Difference between revisions of "031 Review Part 1, Problem 2"
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|Hence, <math style="vertical-align: 0px">A</math> is not diagonalizable and the statement is false. | |Hence, <math style="vertical-align: 0px">A</math> 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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| FALSE | | FALSE | ||
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− | [[031_Review_Part_1|'''<u>Return to | + | [[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: |
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Let |
First, notice that |
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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: |
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FALSE |