Let $V$ be an inner product space of $2\times 2$ matrices where $\langle A,B\rangle =\operatorname{tr}\left(A^{T}B\right)$.
Let $U$ be a basis for all symetric matrices in $V$.
Complete $\left\{\pmatrix{1&0\\0&0} , \pmatrix{0&0\\0&1}\right\}$ to an orthonormal basis of $U$, and find an orthonormal basis for $U^\perp$.
First of all, I took the matrix $\pmatrix{0&1\\1&0}$ and normalized it to $\pmatrix{0&\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&0}$ to get an orthonormal basis for $U$.
Then, to find a basis for $U^\perp$, I figured I'd find a matrix $D =\pmatrix{a&b\\c&d}$ such that $\langle D,X\rangle =0$, X being each and every matrice in the basis of $U$ and I got $D =\pmatrix{0&-1\\1&0}$. and was left to normalize it to $\pmatrix{0&-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&0}$.
My question is, is it the default way to find a basis for $U^\perp$? It seems that I'm kind of missing the point and am just wired to do all the procedures I've been taught without really giving it much thought.. what would happen if the basis of $U^\perp$ would require me to find 2 vectors instead of 1? would I find the first one, $v_{1}$, by the same procedure and then the second one, $v_{2}$, by the same procedure only adding the condition $\langle v_{2},v_{1}\rangle=0$?
Your calculations are correct :) If you have more than one vector in the orthogonal complement, you can apply the method of Gram-Schmidt. https://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process
If you had no idea, then take any basis of $U$ and complete it to a basis of $\mathbb R^{2,2}$, like $$\underbrace{\underbrace{\begin{pmatrix} 1 & 0\\ 0&0\end{pmatrix}, \begin{pmatrix} 0 & 1\\ 1&0\end{pmatrix},\begin{pmatrix} 0 & 0\\ 0&1\end{pmatrix}}_{\text{basis of $U$}}, \begin{pmatrix} 0 & 1\\ 0&0\end{pmatrix}}_{\text{basis of $\mathbb R^{2\times 2}$}}.$$
If you now apply Gram Schmidt's method, you get an orthonormal basis of $U$ and $U^\bot$ automatically.