0
$\begingroup$

The matrix representation of a linear operator T: $\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ is given with respect to the following basis \begin{bmatrix}-5 &2 & \\ 2&5 & \end{bmatrix} is the matrix of T when written with respect to the basis $(1,0)$ and $(1,1)$. We take the standard dot product as its inner product.

I'm getting that it is self-adjoint as $\langle Tv_1,v_2 \rangle = \langle v_1,Tv_2 \rangle$ where $v_1 = (1,0)$ and $v_2=(1,1)$ but I was told that it wasn't. Could someone please shed some light on this? Something about the conjugate transpose not being equal?

$\endgroup$
  • 1
    $\begingroup$ There's a good chance that what you were told was actually: Trying with just two particular vectors is not enough to prove that the operator is self-adjoint. $\endgroup$ – Henning Makholm Nov 4 '18 at 12:29
0
$\begingroup$

HINT: let $u:=u_1e_1+u_2e_2$ and $v:=v_1e_1+v_2e_2$ for some basis $e_1,e_2$ of $\Bbb R^2$ and $T$ a linear operator from $\Bbb R^2$ to itself, then from the linearity of $T$ and the definition of self-adjoint

$$\forall u,v\in\Bbb R^2:\langle Tu,v\rangle=\langle u,Tv\rangle \iff \forall j,k\in\{1,2\}:\langle Te_k,e_j\rangle=\langle e_k,Te_j\rangle$$

So you only need to check if the RHS of above holds or not holds for some basis of $\Bbb R^2$ and $T$ represented using the standard orthonormal basis of $\Bbb R^2$, because the standard inner product is defined using the standard orthonormal basis.

$\endgroup$

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.