Difference between revisions of "031 Review Part 1, Problem 2"

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

Revision as of 13:47, 9 October 2017

True or false: If a matrix    is diagonalizable, then the matrix    must be diagonalizable as well.

Solution:  
Let  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= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}.}  
First, notice 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^2= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix},}  
which is diagonalizable.
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 a diagonal matrix, the eigenvalues 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}   are the entries on the diagonal.
Therefore, the only eigenvalue 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}   is  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 0.}   Additionally, there is only one linearly independent eigenvector.
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}   is not diagonalizable and the statement is false.
Final Answer:  
       FALSE

Return to Sample Exam

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