Quick question regarding matrices of bilinear forms (finding an orthogonal basis) Ok, I've got the form $\langle A,B \rangle = tr(AB)$ and the vector space of real $2 \times 2$ matrices. The question wants me to determine the matrix of the form with respect to the standard basis of matrix units. I've already done this and got the matrix 
$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$
Now it wants me to find an orthogonal basis for the above form, and I'm unsure whether I should be using Gram-Schmidt on $\left\{v_1, v_2, v_3, v_4 \right\}$ where
$v_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$, $v_2 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$, $v_3 = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}$, $v_4 = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$
Or on the standard basis of matrix units. 
Any clarification would be greatly appreciated. 
 A: Are you sure about the definition of the form? Gram-Schmidt only works for inner product spaces. In fact, taking the basis $\{e_1,e_4,\frac{1}{\sqrt2}(e_2+e_3),\frac{1}{\sqrt2}(e_2-e_3)\}$, the matrix will be
$$\begin{pmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&-1\\
\end{pmatrix}$$
and the signature is invariant under base change. Hence it is impossible to find an orthogonal basis. It works, however, if you take $tr(AB^t)$ instead. In this case the standard basis is already orthonormal.
Edit (to address confusion of OP):
Most authors will define an orthogonal basis as a basis $\{b_1,\cdots, b_n\}$ such that $\left<b_i,b_j\right>=\delta_{ij}$. If you use this definition, then the argument above shows that there is no orthogonal basis.
However, there is a weaker definition where you only require $\left<b_i,b_j\right>=\pm\delta_{ij}$. Then, the basis I characterised above is an orthogonal basis. This weaker definition is non-standard in some sense and should be explicitly indicated. The 'usual' Gram-Schmidt doesn't work here, but I am sure that there is a suitable adaptation as mentioned by @rschwieb. Generally any basis should work (to answer the initial question), but I assume that this modified Gram-Schmidt needs non-isotrope base-vectors to work (which neither the suggested basis nor the standard basis are). 
Finally, if you modify the form to $tr(AB^t)$, you will also get a different matrix representing this form. In fact it will be the identity matrix. Thus the standard basis is already orthogonal.
