Question

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

3. $\left[\begin{array}{cc}6& 2\\ 2& 3\end{array}\right]$

Short answer

The orthonormal Eigenbasis of symmetric matrix is$\frac{1}{\sqrt{5}}\left[\begin{array}{c}1\\ -2\end{array}\right]and\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\end{array}\right]$

Step-by-step solution

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.

$A=\left[\begin{array}{cc}6& 2\\ 2& 3\end{array}\right]$

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

$$\begin{gathered} \det \left(A-\lambda I_2\right)=0 \\ \Rightarrow det\left(\left[\begin{array}{cc}6& 2\\ 2& 3\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]\right)=0 \\ \Rightarrow det\left(\left[\begin{array}{cc}6-\lambda & 2\\ 2& 3-\lambda \end{array}\right]\right)=0 \\ \Rightarrow (6-\lambda )(3-\lambda )-2\times 2=0 \\ \Rightarrow {\lambda }^2-9\lambda +14=0 \\ \Rightarrow (\lambda -7)(\lambda -2)=0 \\ \Rightarrow \lambda =2,7 \end{gathered}$$

Therefore the eigenvalues of the given matrix A are 2 and 7.

Now from Definition 7.3 .1 for eigenvalue $\lambda =2$we get the eigenspaces as follows.

$$\begin{gathered} E_2=ker(A-2I_2) \\ =ker\left(\left[\begin{array}{cc}6& 2\\ 2& 3\end{array}\right]-\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]\right) \\ =ker\left[\begin{array}{cc}6-2& 2-0\\ 2-0& 3-2\end{array}\right] \\ {E}_2=ker\left[\begin{array}{cc}4& 2\\ 2& 1\end{array}\right] \end{gathered}$$

Now finding the kernel of the matrix (A-2I₂) as represented below.

$\Rightarrow \left[\begin{array}{cc}4& 2\\ 2& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

Now subtracting 1/2 times the 1^st row from the 2^nd row to get a new 2^nd row as represented below.

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

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

$4x_1+2x_2=0\Rightarrow 2x_1+x_2=0\Rightarrow x_2=-2x_1$

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

$\Rightarrow \left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}{x}_1\\ -2x_1\end{array}\right]=x_1\left[\begin{array}{c}1\\ -2\end{array}\right],x_1\in R$

Therefore

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

Find the value of eigenvalues

Eigenvalue $\lambda =7$ we get the eigenspaces as follows.

$$\begin{gathered} E_7=ker(A-7I_2) \\ =ker\left(\left[\begin{array}{cc}6& 2\\ 2& 3\end{array}\right]-\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]\right) \\ =ker\left[\begin{array}{cc}6-7& 2-0\\ 2-0& 3-7\end{array}\right] \\ {E}_7=ker\left[\begin{array}{cc}-1& 2\\ 2& -4\end{array}\right] \end{gathered}$$

Now finding the kernel of the matrix (A-7I₂) as represented below.

$$\begin{gathered} \Rightarrow \left[\begin{array}{cc}-1& 2\\ 2& -4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \\ \left(R_2+2R_1\to R_2\right)\Rightarrow \left[\begin{array}{cc}-1& 2\\ 2& -4\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \end{gathered}$$

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

$-1x_1+2x_2=0\Rightarrow x_1=2x_2$

Therefore, the eigenvector of the given matrix A corresponding to the eigenvalue $\lambda =7$can be represented as shown below.

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

Therefore

$\Rightarrow E_7=span\left[\begin{array}{c}2\\ 1\end{array}\right]$

Plot a graph of eigenspaces

This represents of the eigenspace graph as follows.

We can see that the Eigenspaces E₂ and E₇ are perpendicular to each other. Therefore we can find an orthonormal eigenbasis of the symmetric matrix A by dividing the eigenvectors by their magnitudes as represented below.

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

$\frac{1}{\sqrt{5}}\left[\begin{array}{c}1\\ -2\end{array}\right]and\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\end{array}\right]$

Therefore, the orthonormal eigenbasis of symmetric matrix is$\frac{1}{\sqrt{5}}\left[\begin{array}{c}1\\ -2\end{array}\right]and\frac{1}{\sqrt{5}}\left[\begin{array}{c}2\\ 1\end{array}\right]$