Difference between revisions of "031 Review Part 3, Problem 11"
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Foundations: | !Foundations: | ||
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+ | |'''1.''' An eigenvector <math style="vertical-align: 0px">\vec{x}</math> of a matrix <math style="vertical-align: 0px">A</math> corresponding to the eigenvalue <math style="vertical-align: 0px">\lambda</math> is a nonzero vector such that | ||
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+ | ::<math>A\vec{x}=\lambda\vec{x}.</math> | ||
+ | |- | ||
+ | |'''2.''' By the Rank Theorem, if <math style="vertical-align: 0px">A</math> is a <math style="vertical-align: 0px">m\times n</math> matrix, then | ||
+ | |- | ||
+ | | | ||
+ | ::<math>\text{rank }A+\text{dim Col }A=n.</math> | ||
|} | |} | ||
Revision as of 21:10, 10 October 2017
Suppose is a basis of the eigenspace corresponding to the eigenvalue 0 of a matrix
(a) Is an eigenvector of If so, find the corresponding eigenvalue.
If not, explain why.
(b) Find the dimension of
Foundations: |
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1. An eigenvector of a matrix corresponding to the eigenvalue is a nonzero vector such that |
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2. By the Rank Theorem, if is a matrix, then |
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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) |