Question

Find the eigenvalues and eigenvectors of A geometrically over the real numbers ℝ. (If an eigenvalue does not exist, enter DNE. If an eigenvector does not exist, enter DNE in any single blank.) A = 0 1 1 0 (reflection in the line y = x) λ1 = has eigenspace span (smaller λ-value) λ2 = has eigenspace span (larger λ-value)

Answer 1

Answer:

∴The eigen values are -1,1.

The eigen vector for 1 is    $\left[\begin{array}{c}1\\-1\end{array}\right]$ .

The eigen vector for $\lambda$= - 1 is    $\left[\begin{array}{c}1\\1\end{array}\right]$ .

Step-by-step explanation:

Given matrix is

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

To find the eigen values of the given matrix, we set

$det(A-\lambda I)=0$

$\left|\begin{array}{cc}0&1\\1&0\end{array}\right|- \lambda \left|\begin{array}{cc}1&0\\0&1\end{array}\right|=0$

$\left|\begin{array}{cc}0&1\\1&0\end{array}\right|- \left|\begin{array}{cc} \lambda&0\\0& \lambda\end{array}\right|=0$

$\left|\begin{array}{cc} -\lambda&1\\1& -\lambda\end{array}\right|=0$

$\Rightarrow \lambda^2-1=0$

$\Rightarrow \lambda^2=1$

$\Rightarrow \lambda=\pm1$

∴The eigen values are -1,1.

For $\lambda=1$

Let the eigen vector for $\lambda=1$ is

$v_1=\left[\begin{array}{c}x_1\\x_2\end{array}\right]$

∴$(A-\lambda I)v_1=O$

$\left[\begin{array}{cc} 1&1\\1&1\end{array}\right] \left[\begin{array}{c}x_1\\x_2\end{array}\right]=O$

$\Rightarrow \left[\begin{array}{c}x_1+x_2\\x_1+x_2\end{array}\right]=O$

$\therefore x_1+x_2=0$

$\Rightarrow x_2=-x_1$

let $x_1=1$

$\therefore x_2=-1$

The eigen vector for 1 is    $\left[\begin{array}{c}1\\-1\end{array}\right]$ .

For $\lambda=-1$

Let the eigen vector for $\lambda=1$ is

$v_2=\left[\begin{array}{c}x_3\\x_4\end{array}\right]$

∴$(A-\lambda I)v_2=O$

$\left[\begin{array}{cc} -1&1\\1&-1\end{array}\right] \left[\begin{array}{c}x_3\\x_4\end{array}\right]=O$

$\Rightarrow \left[\begin{array}{c}-x_3+x_4\\x_3-x_4\end{array}\right]=O$

$\therefore- x_3+x_4=0$

and

$\therefore x_3-x_4=0$

From the above equations, we get

$x_3=x_4$

Let $x_3=1$

Then, $x_4=1$

The eigen vector for $\lambda$= - 1 is    $\left[\begin{array}{c}1\\1\end{array}\right]$ .