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;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 +
|-
 +
|An eigenvector &nbsp;<math style="vertical-align: 0px">\vec{x}</math>&nbsp; of a matrix &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; corresponding to the eigenvalue &nbsp;<math style="vertical-align: 0px">\lambda</math>&nbsp; is a nonzero vector such that
 
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|-
 
|
 
|
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::<math>A\vec{x}=\lambda\vec{x}.</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: 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;
 
|-
 
|-
−
|
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|Now, we have
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 +
\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}}\\
 +
&&\\
 +
& = & \displaystyle{A\cdot A \cdot 2\vec{x}-A\cdot 2\vec{x}+\vec{x}}\\
 +
&&\\
 +
& = & \displaystyle{2A\cdot A\vec{x}-2A\vec{x}+\vec{x}}\\
 +
&&\\
 +
& = & \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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& = & \displaystyle{5\vec{x}}.
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\end{array}</math>
 +
|-
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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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|}
 
|}
  
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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
 +
|-
 +
|
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&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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&&\\
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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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|-
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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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|-
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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  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 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}   corresponding to the eigenvalue 2, then  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.}   What is the corresponding eigenvalue?

(b) Show that 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 \vec{y}}   is an eigenvector of the matrix    corresponding to the eigenvalue 3 and    is invertible, then    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}.}   What is the corresponding eigenvalue?

Foundations:  
An eigenvector  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}}   of a 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}   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 \lambda}   is a nonzero vector such 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\vec{x}=\lambda\vec{x}.}


Solution:

(a)

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{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}   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 2,}   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{x}\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{x}=2\vec{x}.}
Step 2:  
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{(A^3-A^2+I)\vec{x}} & = & \displaystyle{A^3\vec{x}-A^2\vec{x}+I\vec{x}}\\ &&\\ & = & \displaystyle{A\cdot A\cdot A\vec{x}-A\cdot A\vec{x}+\vec{x}}\\ &&\\ & = & \displaystyle{A\cdot A \cdot 2\vec{x}-A\cdot 2\vec{x}+\vec{x}}\\ &&\\ & = & \displaystyle{2A\cdot A\vec{x}-2A\vec{x}+\vec{x}}\\ &&\\ & = & \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}}\\ &&\\ & = & \displaystyle{(4)\cdot 2\vec{x}-4\vec{x}+\vec{x}}\\ &&\\ & = & \displaystyle{5\vec{x}}. \end{array}}
Hence, 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{x}\ne \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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