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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.

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  • $\begingroup$ What's the inner product that you are working with here? $\endgroup$ Oct 19, 2017 at 16:39
  • $\begingroup$ using the inner product (A, B) = tr(B∗A) which just works out to be the dot product for complex. $\endgroup$
    – mmm
    Oct 19, 2017 at 16:49
  • $\begingroup$ First find a basis starting with the identity matrix. Then apply Gram Schmidt, Then the elements starting from the second element will form a basis for the complement. $\endgroup$
    – copper.hat
    Oct 19, 2017 at 16:50
  • $\begingroup$ Also, you know the dimension is 4 so you need only find 3 orthogonal elements. These are not too hard to guess using the inner product formula. $\endgroup$
    – copper.hat
    Oct 19, 2017 at 16:51
  • $\begingroup$ okay, so after applying gram-schmidt to my basis above I now have 1/root2[1 0 0 -1], [0 1 0 0], [0 0 1 0]. I've disregarded the first element of that basis and the last 4 ones (they are redundant?). do these form an orthonormal basis for W⊥? $\endgroup$
    – mmm
    Oct 19, 2017 at 16:58

1 Answer 1

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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 $W^\perp$.

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