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

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           2 & -6  
 
           2 & -6  
 
         \end{bmatrix}^k</math>&nbsp; by diagonalizing the matrix.
 
         \end{bmatrix}^k</math>&nbsp; by diagonalizing the matrix.
−
 
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!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>
 
|}
 
|}
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 +
|-
 +
|To diagonalize this matrix, we need to find the eigenvalues and a basis for each eigenspace.
 +
|-
 +
|First, we find the eigenvalues of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; by solving &nbsp;<math style="vertical-align: -5px">\text{det }(A-\lambda I)=0.</math>
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\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}</math>
 +
|-
 +
|Therefore, setting
 
|-
 
|-
 
|
 
|
 +
::<math>(\lambda+2)(\lambda+3)=0,</math>&nbsp;
 +
|-
 +
|we find that the eigenvalues of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; are &nbsp;<math style="vertical-align: 0px">-2</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">-3.</math>
 
|}
 
|}
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 +
|-
 +
|Now, we find a basis for each eigenspace by solving &nbsp;<math style="vertical-align: -6px">(A-\lambda I)\vec{x}=\vec{0}</math>&nbsp; for each eigenvalue &nbsp;<math style="vertical-align: 0px">\lambda.</math>
 +
|-
 +
|For the eigenvalue &nbsp;<math style="vertical-align: -4px">\lambda=-2,</math>&nbsp; we have
 +
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp;<math>\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}</math>
 +
|-
 +
|We see that &nbsp;<math style="vertical-align: -3px">x_2</math>&nbsp; is a free variable. So, a basis for the eigenspace corresponding to &nbsp;<math style="vertical-align: 0px">-2</math>&nbsp; is 
 +
|-
 +
|
 +
::<math>\bigg\{\begin{bmatrix}
 +
          2  \\
 +
          1 \\ 
 +
        \end{bmatrix}\bigg\}.</math>
 +
 +
|}
 +
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Step 3: &nbsp;
 +
|-
 +
|For the eigenvalue &nbsp;<math style="vertical-align: -4px">\lambda=-3,</math>&nbsp; we have
 +
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp;<math>\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}</math>
 +
|-
 +
|We see that &nbsp;<math style="vertical-align: -3px">x_2</math>&nbsp; is a free variable. So, a basis for the eigenspace corresponding to &nbsp;<math style="vertical-align: 0px">-3</math>&nbsp; is 
 +
|-
 +
|
 +
::<math>\bigg\{\begin{bmatrix}
 +
          \frac{3}{2}  \\
 +
          1 \\
 +
        \end{bmatrix}\bigg\}.</math>
 +
|}
 +
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Step 4: &nbsp;
 +
|-
 +
|To diagonalize our matrix, we use the information from the steps above.
 +
|-
 +
|
 +
Using the Diagonalization Theorem, we have &nbsp;<math style="vertical-align: -1px">A=PDP^{-1}</math>&nbsp; where
 +
|-
 +
|
 +
::<math>D=\begin{bmatrix}
 +
          -2 & 0 \\
 +
          0 & -3 
 +
        \end{bmatrix},P=\begin{bmatrix}
 +
          2 & \frac{3}{2} \\
 +
          1 & 1 
 +
        \end{bmatrix}.</math>
 +
|}
 +
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Step 5: &nbsp;
 +
|-
 +
|Notice that
 
|-
 
|-
 
|
 
|
 +
&nbsp; &nbsp; &nbsp; &nbsp;<math>\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}</math>
 
|}
 
|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\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}</math>
 
|&nbsp;&nbsp; &nbsp; &nbsp;   
 
|&nbsp;&nbsp; &nbsp; &nbsp;   
 
|}
 
|}
−
[[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]
+
[[031_Review_Part_3|'''<u>Return to Review Problems</u>''']]

Latest revision as of 12:59, 15 October 2017

Find a formula for  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} 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    matrix    is diagonalizable if and only if    has    linearly independent eigenvectors.
In fact,  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 A=PDP^{-1},}   with  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}   a diagonal matrix, if and only if the columns of  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 P}   are  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 n}   linearly
independent eigenvectors of  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 A.}   In this case, the diagonal entries of  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}   are eigenvalues of  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 A}   that
correspond, respectively , to the eigenvectors in  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 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  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 A}   by solving  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 \text{det }(A-\lambda I)=0.}
       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{\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
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 (\lambda+2)(\lambda+3)=0,}  
we find that the eigenvalues of  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 A}   are  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 -2}   and  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 -3.}
Step 2:  
Now, we find a basis for each eigenspace by solving  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 (A-\lambda I)\vec{x}=\vec{0}}   for each eigenvalue  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 \lambda.}
For the eigenvalue  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 \lambda=-2,}   we have

       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+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  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 x_2}   is a free variable. So, a basis for the eigenspace corresponding to  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 -2}   is
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 \bigg\{\begin{bmatrix} 2 \\ 1 \\ \end{bmatrix}\bigg\}.}
Step 3:  
For the eigenvalue  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 \lambda=-3,}   we have

       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+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  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 x_2}   is a free variable. So, a basis for the eigenspace corresponding to  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 -3}   is
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 \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  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 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}}       

Return to Review Problems

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