0
$\begingroup$

Let $\mathbf{A}, \mathbf{B} \in \mathbb{C}^{n \times n}$ be Hermitian and invertible. Show that exists eigenvalues $\lambda$ which satisfy $\mathbf{Av} = \lambda \mathbf{Bv}$ are not real.

My solution and question:

From the property of Hermitian we know that all eigenvalues of Hermitian matrix are real. (Omit the proof)

Then $\forall x \in \mathbb{C}^n$ we have $x^HAx = x^HP^HDPx = (Px)^HD(Px) = y^HDy = \lambda_1|y_1|^2 + \cdots + \lambda_n |y_n|^2$ which is a real number, where $D$ is diagonal matrix and the second inequality is derived form eigendecomposition.

Back to the question, $Av = \lambda Bv \Rightarrow v^HAv = \lambda v^HBv$.

Since $\mathbf{A}$ and $\mathbf{B} $ are Hermitian, the quadratic form should real. Therefore, the $\lambda = \frac{v^HAv}{v^HBv}$ is real.

Is the original problem wrong? Could anyone help me out? Thanks in advance!

$\endgroup$
0
$\begingroup$

Your considerations are fine. The original problem wrong, as can also seen by taking $\mathbf{B} =\mathbf{I} $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.