Difference between revisions of "031 Review Part 3, Problem 1"

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Step 2:  
 
!Step 2:  
 +
|-
 +
|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}
 +
          2 & 0 & -2 \\
 +
          1 & 3  & 2 \\
 +
          0 & 0 & 3
 +
        \end{bmatrix}-\begin{bmatrix}
 +
          2 & 0 & 0 \\
 +
          0 & 2  & 0 \\
 +
          0 & 0 & 2
 +
        \end{bmatrix}}\\
 +
&&\\
 +
& = & \displaystyle{\begin{bmatrix}
 +
          0 & 0 & -2 \\
 +
          1 & 1  & 2 \\
 +
          0 & 0 & 1
 +
        \end{bmatrix}}\\
 +
&&\\
 +
& \sim & \displaystyle{\begin{bmatrix}
 +
          1 & 1 & 0 \\
 +
          0 & 0  & 1 \\
 +
          0 & 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: -1px">2</math>&nbsp; is 
 +
|-
 +
|
 +
::<math>\bigg\{\begin{bmatrix}
 +
          -1  \\
 +
          1 \\
 +
          0 
 +
        \end{bmatrix}\bigg\}.</math>
 +
|}
 +
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Step 3: &nbsp;
 +
|-
 +
|For the eigenvalue &nbsp;<math>\lambda=3,</math>&nbsp; we have
 +
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 +
\displaystyle{A-3I} & = & \displaystyle{\begin{bmatrix}
 +
          2 & 0 & -2 \\
 +
          1 & 3  & 2 \\
 +
          0 & 0 & 3
 +
        \end{bmatrix}-\begin{bmatrix}
 +
          3 & 0 & 0 \\
 +
          0 & 3  & 0 \\
 +
          0 & 0 & 3
 +
        \end{bmatrix}}\\
 +
&&\\
 +
& = & \displaystyle{\begin{bmatrix}
 +
          -1 & 0 & -2 \\
 +
          1 & 0  & 2 \\
 +
          0 & 0 & 0
 +
        \end{bmatrix}}\\
 +
&&\\
 +
& \sim & \displaystyle{\begin{bmatrix}
 +
          1 & 0 & 2 \\
 +
          0 & 0  & 0 \\
 +
          0 & 0 & 0
 +
        \end{bmatrix}.}
 +
\end{array}</math>
 +
|-
 +
|We see that &nbsp;<math>x_2</math>&nbsp; and &nbsp;<math>x_3</math>&nbsp; are free variables. So, a basis for the eigenspace corresponding to &nbsp;<math>3</math>&nbsp; is 
 +
|-
 +
|
 +
::<math>\bigg\{\begin{bmatrix}
 +
          0  \\
 +
          1 \\
 +
          0 
 +
        \end{bmatrix},\begin{bmatrix}
 +
          -2  \\
 +
          0 \\
 +
          1 
 +
        \end{bmatrix}\bigg\}.</math>
 +
|}
 +
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Step 4: &nbsp;
 +
|-
 +
|Since &nbsp;<math>A</math>&nbsp; has &nbsp;<math>3</math>&nbsp; linearly independent eigenvectors,
 +
|-
 +
|<math>A</math>&nbsp; is diagonalizable by the Diagonalization Theorem.
 +
|-
 +
|Using the Diagonalization Theorem, we can diagonalize &nbsp;<math>A</math>&nbsp; using the information from the steps above.
 +
|-
 +
|So, we have
 
|-
 
|-
 
|
 
|
 +
::<math>D=\begin{bmatrix}
 +
          2 & 0 & 0 \\
 +
          0 & 3  & 0 \\
 +
          0 & 0 & 3
 +
        \end{bmatrix},P=\begin{bmatrix}
 +
          -1 & 0 & -2 \\
 +
          1 & 1  & 0 \\
 +
          0 & 0 & 1
 +
        \end{bmatrix}.</math>
 
|}
 
|}
  
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|-
 
|-
−
|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp;  
+
|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: 0px">A</math>&nbsp; is diagonalizable and &nbsp;<math>D=\begin{bmatrix}
 +
          2 & 0 & 0 \\
 +
          0 & 3  & 0 \\
 +
          0 & 0 & 3
 +
        \end{bmatrix},P=\begin{bmatrix}
 +
          -1 & 0 & -2 \\
 +
          1 & 1  & 0 \\
 +
          0 & 0 & 1
 +
        \end{bmatrix}.</math>
 
|}
 
|}
 
[[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]
 
[[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 17:05, 13 October 2017

(a) Is the matrix  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= \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix}}   diagonalizable? If so, explain why and diagonalize it. If not, explain why not.

