91Ó°ÊÓ

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

Given,

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

This matrix is lower triangular.

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

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

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

so there exists an eigen basis in which the diagonalization of A is

$B=\left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& -1\end{array}\right]$

Now, we solve for $\mathrm{λ}=0$

$Ax=0$

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

$x_1-x_2=0, x_2+x_3=0$

$\mathrm{λ}=-1$solved,

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

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

$$\begin{gathered} E_{-1}=span\left(\left[0 \\ -2 \\ 1\right]\right) \end{gathered}$$

Finally,

$\mathrm{λ}=1$we solved,

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

$$\begin{gathered} \left[-1       0      0 \\ 1    -2        0 \\ 0        1       0\right]  \left[x_1 \\  x_2 \\ {x}_3\right]=\left[0 \\ 0 \\ 0\right] \\ {x}_1=0,x_1-2x_2=0,x_2=0 \end{gathered}$$

$x_1=x_2=0$

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

$S=\left[\begin{array}{ccc}0& -1& 0\\ 0& -1& -2\\ 1& 1& 1\end{array}\right]$

And therefore the inverse of the matrix S is shown below,

$S^{-1}=\left[\begin{array}{ccc}1/2& 1/2& 1\\ -1& 0& 0\\ 1/2& -1/2& 0\end{array}\right]$

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

$$\begin{gathered} =\left[\begin{array}{ccc}0& -1& 0\\ 0& -1& -2\\ 1& 1& 1\end{array}\right]×\left[\begin{array}{ccc}{1}^t& 0& 0\\ 0& 0& 0\\ 0& 0& {\left(-1\right)}^t\end{array}\right]×\frac{1}{2}\left[\begin{array}{ccc}1& 1& 2\\ -2& 0& 0\\ 1& -1& 0\end{array}\right] \\ =\frac{1}{2}×\left[\begin{array}{ccc}0& -1& 0\\ 0& -1& -2\\ 1& 1& 1\end{array}\right]×\left[\begin{array}{ccc}{1}^t& 0& 0\\ 0& 0& 0\\ 0& 0& {\left(-1\right)}^t\end{array}\right]×\left[\begin{array}{ccc}1& 1& 2\\ -2& 0& 0\\ 1& -1& 0\end{array}\right] \\ =\frac{1}{2}\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& -2{\left(-1\right)}^t\\ 1& 0& {\left(-1\right)}^t\end{array}\right]×\left[\begin{array}{ccc}1& 1& 2\\ -2& 0& 0\\ 1& -1& 0\end{array}\right] \end{gathered}$$

role="math" localid="1668581281656" $=\frac{1}{2}\left[\begin{array}{ccc}0& 0& 0\\ 2\left(-1^{t+1}\right)& 2{\left(-1\right)}^t& 0\\ {\left(-1\right)}^t+1& {\left(-1\right)}^{t+1}+1& 2\end{array}\right]$

Hence,

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