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So we assume $M_{2\times 2}(\Bbb{C}) $ has inner product $\langle A,B\rangle=$trace${(A^T\bar{B})}$

How to find an orthogonal basis for $W=\{\begin{bmatrix}a&b \\c&d\end{bmatrix}|a+b+c=0,a=d\}$ defined under $M_{2\times 2}(\Bbb{C}) $

In my opinion, the subspace is spanned by the vectors that satisfy $a=-b-c,\ b=b,\ c=c,\ d=-b-c$. So they will be $(-1,1,0,-1),(-1,0,1,-1)$. To find orthogonal basis we apply Gram–Schmidt process. But the problem is the order. Let's say $v_2=u_2-\frac{\langle v_1,u_2\rangle}{\langle v_1,v_1\rangle}v_1$. In complex inner product space, $\langle v_1,u_2\rangle\ne\langle u_2,v_1\rangle $. Which order should we put?

In addition, if we want to find any orthogonal projection (i.e. $P_w(\begin{bmatrix}1&0\\0&0\end{bmatrix})$) onto $W$, we just find the orthonormal basis of $W$ and then apply $P_w(x)=\sum^k_{i=1} \langle x,w_i\rangle w_i$.

Am I missing anything?

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  • $\begingroup$ I think you are on track here in terms of how one would approach this problem. But make sure you understand why that recipe for projection works; for example, before asking for the "orthogonal" projection, you must indicate your inner product (which you rightly employ in the definition of $P_W$). $\endgroup$ – knsam Apr 10 '16 at 13:14
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If a+ b+ c= 0, then c= -a- b. If, further, d= a, then any such matrix can be written $\begin{bmatrix}a & b \\ -a- b & a\end{bmatrix}= \begin{bmatrix}a & 0 \\- a & a\end{bmatrix}+ \begin{bmatrix}0 & b \\ -b & 0\end{bmatrix}= a\begin{bmatrix}1 & 0 \\ -1 & 1\end{bmatrix}+ b\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$

So those two matrices form a basis. As for "orthogonality" how are you defining "orthogonal" for matrices? What inner product are you using?

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