Difference between revisions of "031 Review Part 3, Problem 6"

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<span class="exam">(b) Show that if &nbsp;<math style="vertical-align: -3px">\vec{y}</math>&nbsp; is an eigenvector of the matrix &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; corresponding to the eigenvalue 3 and &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible, then &nbsp;<math style="vertical-align: -3px">\vec{y}</math>&nbsp; is an eigenvector of &nbsp;<math style="vertical-align: 0px">A^{-1}.</math>&nbsp; What is the corresponding eigenvalue?
 
<span class="exam">(b) Show that if &nbsp;<math style="vertical-align: -3px">\vec{y}</math>&nbsp; is an eigenvector of the matrix &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; corresponding to the eigenvalue 3 and &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible, then &nbsp;<math style="vertical-align: -3px">\vec{y}</math>&nbsp; is an eigenvector of &nbsp;<math style="vertical-align: 0px">A^{-1}.</math>&nbsp; What is the corresponding eigenvalue?
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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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;  
 +
|-
 +
|Since &nbsp;<math style="vertical-align: 0px">\vec{x}</math>&nbsp; is an eigenvector of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; corresponding to the eigenvalue &nbsp;<math style="vertical-align: -4px">2,</math>&nbsp; we know &nbsp;<math style="vertical-align: -5px">\vec{x}\neq \vec{0}</math>&nbsp; and
 
|-
 
|-
 
|
 
|
 +
::<math>A\vec{x}=2\vec{x}.</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|
+
|Now, we have
 +
|-
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
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\displaystyle{(A^3-A^2+I)\vec{x}} & = & \displaystyle{A^3\vec{x}-A^2\vec{x}+I\vec{x}}\\
 +
&&\\
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& = & \displaystyle{A\cdot A\cdot A\vec{x}-A\cdot A\vec{x}+\vec{x}}\\
 +
&&\\
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& = & \displaystyle{A\cdot A \cdot 2\vec{x}-A\cdot 2\vec{x}+\vec{x}}\\
 +
&&\\
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& = & \displaystyle{2A\cdot A\vec{x}-2A\vec{x}+\vec{x}}\\
 +
&&\\
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& = & \displaystyle{2A\cdot 2\vec{x}-2\cdot 2\vec{x}+\vec{x}}\\
 +
&&\\
 +
& = & \displaystyle{(2\cdot 2)A\vec{x}-4\vec{x}+\vec{x}}\\
 +
&&\\
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& = & \displaystyle{(4)\cdot 2\vec{x}-4\vec{x}+\vec{x}}\\
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&&\\
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& = & \displaystyle{5\vec{x}}.
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\end{array}</math>
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|-
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|Hence, since &nbsp;<math style="vertical-align: -5px">\vec{x}\ne \vec{0},</math>&nbsp; we conclude that &nbsp;<math style="vertical-align: 0px">\vec{x}</math>&nbsp; is an eigenvector of &nbsp;<math style="vertical-align: -2px">A^3-A^2+I</math>&nbsp; corresponding to the eigenvalue &nbsp;<math style="vertical-align: 0px">5.</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;  
 +
|-
 +
|Since &nbsp;<math style="vertical-align: -3px">\vec{y}</math>&nbsp; is an eigenvector of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; corresponding to the eigenvalue &nbsp;<math style="vertical-align: -4px">3,</math>&nbsp; we know &nbsp;<math style="vertical-align: -5px">\vec{y}\neq \vec{0}</math>&nbsp; and
 
|-
 
|-
 
|
 
|
 +
::<math>A\vec{y}=3\vec{y}.</math>
 +
|-
 +
|Also, since &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible, &nbsp;<math style="vertical-align: 0px">A^{-1}</math>&nbsp; exists.
 
|}
 
|}
  
 
{| 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 multiply the equation from Step 1 on the left by &nbsp;<math style="vertical-align: 0px">A^{-1}</math>&nbsp; to obtain
 
|-
 
|-
 
|
 
|
 +
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
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\displaystyle{A^{-1}(A\vec{y})} & = & \displaystyle{A^{-1}(3\vec{y})}\\
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&&\\
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& = & \displaystyle{3(A^{-1}\vec{y}).}
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\end{array}</math>
 +
|-
 +
|Now, we have
 +
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 +
\displaystyle{3(A^{-1}\vec{y})} & = & \displaystyle{A^{-1}(A\vec{y})}\\
 +
&&\\
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& = & \displaystyle{(A^{-1}A)\vec{y}}\\
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&&\\
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& = & \displaystyle{I\vec{y}}\\
 +
&&\\
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& = & \displaystyle{\vec{y}.}
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\end{array}</math>
 +
|-
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|Hence, &nbsp;<math style="vertical-align: -13px">A^{-1}\vec{y}=\frac{1}{3}\vec{y}.</math>
 +
|-
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|Therefore, &nbsp;<math style="vertical-align: -3px">\vec{y}</math>&nbsp; is an eigenvector of &nbsp;<math style="vertical-align: 0px">A^{-1}</math>&nbsp; corresponding to the eigenvalue &nbsp;<math style="vertical-align: -12px">\frac{1}{3}.</math>
 
|}
 
|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
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|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp;  
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|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See solution above.
 
|-
 
|-
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|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp;  
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|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; See solution above.
 
|}
 
|}
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[[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]
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[[031_Review_Part_3|'''<u>Return to Review Problems</u>''']]

Latest revision as of 13:02, 15 October 2017

(a) Show that if    is an eigenvector of the matrix    corresponding to the eigenvalue 2, then    is an eigenvector of    What is the corresponding eigenvalue?

(b) Show that if    is an eigenvector of the matrix    corresponding to the eigenvalue 3 and    is invertible, then    is an eigenvector of    What is the corresponding eigenvalue?

Foundations:  
An eigenvector    of a matrix    corresponding to the eigenvalue    is a nonzero vector such that


Solution:

(a)

Step 1:  
Since    is an eigenvector of    corresponding to the eigenvalue    we know  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\vec {x}}\neq {\vec {0}}}   and
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle A{\vec {x}}=2{\vec {x}}.}
Step 2:  
Now, we have
       
Hence, since  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\vec {x}}\neq {\vec {0}},}   we conclude 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 \vec{x}}   is an eigenvector 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^3-A^2+I}   corresponding to 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 5.}

(b)

Step 1:  
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 \vec{y}}   is an eigenvector 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}   corresponding to 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 3,}   we know  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 \vec{y}\neq \vec{0}}   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 A\vec{y}=3\vec{y}.}
Also, 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 invertible,  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^{-1}}   exists.
Step 2:  
Now, we multiply the equation from Step 1 on the left by  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^{-1}}   to obtain

       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^{-1}(A\vec{y})} & = & \displaystyle{A^{-1}(3\vec{y})}\\ &&\\ & = & \displaystyle{3(A^{-1}\vec{y}).} \end{array}}

Now, 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{3(A^{-1}\vec{y})} & = & \displaystyle{A^{-1}(A\vec{y})}\\ &&\\ & = & \displaystyle{(A^{-1}A)\vec{y}}\\ &&\\ & = & \displaystyle{I\vec{y}}\\ &&\\ & = & \displaystyle{\vec{y}.} \end{array}}

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^{-1}\vec{y}=\frac{1}{3}\vec{y}.}
Therefore,  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 \vec{y}}   is an eigenvector 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^{-1}}   corresponding to 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 \frac{1}{3}.}


Final Answer:  
   (a)     See solution above.
   (b)     See solution above.

Return to Review Problems

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