91Ó°ÊÓ

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

Given,

$$\begin{gathered} A=\left[0.8           0.6 \\ 0.2           0.4\right] \end{gathered}$$

$\det \left(A-\mathrm{λ}I\right)=0$

$$\begin{gathered} \left|0.8-\mathrm{λ}    0.6 \\ 0.2            0.4-\mathrm{λ}\right|=0 \\ \left(0.8-\mathrm{λ} \right)\left(0.4-\mathrm{λ}\right)-0.12=0 \end{gathered}$$

${\mathrm{λ}}^2-1.2\mathrm{λ}+0.2=0$

$$\begin{gathered} 5{\mathrm{λ}}^2-6\mathrm{λ}+1=0 \\ \left(5\mathrm{λ}-1\right)\left(\mathrm{λ}-1\right)=0 \end{gathered}$$

${\mathrm{λ}}_1=\frac{1}{5},    {\mathrm{λ}}_2=1$

Now, we solve for$\mathrm{λ}=\frac{1}{5}=0.2,$

$\left(A-\frac{1}{5}I \right)x=0  $

$$\begin{gathered} \left[\frac{3}{5}      \frac{3}{5} \\ \frac{1}{5}      \frac{1}{5}\right]\left[ x_1 \\  x_2\right]=\left[0 \\ 0\right] \end{gathered}$$

$$\begin{gathered} E_{\frac{1}{5}}=span\left(\left[1 \\ -1\right]\right) \end{gathered}$$

We solve,

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

$$\begin{gathered} \left[\frac{-1}{5}      \frac{3}{5} \\ \frac{1}{5}      \frac{-3}{5}\right]\left[ x_1 \\  x_2\right]=\left[0 \\ 0\right] \end{gathered}$$

$x_1-3x_2=0$

$$\begin{gathered} E_{\frac{1}{4}}=span\left(\left[3 \\ 1\right]\right) \end{gathered}$$

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

And therefore, the inverse of this matrix is found to be,

$s^{-1}=\frac{1}{4}\left|\begin{array}{cc}1& -3\\ 1& 1\end{array}\right|$

Diagonalization of A Is

$$\begin{gathered} B=\left[0.2      0 \\ 0      1\right] \end{gathered}$$

We compute:

$$\begin{gathered} A^t=S{B}^t{S}^{-1}=\left[\begin{array}{cc}1& 3\\ -1& 1\end{array}\right]×\left[\begin{array}{cc}{\left(0.2\right)}^t& 0\\ 0& 1\end{array}\right]×\frac{1}{4}\left[\begin{array}{cc}1& -3\\ 1& 1\end{array}\right] \\ =\frac{1}{4}\left[\begin{array}{cc}1& 3\\ -1& 1\end{array}\right]×\left[\begin{array}{cc}{\left(0.2\right)}^t& 0\\ 0& 1\end{array}\right]×\left[\begin{array}{cc}1& -3\\ 1& 1\end{array}\right] \\ =\frac{1}{4}\left[\begin{array}{cc}1& 3\\ -1& 1\end{array}\right]×\left[\begin{array}{cc}{\left(0.2\right)}^t& -3×{\left(0.2\right)}^t\\ 1& 1\end{array}\right] \\ =\frac{1}{4}\left[\begin{array}{cc}{\left(0.2\right)}^t+3& -3×{\left(0.2\right)}^t+3\\ {\left(0.2\right)}^t+1& 3×{\left(0.2\right)}^t+1\end{array}\right] \end{gathered}$$

Hence,

$A^t=\frac{1}{4}\left[\begin{array}{cc}{\left(0.2\right)}^t+3& -3×{\left(0.2\right)}^t+3\\ -{\left(0.2\right)}^t+1& 3×{\left(0.2\right)}^t+1\end{array}\right]$