91Ó°ÊÓ

A function is defined as a relationship between a set of inputs that each have one output.

Given,

$$\begin{gathered} A=\left[1         1      1 \\ 0        0     1 \\ 0        0      2\right] \end{gathered}$$

This matrix is lower triangular.

So its eigenvalues are the entries on its main diagonal, which are

${\mathrm{λ}}_1=1,{\mathrm{λ}}_2=0,{\mathrm{λ}}_3=2$

We have three distinct real eigenvalues of a$3×3$matrix.

so there exists an eigenbasis in which the diagonalization of A is

$$\begin{gathered} B=\left[1         0      0 \\ 0        0     0 \\ 0        1      2\right] \end{gathered}$$

$\mathrm{λ}=1$, we solved,

$\left(A-I\right)x=0$

$$\begin{gathered} \left[1         1      1 \\ 0    0       2 \\ 0        0       0\right]  \left[ x_1 \\  x_2 \\ {x}_3\right]=\left[0 \\ 0 \\ 0\right]    \end{gathered}$$

$$\begin{gathered} x_2+x_3=0 \\ 2x_3=0 \end{gathered}$$

$X=x\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]$

$$\begin{gathered} E_1=\text{span}\left(\left[1 \\ 0 \\ 0\right]\right) \end{gathered}$$

$ \mathrm{λ}=0$,we solved,

$Ax=0$

$$\begin{gathered} \left[0         1      1 \\ 0        0       1 \\ 0        1       2\right]  \left[ x_1 \\  x_2 \\ {x}_3\right]=\left[0 \\ 0 \\ 0\right] \end{gathered}$$

$$\begin{gathered} x_1+x_2+x_3=0,  x_3=0, 2x_3=0 \\ {x}_1+x_2=0,x_3=0 \end{gathered}$$

$$\begin{gathered} E_0=\text{span}\left(\left[-1 \\ 1 \\ 0\right]\right) \end{gathered}$$

$\mathrm{λ}=2$, we solved,

$\left(A-2I\right)x=0$

$$\begin{gathered} \left[-1         1         1 \\ 0        -2       1 \\ 0            0        0\right]  \left[ x_1 \\  x_2 \\ {x}_3\right]=\left[0 \\ 0 \\ 0\right]    \end{gathered}$$

$-x_1+x_2+x_3=0,  -2x_2+x_3=0$

$$\begin{gathered} E_2=\text{span}\left(\left[3/2 \\ 1/2 \\ 0\right]\right) \end{gathered}$$

$$\begin{gathered} s=\left[-1           1         3/2 \\ 1     0         1/2 \\ 0         0         1\right] \end{gathered}$$

$S^{-1}AS=B$

localid="1664282087514" $$\begin{gathered} =\left[\begin{array}{ccc}0& 1& -\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\\ 1& 1& -2\\ 0& 0& 1\end{array}\right]×\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 1\\ 0& 0& 2\end{array}\right]×\left[\begin{array}{ccc}-1& 1& 3/2\\ 1& 0& 1/2\\ 0& 0& 1\end{array}\right] \\ B=\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 2& 2\end{array}\right] \end{gathered}$$

Now,

$$\begin{gathered} S^{-1}AS=B \\ A=SB{S}^{-1} \\ {A}^t=S{B}^t{S}^{-1} \end{gathered}$$

So,

$$\begin{gathered} A^t=S{B}^t{S}^{-1} \\ =\left[\begin{array}{ccc}-1& 1& 3/2\\ 1& 0& 1/2\\ 0& 0& 1\end{array}\right]×\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 2& 2\end{array}\right]×\left[\begin{array}{ccc}0& 1& -1/2\\ 1& 1& -2\\ 0& 0& 1\end{array}\right] \\ =\left[\begin{array}{ccc}-1& 1& 3/2\\ 1& 1& 1/2\\ 0& 0& 1\end{array}\right] \\ =\left[\begin{array}{ccc}1& 1& -2+3×2^{t-1}\\ 0& 0& 2^{t-1}\\ 0& 0& 2^t\end{array}\right] \end{gathered}$$

Hence,

$A^t=\left[\begin{array}{ccc}1& 1& -2+3.2^{t-1}\\ 0& 0& 2^{t-1}\\ 0& 0& 2^t\end{array}\right]$