Finding the associated matrix to an orthogonal projection

Question

Considering the usual inner product let $~Ps:\mathbb R^3\to\mathbb R^3~$ be the orthogonal projection over the plane $~S~$ defined by $~(x,y,z)~$ such that $~x−2y+z=0~$.Find the associated matrix of $~Ps~$ with the canonical basis of $~\mathbb R^3~$.

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So I have been doing plenty of orthogonal projections of vectors over certain subspace until I was hit with this, I'm so confused over how to solve it and I'm struggling in finding a standard procedure in how to do it, I'm thinking of finding the projection of the $~3~$ vectors of the canonical basis but I'm stuck in how to proceed next.

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Although it's in Spanish, here is the link to where the problem came from Ejercicio 3 parte 1

Answer 1

A general idea - There is a plane in $R^3$, meaning - the matrix which would represent that plane would be $[A]$ - a 3x2 matrix with rank $2$ i.e $2$ linearly independent columns.

Now, consider the plane is spanned by the columns of [A] = [$a_1 a_2$] where $a_i$ = columns of $[A]$. By hit and trial we see one vector would be $a_1$ = [2 1 0$]^T$ that is in the plane S.

To get another vector which is orthogonal to this vector and in the plane S, consider $a_2$ = [j k l$]^T$. Satisfying the orthogonality condition we get $\to$ $2j+k=0$ and satisfying that it lies in the plane we get $j-2k+l=0$. So we see a vector satisfying these two conditions is $a_2$ = [1 -2 -5$]^T$. So that is our matrix $[A]_{3*2} = $.

$$ \begin{matrix} 2 & 1 &\\ 1 & -2 \\ 0 & -5 \\ \end{matrix} $$

To get the orthogonal projection matrix onto this plane - we compute its orthogonal projector which would be a 3x3 matrix as we are projecting on a plane in $R^3$; $$[P] = A(A^TA)^{-1}A^T$$. Computing this we get $- [P] =$ $$ \begin{matrix} 5/6 & 1/3 & -1/6\\ 1/3 & 1/3 & 1/3 \\ -1/6 & 1/3 & 5/6\\ \end{matrix} $$

This is the orthogonal projector matrix onto the plane S $\to$ $x−2y+z=0$. So the orthogonal projection of any vector v $\to$ $v^{"}$ would be $v^{"} = [P][v]$.

Answer 2

I'm thinking of finding the projection of the 3 vectors of the canonical basis

That's what I'd do.

With $X' = Ps · X$ then each row of $Ps$ is the factors for each coordinate of $X$

But it's only true if the origin of the basis in the plane is the projection of the origin of the $R^3$ basis

Answer 3

Here’s a way to approach this: If $P_S$ is the orthogonal projection onto a subspace $S$, then $I-P_S$ is the orthogonal projection onto $S^\perp$. Writing the defining equation of your $S$ as $(1,-2,1)\cdot(x,y,1)=0$, you can see that it is the orthogonal complement of the span of $\mathbf n = (1,-2,1)^T$. From what you’ve written in your question, I assume that you know how to compute the projection matrix onto a single vector $\mathbf n$; just subtract that from the identity matrix to get the projection onto $S$.

There are other viable ways to approach this, of course, including your idea of computing the projections of the standard basis vectors, but the above requires the least work. You could, for example, find an orthogonal basis for $S$. The orthogonal projection onto $S$ is then the sum of the individual projections onto the basis vectors. Or, once you’ve found any basis for $S$, assemble those column vectors into a matrix $A$ and use the formula $P_S = A(A^TA)^{-1}A^T$.