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Math Help - Find the formula for the orthogonal projection on a subspace spanned by 2x2 matrices?

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

    Consider M_{2\times2}(R) with respect to the Frobenius inner product defined by <A,B>=tr( B^*A).Find the formula for the orthogonal projection P: M_{2\times2}(R) \rightarrow M_{2\times2}(R) on W, where
    W=span{ \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}={ \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( \left(\begin{array}{cc}a&b\\c&d\end{array}\right)=< \left(\begin{array}{cc}a&b\\c&d\end{array}\right), \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) + < \left(\begin{array}{cc}a&b\\c&d\end{array}\right), \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)
    and I end up with
    P( \left(\begin{array}{cc}a&b\\c&d\end{array}\right)= \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.
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  2. #2
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    Re: Find the formula for the orthogonal projection on a subspace spanned by 2x2 matri

    Quote Originally Posted by chocaholic View Post
    Consider M_{2\times2}(R) with respect to the Frobenius inner product defined by \langle A,B\rangle =\text{tr}(B^*A). Find the formula for the orthogonal projection P:M_{2\times2}(R)\rightarrow M_{2\times2}(R) on W, where 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 \{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 \langle X,X\rangle = \langle Y,Y\rangle = 1 and \langle X,Y\rangle = 0.

    Quote Originally Posted by chocaholic View Post
    In R B* is just the transpose of matrix B so:
    Next I said:
    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
    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.
    Last edited by Opalg; September 3rd 2011 at 12:52 AM. Reason: corrected error
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