Thread: Find the formula for the orthogonal projection on a subspace spanned by 2x2 matrices?

1. Find the formula for the orthogonal projection on a subspace spanned by 2x2 matrices?

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.

2. Re: Find the formula for the orthogonal projection on a subspace spanned by 2x2 matri

Originally Posted by chocaholic
Consider $\displaystyle M_{2\times2}(R)$ with respect to the Frobenius inner product defined by $\displaystyle \langle A,B\rangle =\text{tr}(B^*A).$ Find the formula for the orthogonal projection $\displaystyle P:M_{2\times2}(R)\rightarrow M_{2\times2}(R)$ on W, where $\displaystyle W=\text{span}\Bigl\{\frac12\begin{pmatrix}1&1\\-1&-1\end{pmatrix},\frac12\begin{pmatrix}1&1\\1&1\end{ pmatrix}\Bigr\}$.

My attempt:

The first step is to calculate an orthonormal basis for W and using the Gram Schmidt process I found that $\displaystyle \{X,Y\}=\Bigl\{\frac{1}{2}\left(\begin{array}{cc}1 &1\\-1&-1\end{array}\right),\frac{1}{2}\left(\begin{array} {cc}1&1\\1&1\end{array}\right)\Bigr\}$ is already an orthonormal basis. Is there a way to check this without using the Gram Schmidt process?
You could just check that $\displaystyle \langle X,X\rangle = \langle Y,Y\rangle = 1$ and $\displaystyle \langle X,Y\rangle = 0.$

Originally Posted by chocaholic
In R B* is just the transpose of matrix B so:
Next I said:
$\displaystyle P\begin{pmatrix}a&b\\c&d\end{pmatrix} = \Bigl\langle \begin{pmatrix}a&b\\c&d\end{pmatrix},\frac12\begin {pmatrix}1&1\\-1&-1\end{pmatrix}\Bigr\rangle \frac12\begin{pmatrix}1&1\\-1&-1\end{pmatrix} + \Bigl\langle \begin{pmatrix}a&b\\c&d\end{pmatrix},\frac12\begin {pmatrix}1&1\\1&1\end{pmatrix}\Bigr\rangle \frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}$
and I end up with
$\displaystyle P\begin{pmatrix}a&b\\c&d\end{pmatrix} = \frac{1}{4}\begin{pmatrix}2a+2b&2a+2b\\2c+2d&2c+2d \end{pmatrix}\right).$

Is the method correct? If not where am I going wrong? Is the answer correct? If not where did I go wrong?
That looks completely correct to me.