Difference between revisions of "031 Review Part 3, Problem 10"

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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations:    
 
!Foundations:    
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|An eigenvector &nbsp;<math style="vertical-align: 0px">\vec{x}</math>&nbsp; of a matrix &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is a nonzero vector such that
 
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::<math>A\vec{x}=\lambda\vec{x}</math>
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|for some scalar &nbsp;<math style="vertical-align: 0px">\lambda.</math>
 
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Revision as of 20:57, 10 October 2017

Show that if    is an eigenvector of the matrix product    and    then    is an eigenvector of  


Foundations:  
An eigenvector    of a matrix    is a nonzero vector such that
for some scalar  


Solution:

Step 1:  
Step 2:  


Final Answer:  
      

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