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Basis of eigenspace

  • Thread starter roam
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  • #1
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Homework Statement



[tex]A = \left[\begin{array}{ccccc} 2&2&-1 \\ 1&3&-1 \\ 0&2&1 \end{array}\right][/tex]

Given that [tex]\lambda = 3[/tex] is an eigenvalue of A, find a basis for the eigenspace corresponding to the eigenvalue 3.


Homework Equations





The Attempt at a Solution



Is this question asking for the corresponding eigenvector to the eigenvalue [tex]\lambda = 3[/tex] is??

I already found that the corresponding eigenvector to the eigenvalue 3 is: {1, 1, 1}

So, what do I need to write as a basis for the eigenspace corresponding to [tex]\lambda = 3[/tex] ?
 

Answers and Replies

  • #2
Pengwuino
Gold Member
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Remember, an eigenvector is defined by the property [tex]Ax = \lambda x[/tex].

However, for example, it is also true that [tex]A(3x) = \lambda (3x)[/tex]. Thus every constant multiple of your eigenvector is also an eigenvector of A. Your basis is the eigenvector.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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Homework Statement



[tex]A = \left[\begin{array}{ccccc} 2&2&-1 \\ 1&3&-1 \\ 0&2&1 \end{array}\right][/tex]

Given that [tex]\lambda = 3[/tex] is an eigenvalue of A, find a basis for the eigenspace corresponding to the eigenvalue 3.


Homework Equations





The Attempt at a Solution



Is this question asking for the corresponding eigenvector to the eigenvalue [tex]\lambda = 3[/tex] is??

I already found that the corresponding eigenvector to the eigenvalue 3 is: {1, 1, 1}

So, what do I need to write as a basis for the eigenspace corresponding to [tex]\lambda = 3[/tex] ?
I presume that what you found was that the equations reduce to x= z and y= z. That is, that any eigenvector can be written as, say, {x, x, x}= x{1, 1, 1}. Penguino said "Your basis is the eigenvector." I would say, rather, that the basis is the singleton set containing that vector" but I doubt the distinction is important.
 
  • #4
1,266
11
I presume that what you found was that the equations reduce to x= z and y= z. That is, that any eigenvector can be written as, say, {x, x, x}= x{1, 1, 1}. Penguino said "Your basis is the eigenvector." I would say, rather, that the basis is the singleton set containing that vector" but I doubt the distinction is important.

Yes, that's right. Thanks very much :)
 

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