Question

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

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

Short answer

The orthonormal Eigenbasis of symmetric matrix is$\left[\begin{array}{c}0\\ 1\end{array}\right]and\left[\begin{array}{c}1\\ 0\end{array}\right]$.

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.

$A=\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]$

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

$$\begin{gathered} \det \left(A-\lambda I_2\right)=0 \\ det\left(\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]\right)=0 \\ det\left(\left[\begin{array}{cc}1-\lambda & 0\\ 0& 2-\lambda \end{array}\right]\right)=0 \\ (1-\lambda )(2-\lambda )=0 \\ \lambda =1,2 \end{gathered}$$

Therefore, the eigenvalues of the given matrix A are 1 and 2. (The eigenvalues of a triangular matrix are its diagonal entries.)

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

$$\begin{gathered} E_1=\ker \left(A-1I_2\right) \\ =ker\left(\left[\begin{array}{cc}1& 0\\ 0& 2\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right) \\ =ker\left(\left[\begin{array}{cc}1-1& 0-0\\ 0-0& 2-1\end{array}\right]\right) \\ =ker\left[\begin{array}{cc}0& 0\\ 0& 1\end{array}\right] \end{gathered}$$

Now finding the kernel of the matrix (A-1/2) as represented below.

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

From the above equation, we get x₂=0 and x₁ as a free variable. Therefore, the eigenvector of the given matrix A corresponding to the eigenvalue $\lambda =1$can be represented as shown below.

$\Rightarrow \left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}{x}_1\\ 0\end{array}\right]=x_1\left[\begin{array}{c}1\\ 0\end{array}\right],x_1\in R$

Therefore

$\Rightarrow E_1=span\left[\begin{array}{c}1\\ 0\end{array}\right]$

Find the value of eigenvalues

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

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

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

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

From the above equation, we get x₁= 0 and x₂ as a free variable. Therefore, the eigenvector of the given matrix A corresponding to the eigenvalue $\lambda =2$can be represented as shown below.

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

Therefore

$\Rightarrow E_2=span\left[\begin{array}{c}0\\ 1\end{array}\right]$

Plot a graph of eigenspaces

This represents of the eigenspace graph as follows.

We can see that the eigenspaces E₁and E₂are 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.

$$\begin{gathered} \frac{1}{\sqrt{0^2+1^2}}\left[\begin{array}{c}0\\ 1\end{array}\right]and\frac{1}{\sqrt{1^2+0^2}}\left[\begin{array}{c}1\\ 0\end{array}\right] \\ \Rightarrow \left[\begin{array}{c}0\\ 1\end{array}\right]and\left[\begin{array}{c}1\\ 0\end{array}\right] \end{gathered}$$

Therefore, the orthonormal eigenbasis of symmetric matrix is $\left[\begin{array}{c}0\\ 1\end{array}\right]and\left[\begin{array}{c}1\\ 0\end{array}\right]$.