Exercise 6.4.2

Use one orthonormal basis $\beta$ to check ${[T]}_{\beta }$ is normal, self-adjoint, or neither. Usually, we’ll take $\beta$ to be the standard basis. To find an orthonormal basis of eigenvectors of $T$ for $V$, just find an orthonormal basis for each eigenspace and take the union of them as the desired basis.

1.

Pick $\beta$ to be the standard basis and get that

$${[T]}_{\beta }=\left(\begin{array}{cc}\hfill 2\hfill & \hfill -2\hfill \\ \hfill -2\hfill & \hfill 5\hfill \end{array}\right).$$

So it’s self-adjoint. And the basis is

$$\left\{\frac{1}{\sqrt{5}}(1,-2),\frac{1}{\sqrt{5}}(2,1)\right\}.$$

2.

Pick $\beta$ to be the standard basis and get that

$${[T]}_{\beta }=\left(\begin{array}{ccc}\hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 5\hfill & \hfill 0\hfill \\ \hfill 4\hfill & \hfill -2\hfill & \hfill 5\hfill \end{array}\right).$$

So it’s neither normal nor self-adjoint.

3.

Pick $\beta$ to be the standard basis and get that

$${[T]}_{\beta }=\left(\begin{array}{cc}\hfill 2\hfill & \hfill i\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right).$$

So it’s normal but not self-adjoint. And the basis is

$$\left\{(\frac{1}{\sqrt{2}},-\frac{1}{2}+\frac{1}{2}i),(\frac{1}{\sqrt{2}},\frac{1}{2}-\frac{1}{2}i)\right\}.$$

4.

Pick an orthonormal basis $\beta =\{1,\sqrt{3}(2t-1),\sqrt{6}(6t^2-6t+1)\}$ by Exercise 6.2.2(c) and get that

$${[T]}_{\beta }=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 2\sqrt{3}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 6\sqrt{2}\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right).$$

So it’s neither normal nor self-adjoint.

5.

Pick $\beta$ to be the standard basis and get that

$${[T]}_{\beta }=\left(\begin{array}{cccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right).$$

So it’s self-adjoint. And the basis is

$$\left\{(1,0,0,0),\frac{1}{\sqrt{2}}(0,1,1,0),(0,0,0,1),\frac{1}{\sqrt{2}}(0,1,-1,0)\right\}$$

6.

Pick $\beta$ to be the standard basis and get that

$${[T]}_{\beta }=\left(\begin{array}{cccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right).$$

So it’s self-adjoint. And the basis is

$$\left\{\frac{1}{\sqrt{2}}(1,0,-1,0),\frac{1}{\sqrt{2}}(0,1,0,-1),\frac{1}{\sqrt{2}}(1,0,1,0),\frac{1}{\sqrt{2}}(0,1,0,1)\right\}$$