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

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(Created page with "<span class="exam">Consider the matrix  <math style="vertical-align: -31px">A= \begin{bmatrix} 1 & -4 & 9 & -7 \\ -1 & 2 & -4 & 1 \\...")
 
 
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<span class="exam">Consider the matrix &nbsp;<math style="vertical-align: -31px">A=     
+
<span class="exam">(a) Is the matrix &nbsp;<math style="vertical-align: -18px">A=     
 
     \begin{bmatrix}
 
     \begin{bmatrix}
−
           1 & -4 & 9 & -7 \\
+
           3 & 1 \\
−
          -1 & 2  & -4 & 1 \\
+
           0 & 3
−
           5 & -6 & 10 & 7
+
         \end{bmatrix}</math>&nbsp; diagonalizable? If so, explain why and diagonalize it. If not, explain why not.
−
         \end{bmatrix}</math>&nbsp; and assume that it is row equivalent to the matrix  
+
       
−
 
+
<span class="exam">(b) Is the matrix &nbsp;<math style="vertical-align: -31px">A=     
−
::<math>B=     
 
 
     \begin{bmatrix}
 
     \begin{bmatrix}
−
           1 & 0 & -1 & 5 \\
+
           2 & 0 & -2 \\
−
           0 & -2 & 5 & -6 \\
+
           1 & 3 & 2 \\
−
           0 & 0 & 0 & 0
+
           0 & 0 & 3
−
         \end{bmatrix}.</math>     
+
         \end{bmatrix}</math>&nbsp; diagonalizable? If so, explain why and diagonalize it. If not, explain why not.
−
   
 
−
<span class="exam">(a) List rank &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">\text{dim Nul }A.</math>
 
−
 
 
−
<span class="exam">(b) Find bases for &nbsp;<math style="vertical-align: 0px">\text{Col }A</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">\text{Nul }A.</math>&nbsp; Find an example of a nonzero vector that belongs to &nbsp;<math style="vertical-align: -5px">\text{Col }A,</math>&nbsp; as well as an example of a nonzero vector that belongs to &nbsp;<math style="vertical-align: 0px">\text{Nul }A.</math>
 
−
 
 
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 +
|-
 +
|Recall:
 +
|-
 +
|'''1.''' The eigenvalues of a triangular matrix are the entries on the diagonal.
 +
|-
 +
|'''2.''' By the 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.
 
|}
 
|}
  
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|
+
|To answer this question, we examine the eigenvalues and eigenvectors of &nbsp;<math style="vertical-align: 0px">A.</math>
 +
|-
 +
|Since &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is a triangular matrix, the eigenvalues are the entries on the diagonal.
 +
|-
 +
|Hence, the only eigenvalue of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is &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 the eigenspace corresponding to &nbsp;<math style="vertical-align: 0px">3</math>&nbsp; by solving &nbsp;<math style="vertical-align: -5px">(A-3I)\vec{x}=\vec{0}.</math>
 +
|-
 +
|We have
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\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}</math>
 +
|-
 +
|Solving this system, we see &nbsp;<math style="vertical-align: -4px">x_1</math>&nbsp; is a free variable and &nbsp;<math style="vertical-align: -4px">x_2=0.</math>
 +
|-
 +
|Therefore, a basis for this eigenspace is
 
|-
 
|-
 
|
 
|
 +
::<math>\bigg\{\begin{bmatrix}
 +
          1  \\
 +
          0 
 +
        \end{bmatrix}\bigg\}.</math>
 +
 +
|}
 +
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Step 3: &nbsp;
 +
|-
 +
|Now, we know that &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; only has one linearly independent eigenvector.
 +
|-
 +
|By the Diagonalization Theorem, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; must have &nbsp;<math style="vertical-align: 0px">2</math>&nbsp; linearly independent eigenvectors to be diagonalizable.
 +
|-
 +
|Hence, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is not diagonalizable.
 +
 
|}
 
|}
  
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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;  
 +
|-
 +
|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>
 +
|-
 +
|Using cofactor expansion, we have
 
|-
 
|-
 
|
 
|
 +
&nbsp; &nbsp; &nbsp; &nbsp;<math>\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}</math>
 +
|-
 +
|Therefore, setting
 +
|-
 +
|
 +
::<math>(3-\lambda)(2-\lambda)(3-\lambda)=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">3</math>&nbsp; and &nbsp;<math style="vertical-align: 0px">2.</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}
 +
          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 style="vertical-align: -4px">\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 style="vertical-align: -3px">x_2</math>&nbsp; and &nbsp;<math style="vertical-align: -3px">x_3</math>&nbsp; are free variables. So, a basis for the eigenspace corresponding to &nbsp;<math style="vertical-align: -1px">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 style="vertical-align: 0px">A</math>&nbsp; has &nbsp;<math style="vertical-align: 0px">3</math>&nbsp; linearly independent eigenvectors, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is diagonalizable by the Diagonalization Theorem.
 +
|-
 +
|Using the Diagonalization Theorem, we can diagonalize &nbsp;<math style="vertical-align: 0px">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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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
−
|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp;  
+
|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; <math style="vertical-align: 0px">A</math>&nbsp; is not diagonalizable.
 +
 
 
|-
 
|-
−
|&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 Review Problems</u>''']]

Latest revision as of 13:51, 15 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  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.


Solution:

(a)

Step 1:  
To answer this question, we examine the eigenvalues and 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.}
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}   is a triangular matrix, the eigenvalues are the entries on the diagonal.
Hence, the only eigenvalue 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}   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 3.}
Step 2:  
Now, we find 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}   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-3I)\vec{x}=\vec{0}.}
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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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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