Show that the standard ordered basis $E$ for a $2\times 2$ matrix is orthogonal

Question

It says the inner product is defined by

$$\langle A,B\rangle = \mathrm{tr}(B^T A)$$

I know that if the cross product of $2$ vectors is $0$, than it means that the two basis are perpindicular, but how do I go about showing $4$ stardard basis matrices make up a perpendicular basis? The basis of this has $4$ different things.

Answer 1

Yeah, and it is 4-dimensional. So you have to show that $\| e_i \| = 1$ , $(e_i,e_j)=\delta_{ij}$ ($i=1,2,3,4$).

Where $e_1=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and so on

So for example $(e_1, e_2) = \operatorname{tr}(e_1^t e_2) = \operatorname{tr}\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} = 0$

Answer 2

I think the standard basis you refer to is made of $$ E_1=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}\,,\quad E_2=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}\,,\quad E_3=\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}\,,\quad E_4=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix} $$ To show it's orthogonal with respect to the inner product $\langle A,B\rangle=\operatorname{tr}(B^TA)$ you have just to do the computations; for instance $$ \langle E_1,E_2\rangle= \operatorname{tr}\left( \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix} \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} \right)= \operatorname{tr}\begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}=0 $$ Do the other cases; there's no other way.