- #1

- 213

- 0

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]?

Last edited: