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

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|Hence, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is not diagonalizable and the statement is false.
 
|Hence, &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is not diagonalizable and the statement is false.
 
|}
 
|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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|&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 Review Problems</u>''']]

Revision as of 12:01, 15 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

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