.

Now from Definition 7.3.1 for eigenvalue $\lambda =-1$we get the eigenspaces as follows.

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

Now finding the kernel of the matrix (A+1I₃) as represented below.

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

Now subtracting the 2^nd row from the 3^rd row to get a new 3^rd row as represented below.

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

Now subtracting the 1^st row from the 3^rd row to get a new 3^rd row as represented below.

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

Find the value of eigenvalues

From equation (1) we get the solutions as represented below.

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

Therefore, the eigenvector of the given matrix A corresponding to the eigenvalue $\lambda =-1$can be represented as shown below.

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

Therefore

$\Rightarrow E_{-1}=span\left[\begin{array}{c}-1\\ -1\\ 1\end{array}\right]$

Similarly from Definition 7.3.1 for eigenvalue $\lambda =0$we get the eigenspaces as follows.

$$\begin{gathered} E_0=ker(A-0I_3) \\ {E}_0=ker\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 1\\ 1& 1& 1\end{array}\right] \end{gathered}$$

Now finding the kernel of the matrix $(A-0I_3)$as represented below.

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

From the above equation we get the solutions as represented below.

$$\begin{gathered} 0x_1+0x_2+x_3=0\Rightarrow x_3=0 \\ {x}_1+x_2+x_3=0\Rightarrow x_1+x_2+0=0\Rightarrow x_1=-x_2 \end{gathered}$$

Therefore, the eigenvector of the given matrix A corresponding to the eigenvalue $\lambda =0$can be represented as shown below.

$\Rightarrow \left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}-x_2\\ x_2\\ 0\end{array}\right]=x_2\left[\begin{array}{c}-1\\ 1\\ 0\end{array}\right],x_2\in R$

Therefore

$\Rightarrow E_0=span\left[\begin{array}{c}-1\\ 1\\ 0\end{array}\right]$

Similarly from Definition 7.3.1 for eigenvalue $\lambda =2$ we get the eigenspaces as follows.

$$\begin{gathered} E_2=ker(A-2I_3) \\ =ker\left(\left[\begin{array}{ccc}0& 0& 1\\ 0& 0& 1\\ 1& 1& 1\end{array}\right]-\left[\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]\right) \\ =ker\left[\begin{array}{ccc}0-2& 0& 1-0\\ 0& 0-2& 1-0\\ 1-0& 1-0& 1-2\end{array}\right] \\ =ker\left[\begin{array}{ccc}-2& 0& 1\\ 0& -2& 1\\ 1& 1& -1\end{array}\right] \end{gathered}$$

Determine the kernel of the matrix

Now finding the kernel of the matrix $(A-2I_3)$as represented below.

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

Now adding 1/2 times the 1^st row to the 3^rd row to get a new 3^rd row as represented below.

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

Now adding 1/2 times the 2^nd row to the 3^rd row to get a new 3^rd row as represented below.

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

From the above equation we get the solutions as represented below.

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

Therefore, the eigenvector of the given matrix A corresponding to the eigenvalue $\lambda =2$can be represented as shown below.

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

Therefore

$\Rightarrow E_2=span\left[\begin{array}{c}1\\ 1\\ 2\end{array}\right]$

Plot the graph of the eigenspaces

The graph represent the eigenspaces and as shown below.

We can see that the eigenspaces E_-1, E₀ and E₂ are perpendicular to each other. Therefore we can find an orthonormal eigenbasis of the symmetric matrix A by dividing the eigenvectors by their lengths as represented below.

$$\begin{gathered} \frac{1}{\sqrt{{(-1)}^2+{(-1)}^2+1^2}}\left[\begin{array}{c}-1\\ -1\\ 1\end{array}\right]\Rightarrow \frac{1}{\sqrt{3}}\left[\begin{array}{c}1\\ -1\\ 1\end{array}\right] \\ \frac{1}{\sqrt{{(-1)}^2+{\left(1\right)}^2+\left(\right)}}\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]\Rightarrow \frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right] \\ \frac{1}{\sqrt{{\left(1\right)}^2+{\left(1\right)}^2+{\left(2\right)}^2}}\left[\begin{array}{c}1\\ 1\\ 2\end{array}\right]\Rightarrow \frac{1}{\sqrt{6}}\left[\begin{array}{c}1\\ 1\\ 2\end{array}\right] \end{gathered}$$

Therefore, the orthonormal Eigenbasis of symmetric matrix is

$\frac{1}{\sqrt{3}}\left[\begin{array}{c}-1\\ -1\\ 1\end{array}\right],\frac{1}{\sqrt{2}}\left[\begin{array}{c}-1\\ 1\\ 0\end{array}\right],\frac{1}{\sqrt{6}}\left[\begin{array}{c}1\\ 1\\ 2\end{array}\right]$