Difference between revisions of "031 Review Part 3, Problem 1"
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!Foundations: | !Foundations: | ||
| + | |- | ||
| + | |Recall: | ||
| + | |- | ||
| + | |'''1.''' The eigenvalues of a triangular matrix are the entries on the diagonal. | ||
| + | |- | ||
| + | |'''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. | ||
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Revision as of 13:10, 13 October 2017
(a) Is the matrix 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} 3 & 1 \\ 0 & 3 \end{bmatrix}} diagonalizable? If so, explain why and diagonalize it. If not, explain why not.
(b) Is the matrix 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} 2 & 0 & -2 \\ 1 & 3 & 2 \\ 0 & 0 & 3 \end{bmatrix}} diagonalizable? If so, explain why and diagonalize it. If not, explain why not.
| Foundations: |
|---|
| Recall: |
| 1. The eigenvalues of a triangular matrix are the entries on the diagonal. |
| 2. By the Diagonalization Theorem, an 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 n\times n} matrix 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 diagonalizable |
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Solution:
(a)
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(b)
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| Final Answer: |
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| (a) |
| (b) |