031 Review Part 3, Problem 5

Find a formula for  ${\begin{bmatrix}1&-6\\2&-6\end{bmatrix}}^{k}$  by diagonalizing the matrix.

Foundations:
Recall:
1. To diagonalize a matrix, you need to know the eigenvalues of the matrix.
2. Diagonalization Theorem
An  $n\times n$  matrix  $A$  is diagonalizable if and only if  $A$  has  $n$  linearly independent eigenvectors.
In fact,  $A=PDP^{-1},$  with  $D$  a diagonal matrix, if and only if the columns of  $P$  are  $n$  linearly
independent eigenvectors of  $A.$  In this case, the diagonal entries of  $D$  are eigenvalues of  $A$  that
correspond, respectively , to the eigenvectors in  $P.$

Solution:

Step 1:
To diagonalize this matrix, we need to find the eigenvalues and a basis for each eigenspace.
First, we find the eigenvalues of  $A$  by solving  ${\text{det }}(A-\lambda I)=0.$
${\begin{array}{rcl}\displaystyle {{\text{det }}(A-\lambda I)}&=&\displaystyle {{\text{det }}{\bigg (}{\begin{bmatrix}1&-6\\2&-6\end{bmatrix}}-{\begin{bmatrix}\lambda &0\\0&\lambda \end{bmatrix}}{\bigg )}}\\&&\\&=&\displaystyle {{\text{det }}{\bigg (}{\begin{bmatrix}1-\lambda &-6\\2&-6-\lambda \end{bmatrix}}{\bigg )}}\\&&\\&=&\displaystyle {(1-\lambda )(-6\lambda )+12}\\&&\\&=&\displaystyle {\lambda ^{2}+5\lambda +6}\\&&\\&=&\displaystyle {(\lambda +2)(\lambda +3).}\\\end{array}}$
Therefore, setting
$(\lambda +2)(\lambda +3)=0,$
we find that the eigenvalues of  $A$  are  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -2}   and  $3.$

Step 2:
Now, we find a basis for each eigenspace by solving  $(A-\lambda I){\vec {x}}={\vec {0}}$  for each eigenvalue  $\lambda .$
For the eigenvalue  $\lambda =-2,$  we have
${\begin{array}{rcl}\displaystyle {A+2I}&=&\displaystyle {{\begin{bmatrix}1&-6\\2&-6\end{bmatrix}}+{\begin{bmatrix}2&0\\0&2\end{bmatrix}}}\\&&\\&=&\displaystyle {\begin{bmatrix}3&-6\\2&-4\end{bmatrix}}\\&&\\&\sim &\displaystyle {{\begin{bmatrix}1&-2\\0&0\end{bmatrix}}.}\end{array}}$
We see that  $x_{2}$  is a free variable. So, a basis for the eigenspace corresponding to  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -2}   is
${\bigg \{}{\begin{bmatrix}2\\1\\\end{bmatrix}}{\bigg \}}.$

Step 3:
For the eigenvalue  $\lambda =-3,$  we have
${\begin{array}{rcl}\displaystyle {A+3I}&=&\displaystyle {{\begin{bmatrix}1&-6\\2&-6\end{bmatrix}}-{\begin{bmatrix}3&0\\0&3\end{bmatrix}}}\\&&\\&=&\displaystyle {\begin{bmatrix}4&-6\\2&-3\end{bmatrix}}\\&&\\&\sim &\displaystyle {{\begin{bmatrix}1&{\frac {-3}{2}}\\0&0\end{bmatrix}}.}\end{array}}$
We see that  $x_{2}$  is a free variable. So, a basis for the eigenspace corresponding to  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -3}   is
${\bigg \{}{\begin{bmatrix}{\frac {3}{2}}\\1\\\end{bmatrix}}{\bigg \}}.$

Step 4:
To diagonalize our matrix, we use the information from the steps above.
Using the Diagonalization Theorem, we have  $A=PDP^{-1}$  where
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D=\begin{bmatrix} -2 & 0 \\ 0 & -3 \end{bmatrix},P=\begin{bmatrix} 2 & \frac{3}{2} \\ 1 & 1 \end{bmatrix}.}

Step 5:
Notice that
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{A^k} & = & \displaystyle{(PDP^{-1})^k}\\ &&\\ & = & \displaystyle{PD^kP^{-1}}\\ &&\\ & = & \displaystyle{\begin{bmatrix} 2 & \frac{3}{2} \\ 1 & 1 \end{bmatrix}\bigg(\begin{bmatrix} -2 & 0 \\ 0 & -3 \end{bmatrix}\bigg)^k (-2)\begin{bmatrix} 1 & -\frac{3}{2} \\ -1 & 2 \end{bmatrix}}\\ &&\\ & = & \displaystyle{\begin{bmatrix} 2 & \frac{3}{2} \\ 1 & 1 \end{bmatrix}\begin{bmatrix} (-2)^k & 0 \\ 0 & (-3)^k \end{bmatrix} \begin{bmatrix} -2 & 3 \\ 2 & -4 \end{bmatrix}}\\ &&\\ & = & \displaystyle{\begin{bmatrix} 2 & \frac{3}{2} \\ 1 & 1 \end{bmatrix}\begin{bmatrix} (-2)^{k+1} & 3(-2)^k \\ 2(-3)^k & (-4)(-3)^k \end{bmatrix}} \\ &&\\ & = & \displaystyle{\begin{bmatrix} 2(-2)^{k+1}+3(-3)^k & 6(-2)^k+-6(-3)^k \\ (-2)^{k+1}+2(-3)^k & 3(-2)^k+(-4)(-3)^k \end{bmatrix}.} \\ \end{array}}

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{bmatrix} 2(-2)^{k+1}+3(-3)^k & 6(-2)^k+-6(-3)^k \\ (-2)^{k+1}+2(-3)^k & 3(-2)^k+(-4)(-3)^k \end{bmatrix}}

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