Orthogonal projection onto a subspace

Question

Let $v_1=\begin{pmatrix}1\\ 1\\ 1\\ 0\end{pmatrix}$ and $v_2 = \begin{pmatrix}0\\ 1\\ 1\\ 1\end{pmatrix}$. Find the orthogonal projection of $v = \begin{pmatrix}3\\ 2\\ 1\\ 0\end{pmatrix}$ onto $V= \operatorname{span} (v_1,v_2)$.

If I have $A= \begin{pmatrix}1&0\\ \:1&1\\ \:1&1\\ \:0&1\end{pmatrix}$ and compute the projection matrix $P=A(A^TA)^{-1}A^T = \begin{pmatrix}\frac{3}{5}&\frac{1}{5}&\frac{1}{5}&-\frac{2}{5}\\ \frac{1}{5}&\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\ \frac{1}{5}&\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\ -\frac{2}{5}&\frac{1}{5}&\frac{1}{5}&\frac{3}{5}\end{pmatrix}$ then how does this help? I'm not sure I've entirely understood the idea of orthogonal projection?

Answer 1

If you apply Gram-Schmidt to $\{v_1,v_2\}$, you will get $\{e_1,e_2\}$, with$$e_1=\frac1{\sqrt3}(1,1,1,0)\quad\text{and}\quad e_2=\frac1{\sqrt{15}}(-2,1,1,3).$$Therefore, the orthogonal projection of $v$ onto $\operatorname{span}\bigl(\{v_1,v_2\}\bigr)$ is $\langle v,e_1\rangle e_1+\langle v,e_2\rangle e_2$, which happens to be equal to $=\frac15\left(12,9,9,-3\right)$.

Answer 2

What you had was the projection matrix for projection onto the span of the two vectors. Just multiply it by your given vector! (You might find some of my YouTube lectures, linked in my profile, helpful.)