- #1
interested_learner
- 213
- 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]?
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]?
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