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

Latest revision as of 14:02, 15 October 2017

(a) Show that if ${\vec {x}}$ is an eigenvector of the matrix $A$ 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 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 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 A} is invertible, 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{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}.} 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

@@ Line 1: / Line 1: @@
-<span class="exam">Consider the matrix &nbsp;<math style="vertical-align: -31px">A=
+<span class="exam"> (a) Show that if &nbsp;<math style="vertical-align: 0px">\vec{x}</math>&nbsp; is an eigenvector of the matrix &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; corresponding to the eigenvalue 2, then &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; What is the corresponding eigenvalue?
-    \begin{bmatrix}
-& -4 & 9 & -7 \\
-           -1 & 2  & -4 & 1 \\
-& -6 & 10 & 7
-         \end{bmatrix}</math>&nbsp; and assume that it is row equivalent to the matrix
-::<math>B=
-    \begin{bmatrix}
-& 0 & -1 & 5 \\
-& -2  & 5 & -6 \\
-& 0 & 0 & 0
-         \end{bmatrix}.</math>
-<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>
+<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?
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !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
 |-
 |
+::<math>A\vec{x}=\lambda\vec{x}.</math>
 |}
@@ Line 31: / Line 19: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !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>
 |}
@@ Line 38: / Line 29: @@
 !Step 2: &nbsp;
 |-
-|
+|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}}\\
+&&\\
+& = & \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}</math>
+|-
+|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>
 |}
@@ Line 45: / Line 57: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !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;"
 !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}
+\displaystyle{A^{-1}(A\vec{y})} & = & \displaystyle{A^{-1}(3\vec{y})}\\
+&&\\
+& = & \displaystyle{3(A^{-1}\vec{y}).}
+\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})}\\
+&&\\
+& = & \displaystyle{(A^{-1}A)\vec{y}}\\
+&&\\
+& = & \displaystyle{I\vec{y}}\\
+&&\\
+& = & \displaystyle{\vec{y}.}
+\end{array}</math>
+|-
+|Hence, &nbsp;<math style="vertical-align: -13px">A^{-1}\vec{y}=\frac{1}{3}\vec{y}.</math>
+|-
+|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>
 |}
@@ Line 59: / Line 100: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp;
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; See solution above.
 |-
-|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp;
+|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; See solution above.
 |}
-[[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]
+[[031_Review_Part_3|'''<u>Return to Review Problems</u>''']]

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

Latest revision as of 14:02, 15 October 2017

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