Difference between revisions of "031 Review Part 3, Problem 3"
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|'''2.''' By the Diagonalization Theorem, an <math style="vertical-align: 0px">n\times n</math> matrix <math style="vertical-align: 0px">A</math> is diagonalizable | |'''2.''' By the Diagonalization Theorem, an <math style="vertical-align: 0px">n\times n</math> matrix <math style="vertical-align: 0px">A</math> is diagonalizable | ||
|- | |- | ||
− | |if and only if <math style="vertical-align: 0px">A</math> has <math style="vertical-align: 0px">n</math> linearly independent eigenvectors. | + | | |
+ | :if and only if <math style="vertical-align: 0px">A</math> has <math style="vertical-align: 0px">n</math> linearly independent eigenvectors. | ||
|} | |} | ||
Revision as of 13:09, 13 October 2017
Let
(a) Find a basis for the eigenspace(s) of
(b) Is the matrix diagonalizable? Explain.
Foundations: |
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Recall: |
1. The eigenvalues of a triangular matrix are the entries on the diagonal. |
2. By the Diagonalization Theorem, an matrix is diagonalizable |
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Solution:
(a)
Step 1: |
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Step 2: |
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(b)
Step 1: |
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Step 2: |
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Final Answer: |
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(a) |
(b) |