1
$\begingroup$

I have three matrices and I am trying to determine if they represent a self-adjoint linear transformation in some basis on an inner product space. The matrices are:

$$A_1 = \left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array} \right) A_2 = \left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{array} \right) A_3 = \left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{array} \right) $$

Now for the first one I can see immediately that this only has a 2 dimensional eigenspace so cannot be self-adjoint. For the other two I am quite lost though.

I know that in an orthonormal basis the matrices should be the conjugate transpose of themselves, but here we have no idea what the basis/inner product us, so I'm not sure how we can tell.

Any help is very much appreciated!

$\endgroup$
2
$\begingroup$

You want to use the following fact:

A matrix $A$ represents a self-adjoint linear transformation in some basis if and only if the $A$ is real diagonalizable.

To prove the forward direction, notice that if $A$ represents a self-adjoint linear transformation, then $A = X B X^{-1}$, where $B$ is self-adjoint with respect to the standard inner product. But $B$ is real diagonalizable, and hence $A$ is.

For the reverse direction, if $A = X \Lambda X^{-1}$, where $\Lambda$ is a real diagonal matrix, then $\Lambda$ is self-adjoint in the standard basis.

$\endgroup$
  • $\begingroup$ Thank you for your answer, that makes it all clear $\endgroup$ – Wooster Jun 18 '14 at 7:52

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.