Proving that the orthogonal projection operator E0 (onto the column space) of the matrix A (which is the standard basis) gets back matrix A.

Question

Let $A$ be an × matrix over $ℝ$, where $$ ≥ $$. Let $$: $ℝ^m$→ $ℝ^m$ be the orthogonal projection operator onto the column space of $$, and let $_0$ denote the matrix representation of $$ relative to the standard basis of $ℝ^$. Let $$ ∈ $ℝ^$ and $$ ∈ $ℝ^$ be given.

Proof that $_0$ * $A$ = $A$.

Question: I think I have totally gone off-tangent w.r.t. to my approach to this question. I am stuck and have no idea how to proceed. Could anyone point me in the right direction to prove this?

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My workings so far:

Workings for $m$ = $n$:

1. Assume $_0$ * $A$ = $A$ is true.
2. Suppose $A$ is a matrix where $$ = $$ such that $dim$($$) = $$.
3. Next we assume that $X$ $\neq$ $^{-1}$ or $I$ or $A$.
4. Then $()$ = $$ * $X$ = $_0$ implies $_0$ = $X$ since $A$ is $I$ when $m$ = $n$.
5. Then $_0$ * $A$ = $A$ implies $X$ * $A$ = $A$.
6. Because $X$ $\neq$ $^{-1}$ or $I$ or $A$ we have arrived to a contradiction such that $_0$ * $A$ $\neq$ $A$.
7. Therefore $_0$ * $A$ = $A$ holds true only when $m$ = $n$ and when $X$ = $^{-1}$ or $I$ or $A$.

Workings for $m$ $>$ $n$:

8. Suppose $A$ is a matrix where $$ > $$ such that $dim$($$) < $$.
9. Next we assume that $X$ is $n$ by $n$ with any values such that $$ * $X$ = $_0$ implies $_0$ to be a $m$ by $n$ matrix.
10. Then $_0$ * $A$ = $A$ implies $$ * $X$ * $A$ = $A$.
11. Both $_0$ * $A$ = $A$ and $$ * $X$ * $A$ = $A$ are equivalent statements that cannot hold true because the rules of matrix multiplication do not allow that.
12. $_0$ would be a $m$ by $n$ matrix multiplying by $A$ a $m$ by $n$ matrix.
13. As such we have arrived at a contradiction such that $_0$ * $A$ $\neq$ $A$.

Answer 1

Try first having an intuitive knowledge of why that is true. Graphically it makes sense, because $E_0$ projects to the column space of $A$, and $Ax$ just moves $x$ to $A$'s column space, so it's projecting some element $v$ to the same space it is already in, hence $E_0 A=A$. More formally:

$E_0=A(A^TA)^{-1}A^T$

$E_0 A=A(A^TA)^{-1}(A^TA)$

$E_0 A=A$

As for the proof, some comments. First, you should not start assuming what you want to proof in your proof, as that would lead to circular reasoning (looking at the first step in $m=n$). Second, you are not covering all the cases, you still lack tacking on $dim(A)<m=n$ and $dim(A)=m<n$ (remember that there are four possibilities for $Ax=b$ !). Third, try using some fact about what you are trying to prove, you are proving something specific about projection matrices but have not used a property specific of them! So that means that the proof is either wrong, or that this applies to every matrix.

As a fun analog problem, try proving why for the projection $P_0$ onto the rowspace of $A$ the following is true: $AP_0=A$