Eigenvalues of the power of a matrix

  • #1
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Homework Statement



If [itex]\lambda_i[/itex] are the eigenvalues of a matrix [itex]A^2[/itex], and [itex]A[/itex] is symmetric, then what can you say about the eigenvalues of A?

Homework Equations





The Attempt at a Solution



I know how to prove that if [itex]\sqrt(\lambda_i)[/itex] is an eigenvalue of A, then [itex]\lambda_i [/itex] is an eigenvalue of [itex]A^2[/itex].

I don't know how to prove the converse though.

Thanks!
 

Answers and Replies

  • #2
micromass
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Let [itex]\lambda[/itex] be an eigenvalue of [itex]A^2[/itex]. What can you say about [itex]A^2-\lambda I[/itex]?

Now, use that

[tex]A^2-\lambda I=(A-\sqrt{\lambda}I)(A+\sqrt{\lambda} I)[/tex]
 
  • #3
IniquiTrance
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Let [itex]\lambda[/itex] be an eigenvalue of [itex]A^2[/itex]. What can you say about [itex]A^2-\lambda I[/itex]?

Now, use that

[tex]A^2-\lambda I=(A-\sqrt{\lambda}I)(A+\sqrt{\lambda} I)[/tex]

Then:

[tex]det(A-\sqrt{\lambda}I)*det(A+\sqrt{\lambda} I) = 0[/tex]

Not sure how to proceed...
 
  • #4
micromass
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If xy=0, then what can you say about x or y?
 
  • #5
IniquiTrance
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If xy=0, then what can you say about x or y?

Oh so all we can say is that either [itex]\sqrt{\lambda}[/itex] or [itex]-\sqrt{\lambda}[/itex] is an eigenvalue of A, but not both?
 
  • #6
micromass
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Oh so all we can say is that either [itex]\sqrt{\lambda}[/itex] or [itex]-\sqrt{\lambda}[/itex] is an eigenvalue of A, but not both?

Right.
 
  • #7
IniquiTrance
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Thank you.
 
  • #8
IniquiTrance
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One more question, how do we know that all the eigenvalues of [itex]A^2[/itex] are positive? Since A is symmetric, if [itex]\lambda < 0[/itex] were an eigenevalue of [itex]A^2[/itex] we'd run into a problem...
 
  • #9
micromass
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We can do the same with [itex]\sqrt{\lambda}[/itex] complex numbers, no?
 
  • #10
IniquiTrance
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We can do the same with [itex]\sqrt{\lambda}[/itex] complex numbers, no?

But since A is symmetric, it must have real eigenvalues, no?
 
  • #11
micromass
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But since A is symmetric, it must have real eigenvalues, no?

Yes, sure. So what we do is we argue the same argument where [itex]\lambda[/itex] might be negative and where [itex]\sqrt{\lambda}[/itex] might be complex. Then we can deduce that either [itex]\sqrt{\lambda}[/itex] or [itex]-\sqrt{\lambda}[/itex] are eigenvalues of A. Since A is symmetric, we have that [itex]\sqrt{\lambda}[/itex] must be real, so [itex]\lambda[/itex] was positive to begin with.
 
  • #12
IniquiTrance
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Yes, sure. So what we do is we argue the same argument where [itex]\lambda[/itex] might be negative and where [itex]\sqrt{\lambda}[/itex] might be complex. Then we can deduce that either [itex]\sqrt{\lambda}[/itex] or [itex]-\sqrt{\lambda}[/itex] are eigenvalues of A. Since A is symmetric, we have that [itex]\sqrt{\lambda}[/itex] must be real, so [itex]\lambda[/itex] was positive to begin with.

Ah ok, thanks again!
 
  • #13
Ray Vickson
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Ah ok, thanks again!

No, you can have both. The matrix
[tex] A = \pmatrix{2 & 0 \\0 & -2}[/tex] has
[tex] B = A^2 = \pmatrix{4 & 0 \\0 & 4}, [/tex]
and so both [itex] \sqrt{4}[/itex] and [itex] -\sqrt{4}[/itex] are eigenvalues of [itex]A.[/itex]

RGV
 

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