projection a point into a subspace whose does not have orthonormal basis

Question

I know how we can find the projection of a point into a subspace with orthogonal basis.

But suppose I have a 2 dimension subspace with basis: $$V=\langle(1,2,3),(1,-3,-2)\rangle$$ obviously they are not orthogonal. Now I want to find the length of projection of $p=(4,1,2)$ to this subspace. How I can find it without finding orthogonal basis with Gram-Schmidt algorithm.

Is there any formula?

Answer 1

HINT

• consider a basis with vectors $v_1,v_2 \in V$ and $v_3$ orthogonal to them
• write down the projection matrix in this basis (it is trivial)

$$P_{\mathcal{B}}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0\end{bmatrix}$$

• change to the standard basis to find the projection matrix $A$

$$A=MP_{\mathcal{B}}M^{-1}$$

with $$M=[v_1 \quad v_2 \quad v_3]$$

Notably we have

$$M=\begin{bmatrix} 1 & 1 & 1\\ 2 & -3 & 1\\ 3 & -2 & -1\end{bmatrix}\qquad M^{-1}=\begin{bmatrix} \frac13 & -\frac1{15} & \frac4{15}\\ \frac13 & -\frac4{15} & \frac1{15}\\ \frac13 & \frac13 & -\frac13\end{bmatrix}$$

and then

$$A=MP_{\mathcal{B}}M^{-1}= \left[\begin{array}{rrr}\frac23&-\frac13&\frac13\\-\frac13&\frac23&\frac13\\\frac13&\frac13&\frac23\end{array}\right]$$

and finally $Ap=(3,0,3)\implies |Ap|=3\sqrt 2$.

Answer 2

With a small abuse of notation, form the matrix $V$ that has the elements of the basis as its columns. The order in which they appear is unimportant. Then the matrix $$P=V(V^TV)^{-1}V^T$$ is the matrix of the orthogonal projection onto this subspace. You should be able to verify for yourself that $P^2=P$, so that this is indeed a projection, that it’s the identity map on the subspace $V$ (hint: compute $PV$) and that its null space is the orthogonal complement of the subspace $V$.

In this case, $$V = \left[\begin{array}{rr}1&1\\2&-3\\3&-2\end{array}\right]$$ and so $$P = \left[\begin{array}{rrr}\frac23&-\frac13&\frac13\\-\frac13&\frac23&\frac13\\\frac13&\frac13&\frac23\end{array}\right].$$