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  $-2$  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  $-2$  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  $-3$  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
$D={\begin{bmatrix}-2&0\\0&-3\end{bmatrix}},P={\begin{bmatrix}2&{\frac {3}{2}}\\1&1\end{bmatrix}}.$

Step 5:
Notice that
${\begin{array}{rcl}\displaystyle {A^{k}}&=&\displaystyle {(PDP^{-1})^{k}}\\&&\\&=&\displaystyle {PD^{k}P^{-1}}\\&&\\&=&\displaystyle {{\begin{bmatrix}2&{\frac {3}{2}}\\1&1\end{bmatrix}}{\bigg (}{\begin{bmatrix}-2&0\\0&-3\end{bmatrix}}{\bigg )}^{k}(-2){\begin{bmatrix}1&-{\frac {3}{2}}\\-1&2\end{bmatrix}}}\\&&\\&=&\displaystyle {{\begin{bmatrix}2&{\frac {3}{2}}\\1&1\end{bmatrix}}{\begin{bmatrix}(-2)^{k}&0\\0&(-3)^{k}\end{bmatrix}}{\begin{bmatrix}-2&3\\2&-4\end{bmatrix}}}\\&&\\&=&\displaystyle {{\begin{bmatrix}2&{\frac {3}{2}}\\1&1\end{bmatrix}}{\begin{bmatrix}(-2)^{k+1}&3(-2)^{k}\\2(-3)^{k}&(-4)(-3)^{k}\end{bmatrix}}}\\&&\\&=&\displaystyle {{\begin{bmatrix}2(-2)^{k+1}+3(-3)^{k}&6(-2)^{k}+-6(-3)^{k}\\(-2)^{k+1}+2(-3)^{k}&3(-2)^{k}+(-4)(-3)^{k}\end{bmatrix}}.}\\\end{array}}$

Final Answer:
${\begin{bmatrix}2(-2)^{k+1}+3(-3)^{k}&6(-2)^{k}+-6(-3)^{k}\\(-2)^{k+1}+2(-3)^{k}&3(-2)^{k}+(-4)(-3)^{k}\end{bmatrix}}$

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