Exercise 5.1.3

Calculate the characteristic polynomial of $A$ and find the zeroes of it to solve (i). Find the nonzero vector $v$ such that $(A-\mathit{\lambda I})v=0$ to solve (ii). For (iii) and (iv) just follow the direction given by the textbook.

1.

The characteristic polynomial is $t^2-3t-4$ and the zeroes of it are $4$ and $-1$. For eigenvalue $4$, we may solve (A-4I)x=0. There are infinite solutions. Just pick one from them, say $(2,3)$. Similarly we can find the eigenvector corresponding to $-1$ is $(1,-1)$. Pick

$$\beta =\{(2,3),(1,-1)\}$$

and

$$Q={[I]}_{\beta }^{\alpha }=\left(\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill -1\hfill \end{array}\right),$$

where $\alpha$ is the standard basis for ${ℝ}^2$. Then we know that

$$D=Q^{-1}\mathit{AQ}=\left(\begin{array}{cc}\hfill 4\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill \end{array}\right).$$

2.

The characteristic polynomial is $-\left(t-3\right)\left(t-2\right)\left(t-1\right)$ with $3$ zeroes $1$, $2$, and $3$. The corresponding eigenvector s are $(1,1,-1)$, $(1,-1,0)$, and $(1,0,-1)$. The set of these three vectors are the desired basis. And we also have

$$Q=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 0\hfill & \hfill -1\hfill \end{array}\right)$$

and

$$D=Q^{-1}\mathit{AQ}=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 2\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right).$$

3.

The characteristic polynomial is $\left(t-1\right)\left(t+1\right)$ with two zeroes $-1$ and $1$. The corresponding eigenvectors are $(1,-i-1)$ and $(1,-i+1)$. The set of these two vectors are the desired basis. And we also have

$$Q=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill -i-1\hfill & \hfill -i+1\hfill \end{array}\right)$$

and

$$D=Q^{-1}\mathit{AQ}=\left(\begin{array}{cc}\hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right).$$

4.

The characteristic polynomial is $-{\left(t-1\right)}^{2}t$ with zeroes $0$, $1$, and $1$. The corresponding eigenvectors are $(1,4,2)$, $(1,0,1)$, and $(0,1,0)$. The set of these three vectors are the desired basis. And we also have

$$Q=\left(\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 4\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right)$$

and

$$D=Q^{-1}\mathit{AQ}=\left(\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right).$$