Consider the complex vector space $M_2(\mathbb{C})$. Let $W$ be the subspace spanned by the identity matrix. Find an orthonormal basis for $W^⊥.$
I'm struggling with finding the orthogonal complement of $\operatorname{span}\left(\begin{bmatrix}1 &0\\ 0& 1\end{bmatrix}\right)$ and extending this to a basis.
So far I have: \begin{align*}\begin{bmatrix}1& 0\\ 0& 1\end{bmatrix},\quad & \begin{bmatrix}1& 0\\ 0& -1\end{bmatrix},\quad &\begin{bmatrix}0& 1\\ 0& 0\end{bmatrix}, \quad &\begin{bmatrix}0& 0\\ 1& 0\end{bmatrix},\\ \begin{bmatrix}i& 0\\ 0& i\end{bmatrix},\quad & \begin{bmatrix}i& 0\\ 0& -i\end{bmatrix},\quad &\begin{bmatrix}0& i\\ 0& 0\end{bmatrix}, \quad &\begin{bmatrix}0& 0\\ i& 0\end{bmatrix},\end{align*} but I'm not sure if this forms a basis.
I know that after these steps to proceed with Gram-Schmidt for orthornormal. Thanks.
The eight matrices that you mentioned cannot possibly be a basis of $M_2(\mathbb{C})$, since this is a $4$-dimensional vector space. But$$\left\{\begin{bmatrix}1&0\\0&1\end{bmatrix},\begin{bmatrix}1&0\\0&-1\end{bmatrix},\begin{bmatrix}0&1\\0&0\end{bmatrix},\begin{bmatrix}0&0\\1&0\end{bmatrix}\right\}$$is a basis of $M_2(\mathbb{C})$. The first element is the identity matrix, of course. And the other three are orthogonal to it. Therefore, they form a basis of $W^\perp$.