031 Review Part 3, Problem 3 - Revision history

Kayla Murray at 20:54, 15 October 2017

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Kayla Murray at 23:14, 13 October 2017

2017-10-13T23:14:01Z

Kayla Murray at 23:06, 13 October 2017

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Kayla Murray at 22:56, 13 October 2017

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Kayla Murray at 20:09, 13 October 2017

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Kayla Murray at 20:08, 13 October 2017

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Kayla Murray at 02:23, 10 October 2017

2017-10-10T02:23:09Z

Kayla Murray: Created page with "Consider the matrix $A= \begin{bmatrix} 1 & -4 & 9 & -7 \\ -1 & 2 & -4 & 1 \\..."$

2017-10-10T02:03:10Z

Created page with "Consider the matrix <math style="vertical-align: -31px">A= \begin{bmatrix} 1 & -4 & 9 & -7 \\ -1 & 2 & -4 & 1 \\..."

New page

Consider the matrix  <math style="vertical-align: -31px">A=
\begin{bmatrix}
1 & -4 & 9 & -7 \\
-1 & 2 & -4 & 1 \\
5 & -6 & 10 & 7
\end{bmatrix}</math>  and assume that it is row equivalent to the matrix

::<math>B=
\begin{bmatrix}
1 & 0 & -1 & 5 \\
0 & -2 & 5 & -6 \\
0 & 0 & 0 & 0
\end{bmatrix}.</math>

(a) List rank  <math style="vertical-align: 0px">A</math>  and  <math style="vertical-align: 0px">\text{dim Nul }A.</math>

(b) Find bases for  <math style="vertical-align: 0px">\text{Col }A</math>  and  <math style="vertical-align: 0px">\text{Nul }A.</math>  Find an example of a nonzero vector that belongs to  <math style="vertical-align: -5px">\text{Col }A,</math>  as well as an example of a nonzero vector that belongs to  <math style="vertical-align: 0px">\text{Nul }A.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
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|}

'''Solution:'''

'''(a)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|   '''(a)'''    
|-
|   '''(b)'''    
|}
[[031_Review_Part_3|'''Return to Sample Exam''']]

@@ Line 8: / Line 8: @@
 <span class="exam">(b) Is the matrix &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; diagonalizable? Explain.
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 97: / Line 96: @@
 |&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: 0px">A</math>&nbsp; is not diagonalizable.
 |}
-[[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]
+[[031_Review_Part_3|'''<u>Return to Review Problems</u>''']]

@@ Line 41: / Line 41: @@
 |Now, to find a basis for the eigenspace corresponding to &nbsp;<math style="vertical-align: -4px">5,</math>&nbsp; we need to solve &nbsp;<math style="vertical-align: -5px">(A-5I)\vec{x}=\vec{0}.</math>
 |-
-|To do this, we use row reduction. Thus, we get
+|We have
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
@@ Line 52: / Line 52: @@
           \end{bmatrix}}\\
 &&\\
-& \sim & \displaystyle{\begin{bmatrix}
+& = & \displaystyle{\begin{bmatrix}
 & 1 \\
 & 0

@@ Line 19: / Line 19: @@
 |'''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.
 |}

@@ Line 74: / Line 74: @@
 !Step 1: &nbsp;
 |-
+|From part (a), we know that &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; only has one linearly independent eigenvector.
 |}
@@ Line 80: / Line 80: @@
 !Step 2: &nbsp;
 |-
+|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.
 |}
@@ Line 93: / Line 95: @@
 |-
-|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp;
+|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math style="vertical-align: 0px">A</math>&nbsp; is not diagonalizable.
 |}
 [[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 31: / Line 31: @@
 !Step 1: &nbsp;
 |-
+|Since &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is a triangular matrix, the eigenvalues are the entries on the diagonal.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
 |}
@@ Line 58: / Line 87: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp;
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; The only eigenvalue of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is &nbsp;<math style="vertical-align: 0px">5</math>&nbsp; and the corresponding eigenspace has basis &nbsp;<math style="vertical-align: -20px">\bigg\{\begin{bmatrix}
 |-
 |&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp;
 |}
 [[031_Review_Part_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 13: / Line 13: @@
 !Foundations: &nbsp;
 |-
+|Recall:
 |}

@@ Line 1: / Line 1: @@
-<span class="exam">Consider the matrix &nbsp;<math style="vertical-align: -31px">A=
+<span class="exam">Let &nbsp;<math style="vertical-align: -20px">A=
      \begin{bmatrix}
-& -4 & 9 & -7 \\
+& 1 \\
-            -1 & 2  & -4 & 1 \\
+& 5
-& -6 & 10 & 7
+          \end{bmatrix}.</math>
-::<math>B=
+<span class="exam">(a) Find a basis for the eigenspace(s) of &nbsp;<math style="vertical-align: 0px">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) Is the matrix &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; diagonalizable? Explain.