<s>[INST] Find a compression of the frames over the subspaces 
*

*Find a compression of the frames $B=\{(1,1,0),(1,-1,0),(1,1,1),(0,0,1),(0,1,-1)\}$ over the subspaces (a) $M=\{(x,y,z):x=0\}$, (b) $M=\{(x,y,z):x=y\}$

*Find an orthonormal basis $B$ of $\mathbb{R}^4$ such that the frame $B_1=\{(1,1),(1,-1),(\sqrt{\frac{4}{3}},\sqrt{\frac{2}{3}}),(-\sqrt{\frac{2}{3}},\sqrt{\frac{4}{3}}) \}$ is a compression of B.


Building blocks will be greatly appreciated
 [/INST] $\textbf{Part a:}$
When $M=\{(x,y,z):x=0\}$, $P_MB=\begin{Bmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & -1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 & -1 \end{Bmatrix}$
When $M=\{(x,y,z):x=y\}$ $B_M=\begin{Bmatrix} 1 & 0 \\ 1 & 0 \\ 0 & 1 \end{Bmatrix}$ and $B_{M^\perp}=\begin{Bmatrix} -1 \\ 1 \\ 0 \end{Bmatrix}$ 
$P_{M^\perp}B=\begin{Bmatrix} 0 & 1 & 0 & 0 & \frac{-1}{2} \\ 0 & -1 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 & 0 & 0 \end{Bmatrix}$ and $P_MB=\begin{Bmatrix} 1 & 0 & 1 & 0 & \frac{1}{2} \\ 1 & 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & 1 & -1 \end{Bmatrix}$ 
$\textbf{Part b}$
Let $B_1=\begin{Bmatrix} 1 & 1 & \sqrt{\frac{4}{3}} & -\sqrt{\frac{2}{3}} \\ 1 & -1 & \sqrt{\frac{2}{3}} & \sqrt{\frac{4}{3}}
\end{Bmatrix}$
These vectors in $\mathbb{R}^2$ have a norm of $\sqrt{2}$. Since $v=p_Mv+p_{M^\perp}v \iff \|v\|^2=\|p_Mv \|^2+\|p_{M^\perp}v\|^2 \Rightarrow \|v\| \leq \|p_Mv\|$. This implies that if $v_i \in \mathbb{R}^4$, and $p_Mv=b_i$, then $\|v\| \leq \sqrt{2}$. So we cannot construct an orthonormal basis in $\mathbb{R}^4$. However, we can construct an orthogonal basis.
$\begin{Bmatrix} \begin{Bmatrix} 1 & 1 & \sqrt{\frac{4}{3}} & -\sqrt{\frac{2}{3}} \\ 1 & -1 & \sqrt{\frac{2}{3}} & \sqrt{\frac{4}{3}}
\end{Bmatrix} \\ 0 & 0 & \frac{2+\sqrt{2}}{\sqrt{3}} & \frac{2-\sqrt{2}}{\sqrt{3}} \\
\alpha_1 & \alpha_2 & \alpha_3 & \alpha_4 \end{Bmatrix}$
By Gram-Schmidt, we can find the final orthonormal vector.
\begin{equation*}
\begin{split}
(\alpha_1,\alpha_2,\alpha_3, \alpha_4) 
= & (1,0,0,0)-\frac{1}{2}(1, 1, \sqrt{\frac{4}{3}}, -\sqrt{\frac{2}{3}})-\frac{1}{2}( 1, -1, \sqrt{\frac{2}{3}}, \sqrt{\frac{4}{3}}) \\
= & (1,0,0,0)-(\frac{1}{2}, \frac{1}{2}, \sqrt{\frac{1}{3}}, -\sqrt{\frac{1}{6}})-( \frac{1}{2}, -\frac{1}{2}, \sqrt{\frac{1}{6}}, \sqrt{\frac{1}{3}}) \\
= & (0,0,-\frac{\sqrt{2}+1}{\sqrt{6}},\frac{1-\sqrt{2}}{\sqrt{6}})
\end{split}
\end{equation*}
Hence, the orthonormal basis is 
$\begin{Bmatrix} 1 & 1 & \sqrt{\frac{4}{3}} & -\sqrt{\frac{2}{3}} \\ 1 & -1 & \sqrt{\frac{2}{3}} & \sqrt{\frac{4}{3}} \\ 0 & 0 & \frac{2+\sqrt{2}}{\sqrt{3}} & \frac{2-\sqrt{2}}{\sqrt{3}} \\
0 & 0 & -\frac{\sqrt{2}+1}{\sqrt{6}} & \frac{1-\sqrt{2}}{\sqrt{6}}
\end{Bmatrix}$
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