Difference between revisions of "031 Review Part 1, Problem 2"
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!Solution: | !Solution: | ||
|- | |- | ||
| − | | | + | |Let <math style="vertical-align: -20px">A= |
| + | \begin{bmatrix} | ||
| + | 0 & 1 \\ | ||
| + | 0 & 0 | ||
| + | \end{bmatrix}.</math> | ||
|- | |- | ||
| − | | | + | |First, notice that |
|- | |- | ||
| | | | ||
| − | + | ::<math style="vertical-align: -20px">A^2= | |
| − | \ | + | \begin{bmatrix} |
| − | && | + | 0 & 0 \\ |
| − | & | + | 0 & 0 |
| − | && | + | \end{bmatrix},</math> |
| − | & = & | + | |- |
| − | + | |which is diagonalizable. | |
| + | |- | ||
| + | |Since <math style="vertical-align: 0px">A</math> is a triangular matrix, the eigenvalues of <math style="vertical-align: 0px">A</math> are the entries on the diagonal. | ||
| + | |- | ||
| + | |Therefore, the only eigenvalue of <math style="vertical-align: 0px">A</math> is <math style="vertical-align: -1px">0.</math> Additionally, there is only one linearly independent eigenvector. | ||
| + | |- | ||
| + | |Hence, <math style="vertical-align: 0px">A</math> is not diagonalizable and the statement is false. | ||
|} | |} | ||
| + | |||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | | | + | | FALSE |
|} | |} | ||
| − | [[031_Review_Part_1|'''<u>Return to | + | [[031_Review_Part_1|'''<u>Return to Review Problems</u>''']] |
Latest revision as of 11:17, 15 October 2017
True or false: If a 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^2} is diagonalizable, then 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} must be diagonalizable as well.
| Solution: |
|---|
| Let 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} 0 & 1 \\ 0 & 0 \end{bmatrix}.} |
| First, notice that |
|
| which is diagonalizable. |
| Since 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 a triangular matrix, the eigenvalues of 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} are the entries on the diagonal. |
| Therefore, the only eigenvalue of 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 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 0.} Additionally, there is only one linearly independent eigenvector. |
| Hence, 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 not diagonalizable and the statement is false. |
| Final Answer: |
|---|
| FALSE |