(b) Is the matrix  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= \begin{bmatrix} 2 & 0 & -2 \\ 1 & 3 & 2 \\ 0 & 0 & 3 \end{bmatrix}}   diagonalizable? If so, explain why and diagonalize it. If not, explain why not.


Foundations:  
Recall:
1. The eigenvalues of a triangular matrix are the entries on the diagonal.
2. By the Diagonalization Theorem, an  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\times n}   matrix  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}   is diagonalizable
if and only if  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}   has    linearly independent eigenvectors.


Solution:

(a)

Step 1:  
To answer this question, we examine the eigenvalues and eigenvectors of  
Since    is a triangular matrix, the eigenvalues are the entries on the diagonal.
Hence, the only eigenvalue of    is  
Step 2:  
Now, we find a basis for the eigenspace corresponding to    by solving  
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} 3 & 1 \\ 0 & 3 \end{bmatrix}-\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}}\\ &&\\ & = & \displaystyle{\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}.} \end{array}}
Solving this system, we see  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_1}   is a free variable 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 x_2=0.}
Therefore, a basis for this eigenspace 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} 1 \\ 0 \end{bmatrix}\bigg\}.}
Step 3:  
Now, we know 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 A}   only has one linearly independent eigenvector.
By the Diagonalization Theorem,  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}   must 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 2}   linearly independent eigenvectors to be diagonalizable.
Hence,  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}   is not diagonalizable.

(b)

Step 1:  
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.}
Using cofactor expansion, 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{\text{det }(A-\lambda I)} & = & \displaystyle{\text{det }\Bigg(\begin{bmatrix} 2 & 0 & -2 \\ 1 & 3 & 2 \\ 0 & 0 & 3 \end{bmatrix}-\begin{bmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{bmatrix}\Bigg)}\\ &&\\ & = & \displaystyle{\text{det }\Bigg(\begin{bmatrix} 2-\lambda & 0 & -2 \\ 1 & 3-\lambda & 2 \\ 0 & 0 & 3-\lambda \end{bmatrix}\Bigg)}\\ &&\\ & = & \displaystyle{(-1)^{(2+2)}(3-\lambda)\text{det }\bigg(\begin{bmatrix} 2-\lambda & -2 \\ 0 & 3-\lambda \end{bmatrix}\bigg)}\\ &&\\ & = & \displaystyle{(3-\lambda)(2-\lambda)(3-\lambda).} \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 (3-\lambda)(2-\lambda)(3-\lambda)=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 3}   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 2.}
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} 2 & 0 & -2 \\ 1 & 3 & 2 \\ 0 & 0 & 3 \end{bmatrix}-\begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}}\\ &&\\ & = & \displaystyle{\begin{bmatrix} 0 & 0 & -2 \\ 1 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix}}\\ &&\\ & \sim & \displaystyle{\begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 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} -1 \\ 1 \\ 0 \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} 2 & 0 & -2 \\ 1 & 3 & 2 \\ 0 & 0 & 3 \end{bmatrix}-\begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}}\\ &&\\ & = & \displaystyle{\begin{bmatrix} -1 & 0 & -2 \\ 1 & 0 & 2 \\ 0 & 0 & 0 \end{bmatrix}}\\ &&\\ & \sim & \displaystyle{\begin{bmatrix} 1 & 0 & 2 \\ 0 & 0 & 0 \\ 0 & 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}   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 x_3}   are free variables. 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} 0 \\ 1 \\ 0 \end{bmatrix},\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}\bigg\}.}
Step 4:  
Since  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}   has  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}   linearly independent eigenvectors,
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}   is diagonalizable by the Diagonalization Theorem.
Using the Diagonalization Theorem, we can diagonalize  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}   using the information from the steps above.
So, 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 D=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix},P=\begin{bmatrix} -1 & 0 & -2 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.}


Final Answer:  
   (a)     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}   is not diagonalizable.
   (b)     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}   is diagonalizable 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 D=\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix},P=\begin{bmatrix} -1 & 0 & -2 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.}

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