Consider $\displaystyle M_{2\times2}$(R) with respect to the Frobenius inner product defined by <A,B>=tr($\displaystyle B^*$A).Find the formula for the orthogonal projection P:$\displaystyle M_{2\times2}$(R)$\displaystyle \rightarrow$$\displaystyle M_{2\times2}$(R) on W, where

W=span{$\displaystyle \frac{1}{2}\left(\begin{array}{cc}1&1\\-1&-1\end{array}\right),\frac{1}{2}\left(\begin{array} {cc}1&1\\1&1\end{array}\right)\}$.

My attempt:

The first step is to calculate an orthonormal basis for W and using the Gram Schmidt process I found that {X,Y}={$\displaystyle \frac{1}{2}\left(\begin{array}{cc}1&1\\-1&-1\end{array}\right),\frac{1}{2}\left(\begin{array} {cc}1&1\\1&1\end{array}\right)\}$ is already an orthonormal basis. Is there a way to check this without using the Gram Schmidt process?

In R B* is just the transpose of matrix B so:

Next I said:

P($\displaystyle \left(\begin{array}{cc}a&b\\c&d\end{array}\right)$=<$\displaystyle \left(\begin{array}{cc}a&b\\c&d\end{array}\right)$,$\displaystyle \frac{1}{2}\left(\begin{array}{cc}1&1\\-1&-1\end{array}\right)$>$\displaystyle \frac{1}{2}\left(\begin{array}{cc}1&1\\-1&-1\end{array}\right)$ + <$\displaystyle \left(\begin{array}{cc}a&b\\c&d\end{array}\right)$,$\displaystyle \frac{1}{2}$$\displaystyle \left(\begin{array}{cc}1&1\\1&1\end{array}\right)$>$\displaystyle \frac{1}{2}$$\displaystyle \left(\begin{array}{cc}1&1\\1&1\end{array}\right)$

and I end up with

P($\displaystyle \left(\begin{array}{cc}a&b\\c&d\end{array}\right)$=$\displaystyle \frac{1}{4}\left(\begin{array}{cc}2a+2b&2a+2b\\2c+ 2d&2c+2d\end{array}\right)$

Is the method correct? If not where am I going wrong? Is the answer correct? If not where did I go wrong?

Any input would be appreciated.

Thanks in advance.