Find the matrix of the orthogonal projector on span

Question

I need to find the matrix of the orthogonal projector on $Span([1,1,-1],[1,-1,-1])$. I had used Gram-Schmidt process on it and have $Span([\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}, -\frac{\sqrt{3}}{3}], [ \frac{\sqrt{6}}{6}, -\frac{\sqrt{6}}{3}, -\frac{\sqrt{6}}{6}])$ and the matrix should be $P = \left[\begin{matrix}\frac{\sqrt{3}}{3} & \frac{\sqrt{3}}{3} & -\frac{\sqrt{3}}{3}\\\frac{\sqrt{6}}{6} & -\frac{\sqrt{6}}{3} & -\frac{\sqrt{6}}{6} \\ 0 & 0 & 0 \end{matrix}\right]$. I got this matrix from these reasonings: proj. of any vector $x \in X$ on some subspace $U$, if $U$ has ONB shuld be looking like $P_U (x) = \sum_{i=1}^{k} (x,e_i)e_i$. So in this example I have $P_U (x) = (\frac{\sqrt{3}}{3} x_1 + \frac{\sqrt{3}}{3}x_2 - \frac{\sqrt{3}}{3}x_3)e_1 + (\frac{\sqrt{6}}{6}x_1 -\frac{\sqrt{6}}{3}x_2 -\frac{\sqrt{6}}{6}x_3)e_2 $ So this is equal to $P*x$, where $P$ -- matrix from above, and $x$ -- colomn-vector of our space. -- Where is the mistake? Can somebody help pls?

Answer 1

You didn’t go far enough with your solution. Your matrix $P$ certainly computes the orthogonal projection correctly, but what it computes are its coordinates relative to the orthonormal basis that you found (extended to span all of $\mathbb R^3$). To put it another way, you’re computing the $(x,e_i)$ part of $(x,e_i)e_i$, but neglecting to multiply by $e_i$. Using your matrix, $Px=[(e_1,x),(e_2,x),0]^T$, but what you need to end up with is $(e_1,x)e_1+(e_2,x)e_2$.

With the above in mind, the correct projection matrix is $$P = \begin{bmatrix}e_1&e_2\end{bmatrix} \begin{bmatrix}e_1^T\\e_2^T\end{bmatrix} = e_1e_1^T+e_2e_2^T,$$ that is, it’s the sum of individual projections onto the basis vectors as you’ve noted.

Answer 2

Take $u_1 = v_1 = (1, 1, -1)$ and $v_2 = (1, -1, -1)$.

Applying Gram Schmidt, we have that $u_2 = v_2 - \frac{1}{3}u_1 = (1, -1, -1) - \frac{1}{3}(1, 1, -1) = (\frac{2}{3}, \frac{-4}{3}, -\frac{2}{3})$.

Orthonormalizing, we get $e_1 = \frac{1}{\sqrt{3}}v_1$ and $e_2 = \frac{1}{\sqrt{\frac{8}{3}}} u_2$