91Ó°ÊÓ

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

Given,

$\det \left(A-\mathrm{λ}I\right)=0 $

$$\begin{gathered} \left|0.3-\mathrm{λ}     0.1            0.3 \\ 0.4              0.6-\mathrm{λ}     0.4 \\ 0.3               0.3            0.3\mathrm{λ}\right|=0    \end{gathered}$$

$$\begin{gathered} \frac{1}{{10}^3}\left|3-\text{10}\mathrm{λ}     0.1                  3 \\ 4                    6-1\mathrm{λ}            4 \\ 3                     3            3-10\mathrm{λ} \right|=0 \end{gathered}$$

${\left(3-10\mathrm{λ}\right)}^2\left(6-10\mathrm{λ}\right)+12+36-12\left(3-10\mathrm{λ}\right)-4\left(3-10\mathrm{λ}\right)-9\left(6-10\mathrm{λ}\right)=0$

$$\begin{gathered} -1000{\mathrm{λ}}^3+1200{\mathrm{λ}}^2-200\mathrm{λ}=0 \\ -5{\mathrm{λ}}^3+6{\mathrm{λ}}^2-\mathrm{λ}=0 \end{gathered}$$

$-\mathrm{λ}\left(5\mathrm{λ}-1\right)\left(\mathrm{λ}-1\right)=0$

${\mathrm{λ}}_1=0,{\mathrm{λ}}_2=\frac{1}{5} ,{\mathrm{λ}}_3=1$

We have three distinct real eigenvalues on a matrix$3×3$, so there exists an eigen basis in which the diagonalization of A is

Then,

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

$AX=0$

$$\begin{gathered} \left[0.3               0.1            0.3 \\ 0.4              0.6            0.4 \\ 0.3               0.3            0.3\right]\left[x_1 \\  x_2 \\ {x}_3\right]=\left[0 \\ 0 \\ 0\right]    \end{gathered}$$

$3x_1+x_2+3x_3=0,  2x_1+3x_2+2 x_3=0,   x_1+x_2+x_3=0$

$x_1+x_3=0,x_2=0$

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

Similarly, we solved $\mathrm{λ}=\frac{1}{5}$

$\left(A-\frac{1}{5}I\right)x=0$

$$\begin{gathered} \left[0.1               0.1            0.3 \\ 0.4              0.4            0.4 \\ 0.3               0.3            0.1\right]\left[ x_1 \\  x_2 \\ {x}_3\right]=\left[0 \\ 0 \\ 0\right]  \end{gathered}$$

$$\begin{gathered} x_1+x_2+3x_3=0,  x_1+x_2+x_3=0,   3x_1+3x_2+3x_3=0 \\ {x}_1+x_2=0,x_3=0 \end{gathered}$$

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

Finally,

$\mathrm{λ}=1$solved,

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

$$\begin{gathered} \left[-0.7               0.1            0.3 \\ 0.4              -0.4            0.4 \\ 0.3                  0.3           -0.7\right]\left[ x_1 \\  x_2 \\ {x}_3\right]=\left[0 \\ 0 \\ 0\right]    \end{gathered}$$

$-7x_1+x_2+3x_3=0,  x_1-x_2+x_3=0,   3x_1+3x_2-7x_3=0$

$$\begin{gathered} E_0=span\left(\left[0.3244 \\ 0.8111 \\ 0.4867\right]\right) \end{gathered}$$

We have,

$$\begin{gathered} S=\left[\begin{array}{ccc}0.3244& 1& -1\\ 0.8111& -1& 0\\ 0.4867& 0& 1\end{array}\right] \\ {S}^{-1}=\left[\begin{array}{ccc}0.6& 0.6& 0.6\\ 0.5& -0.5& 0.5\\ -0.3& -0.3& 0.7\end{array}\right] \end{gathered}$$

Finally, we compute,

role="math" localid="1668576625880" $A^t=S{B}^t{S}^{-1}=\left[\begin{array}{ccc}0.3& 1& -1\\ 0.8& -1& 0\\ 0.5& 0& 1\end{array}\right]×\left[\begin{array}{ccc}1& 0& 0\\ 0& {\left(0.2\right)}^t& 0\\ 0& 0& 0\end{array}\right]×\left[\begin{array}{ccc}0.6& 0.6& 0.6\\ 0.5& -0.5& 0.5\\ -0.3& -0.3& 0.7\end{array}\right]$

$=\left[\begin{array}{ccc}0.3& {\left(0.2\right)}^t& 0\\ 0.8& {\left(0.2\right)}^t& 0\\ 0.5& 0& 1\end{array}\right]×\left[\begin{array}{ccc}0.6& 0.6& 0.6\\ 0.5& -0.5& 0.5\\ -0.3& -0.3& 0.7\end{array}\right]$

$=\left[\begin{array}{ccc}0.18+\left(0.5\right){\left(0.2\right)}^t& 0.18-\left(0.5\right){\left(0.2\right)}^t& 0.18+\left(0.5\right){\left(0.2\right)}^t\\ 0.48-\left(0.5\right){\left(0.2\right)}^t& 0.48+\left(0.5\right){\left(0.2\right)}^t& 0.48-\left(0.5\right){\left(0.2\right)}^t\\ 0& 0& 1\end{array}\right]$

Hence,

$A^t=\left[\begin{array}{ccc}0.18+\left(0.5\right){\left(0.2\right)}^t& 0.18-\left(0.5\right){\left(0.2\right)}^t& 0.18+\left(0.5\right){\left(0.2\right)}^t\\ 0.48-\left(0.5\right)\left(0.2\right)& 0.48+\left(0.5\right)\left(0.2\right)& 0.48-\left(0.5\right)\left(0.2\right)\\ 0& 0& 1\end{array}\right]$