Orthogonal projection of $D=\begin{pmatrix}1 & 1\\ -3 & 1\end{pmatrix}$ on $W=\{A\in M_{2\times 2}(\mathbb{R})~|~Trace(A)=0\}$ in product space $V$.

Question

This was an assigned homework problem (Turned in 4/14): $V=M_{2\times 2}(\mathbb{R})$ with the inner product $\langle A,B\rangle = Trace(B^t,A),~D=\begin{pmatrix}1 & 1\\ -3 & 1\end{pmatrix}$, and $W=\{A\in M_{2\times 2}(\mathbb{R})~|~Trace(A)=0\}$.

Using definition of $A$ having trace of zero, I used $\begin{pmatrix}a & b\\c & -a\end{pmatrix}$ to establish a basis for $W, ~ \beta_{W}=\left\{ \begin{pmatrix}1 & 0\\0 & -1\end{pmatrix},\begin{pmatrix}0 & 1\\0 & 0\end{pmatrix},\begin{pmatrix}0 & 0\\1 & 0\end{pmatrix}\right\}$

I tried applying Gram-Schmidt orthogonalization process and I got the same basis that I had for W. I think that I figured out that is because the basis $\beta_W$ is already orthogonal. I tested this by taking inner products of all the basis matrices and getting 0.

I'm struggling with where to go next.

Is it the correct approach to use: $\frac{\langle D,w_1\rangle}{\langle w_1,w_1\rangle}w_1+\frac{\langle D,w_2\rangle}{\langle w_2,w_2\rangle}w_2+\frac{\langle D,w_3\rangle}{\langle w_3,w_3\rangle}w_3=w^{\prime}$

$\frac{tr\left(\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}\begin{pmatrix}1 & 1\\ -3 & 1\end{pmatrix} \right)}{tr\left(\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix} \right)}\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}+\frac{tr\left(\begin{pmatrix}0 & 0\\1 & 0\end{pmatrix}\begin{pmatrix}1 & 1\\ -3 & 1\end{pmatrix} \right)}{tr\left(\begin{pmatrix}0 & 0\\1 & 0\end{pmatrix}\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix} \right)}\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}+\frac{tr\left(\begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}\begin{pmatrix}1 & 1\\ -3 & 1\end{pmatrix} \right)}{tr\left(\begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix} \right)}\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}=w^{\prime}$

$=\frac{tr\begin{pmatrix}1 & 1\\3 & -1\end{pmatrix}}{tr\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}}\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}+\frac{tr\begin{pmatrix}0 & 0\\1 & 1\end{pmatrix}} {tr\begin{pmatrix}0 & 0\\0 & 1\end{pmatrix}}\begin{pmatrix}0 & 1\\ 0 & 0\end{pmatrix}+\frac{tr\begin{pmatrix}-3 & 1\\0 & 0\end{pmatrix}}{tr\begin{pmatrix}1 & 0\\0 & 0\end{pmatrix}}\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}=w^{\prime}$

and $w^{\prime}=0\cdot\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}+1\cdot\begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}-3\cdot\begin{pmatrix}0 & 0\\1 & 0\end{pmatrix}=\begin{pmatrix}0 & 1\\-3 & 0\end{pmatrix}$

I'm not sure how to verify the solution and I'm not sure what the solution means in relation to the given "vector" $D$.

Answer 1

In vector form

$D = [1, 1, -3, 1]^T $

And a basis for $W$ is

$\{ [1, 0, 0, -1]^T , [0, 1, 0, 0]^T , [0, 0, 1, 0]^T \} $

The orthogonal space is spanned by

$N = [1, 0, 0, 1]^T$ (which corresponds to the identity matrix)

Therefore,

$D - Proj(D) = c_1 N $

$ D \cdot N - 0 = c_1 (N \cdot N) $

Hence $c_1 = \dfrac{D \cdot N}{N \cdot N} = \dfrac{1 + 1}{1 + 1} = 1 $

Therefore,

$Proj(D) = D - c_1 N = [1, 1, -3, 1]^T - (1) [1, 0, 0, 1]^T = [0, 1, -3, 0]^T $

which is the matrix

$\begin{bmatrix} 0 && 1 \\ -3 && 0 \end{bmatrix} $