Difference between revisions of "031 Review Part 1, Problem 4"

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!Solution:    
 
!Solution:    
 
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|-
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|First, we switch to the limit to <math style="vertical-align: 0px">x</math> so that we can use L'Hopital's rule.
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|Let &nbsp;<math style="vertical-align: -20px">A=   
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    \begin{bmatrix}
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          1 & 1  \\
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          0 & 1
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        \end{bmatrix}.</math>&nbsp;
 
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|-
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|So, we have
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|First, notice that &nbsp;<math style="vertical-align: -5px">\text{det }A=1\neq 0.</math>
 
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|-
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|
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|Therefore, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is invertible.
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&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
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|-
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\displaystyle{\lim_{x \rightarrow \infty}\frac{3-2x^2}{5x^2 + x +1}} & \overset{L'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{-4x}{10x+1}}\\
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|Since &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is a triangular matrix, the eigenvalues of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; are the entries on the diagonal.
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&&\\
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|-
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& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
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|Therefore, the only eigenvalue of &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is &nbsp;<math style="vertical-align: -1px">1.</math>&nbsp; Additionally, there is only one linearly independent eigenvector.
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&&\\
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|-
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& = & \displaystyle{-\frac{2}{5}}.
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|Hence, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is not diagonalizable and the statement is false.
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\end{array}</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;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|-
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|&nbsp;&nbsp; &nbsp; &nbsp; False
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|&nbsp;&nbsp; &nbsp; &nbsp; FALSE
 
|}
 
|}
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[[031_Review_Part_1|'''<u>Return to Sample Exam</u>''']]
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[[031_Review_Part_1|'''<u>Return to Review Problems</u>''']]

Latest revision as of 12:16, 15 October 2017

True or false: If    is invertible, then    is diagonalizable.

Solution:  
Let   
First, notice that  
Therefore,    is invertible.
Since    is a triangular matrix, the eigenvalues of    are the entries on the diagonal.
Therefore, the only eigenvalue of    is    Additionally, there is only one linearly independent eigenvector.
Hence,    is not diagonalizable and the statement is false.


Final Answer:  
       FALSE

Return to Review Problems

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