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

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!Solution:    
 
!Solution:    
 
|-
 
|-
−
|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}
 +
          0 & 1  \\
 +
          0 & 0
 +
        \end{bmatrix}.</math>&nbsp;
 
|-
 
|-
−
|So, we have
+
|First, notice that
 
|-
 
|-
 
|
 
|
−
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
+
::<math style="vertical-align: -20px">A^2=   
−
\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}
−
&&\\
+
          0 & 0  \\
−
& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
+
          0 & 0
−
&&\\
+
        \end{bmatrix},</math>&nbsp;
−
& = & \displaystyle{-\frac{2}{5}}.
+
|-
−
\end{array}</math>
+
|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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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
−
|&nbsp;&nbsp; &nbsp; &nbsp; False
+
|&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   
First, notice that
 
which is diagonalizable.
Since    is a diagonal 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 Sample Exam

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