Difference between revisions of "031 Review Part 3, Problem 5"

Revision as of 13:19, 13 October 2017

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:

Step 2:

Final Answer:

@@ Line 7: / Line 7: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
+|-
+|Recall:
+|-
+|1. To diagonalize a matrix, you need to know the eigenvalues of the matrix.
+|-
+|2. '''Diagonalization Theorem'''
 |-
 |
+:An &nbsp;<math style="vertical-align: 0px">n\times n</math>&nbsp; matrix &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is diagonalizable if and only if &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; has &nbsp;<math style="vertical-align: 0px">n</math>&nbsp; linearly independent eigenvectors.
+|-
+|
+:In fact, &nbsp;<math style="vertical-align: -4px">A=PDP^{-1},</math>&nbsp; with &nbsp;<math style="vertical-align: 0px">D</math>&nbsp; a diagonal matrix, if and only if the columns of &nbsp;<math style="vertical-align: 0px">P</math>&nbsp; are &nbsp;<math style="vertical-align: 0px">n</math>&nbsp; linearly
+|-
+|
+:independent eigenvectors of &nbsp;<math style="vertical-align: 0px">A.</math>&nbsp; In this case, the diagonal entries of &nbsp;<math style="vertical-align: 0px">D</math>&nbsp; are eigenvalues of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; that
+|-
+|
+:correspond, respectively , to the eigenvectors in &nbsp;<math style="vertical-align: 0px">P.</math>
 |}