Hermitian Matrix question

  • #1
The question was:

If B is Hermitian show that [tex] A=B^2 [/tex] is positive semidefinite.

The answer was:

[tex] B^2 [/tex] has eigenvalues [tex] \lambda_1 ^2, .... \lambda_n^2 [/tex]
(the square of B's eigenvalues) all non negative.

My question is:

Why do we know that [tex] B^2 [/tex] has eigenvalues [tex] \lambda^2 [/tex]?
 
Last edited:

Answers and Replies

  • #2
550
2
What happens if you act [tex]B^2[/tex] on the eigenvectors of [tex]B[/tex]?
 
  • #3
I am not quite sure what you mean by act, but I will assume you mean multiply together. If we take:

[tex] A= \left[ \begin{array} {cc} 1 & 1 \\ 1 & 1 \end {array} \right] [/tex]

Then the eigenvalues are 2 and 0.

Then eigenvectors are:

[tex]B = \left[ \begin{array} {cc} 1 & 1 \\ 1 & -1 \end {array} \right] [/tex]


Then [tex]AB = \left[ \begin{array} {cc} 2 & 2 \\ 0 & 0 \end {array} \right] [/tex]

Wow! The values in the matrix are the eigenvalues. Is this result general? Can we prove it?
How does that help solve the original question?
 
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  • #4
92
0
Let V be any eigenvector of B, and l be any eigenvalue.

B*V= l*V by definition. Starting from that equation you should be able to determine the eigenvalues of B^2. This is what Parlyne meant.

edit: l is the corresponding eigenvalue of V.
 
  • #5
113
1
heres an alternative proof to show that the eigenvalues of A are all nonnegative (and thus A is positive semidefinite). let [itex]x[/itex] be an eigenvector of A, and let [itex]\lambda[/itex] be its corresponding eigenvalue. Then we have the following.

[tex]\lambda \langle x,x \rangle = \langle \lambda x, x \rangle = \langle Ax, x \rangle = \langle B^2 x, x \rangle = \langle B^{*}Bx, x \rangle = \langle Bx, Bx \rangle \geq 0 [/tex]
 
  • #6
Oh gee! of course. it is obvious or should have been. Thank you.
 

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