Show/Prove two conditions when $P$ is an $n \times n$ matrix such that $P^2 = P$ and $P^t = P$. Let $P$ is an $n \times  n$ matrix such that $P^2 = P$ and $P^t = P$. With this I am supposed to do two show/prove two conditions:


*

*Let $E = \operatorname{Col}(P)$. Show that $E^{\perp} = \operatorname{Nul}(P)$. (Hint: This is true for all symmetric matrices, ie. $P^2 = P$ is not necessary.) 

*Show that $P$ is the matrix (with respect to the standard basis) of the orthogonal projection onto $E = \operatorname{Col}(A)$. That is, for each $x$ we have $\hat{x} = Px$. (Hint: $x = Px + (x - Px)$.)


What I know:


*

*$\operatorname{Nul}(A^t) = \operatorname{Col}(A)^{\perp}$

*Since $E=\operatorname{Col}(A)$ then the number of pivots of both $A^t$ and $A$ equals the number of columns of $A$.

*$\dim E + \dim E^{\perp} = \dim\mathbb{R}^n = n $ so $\dim E^{\perp} = \dim\operatorname{Nul}(A) = n - p$


I just don't know how to approach answering this.
 A: If $A$ is any matrix, then $\mathbf{v}\in\mathrm{Null}(A)$ if and only if $A\mathbf{v}=\mathbf{0}$. If you think about how you multiply a matrix by a column vector, you will see that this requires that $\mathbf{v}$ be orthogonal to each row of $A$ (I'm conflating a bit row and column vectors; if your vectors are columns, then transpose the rows of $A$ before taking the inner product with $\mathbf{v}$). Conversely, if $\mathbf{v}$ is orthogonal to every row of $A$, then $A\mathbf{v}=\mathbf{0}$. That is:
$$\mathrm{null}(A) = (\mathrm{Rowspace}(A))^{\perp}.$$
This is true for any matrix.
If $A$ is symmetric, so that $A=A^t$, then the rowspace of $A$ equals its columnspace (because the columnspace of $A$ equals the rowspace of $A^t$). So if $A$ is symmetric, then
$$\mathrm{null}(A) = (\mathrm{Rowspace}(A))^{\perp} = (\mathrm{columnspace}(A))^{\perp}.$$
So for $P$, since $P=P^t$, we have:
$$\mathrm{null}(P) = (\mathrm{columnspace}(P))^{\perp} = E^{\perp},$$
which proves 1.
For part 2, we want to show that (i) $\mathrm{null}(P) = \mathrm{Im}(P)^{\perp}$; (ii) $\mathrm{Im}(P) = E$. The second clause, together with the fact that $P^2=P$, tells us that $P$ is a projection onto $E$. The first tells us that it is in fact an orthogonal projection.
But since $\mathrm{null}(P) = E^{\perp}$, in order to show the first clause we need to show that $\mathrm{Im}(P)=E$; that is, the second clause will give us both the first and second clauses, since we have already proven 1.
So we need to show that the image of $P$ is precisely $E$. But this is immediate: the image of any matrix is precisely its columnspace. 
If you don't already know that $P^2=P$ implies that we have a projection, then we also need to show that $P$ is a projection. If $x\in V$, then I claim that $x-Px\in\mathrm{nullspace}(P)$: indeed,
$$P(x-Px) = Px - P^2x = Px - Px = \mathbf{0},$$
where the equality $Px-P^2x = Px-Px$ is justified because $P^2=P$. 
Now given any element $x\in V$, we can write $x = x + (Px - Px) = Px + (x-Px)$. This shows that we can write $x$ as an element in $\mathrm{Im}(P)$ plus an element in $\mathrm{nullspace}(P)$. That is,
$$V = \mathrm{Im}(P) + \mathrm{nullspace}(P).$$
By the rank-nullity theorem, $\dim(V) =\mathrm{rank}(P) + \mathrm{nullity}(P)$, so we must have $\mathrm{Im}(P)\cap\mathrm{nullspace}(P) = \{\mathbf{0}\}$, so the sum is a direct sum; that is, 
$$V = \mathrm{Im}(P) \oplus \mathrm{nullspace}(P)$$
and for every $x\in V$, to write $x$ as a sum of a vector in $\mathrm{Im}(P)$ and a vector in $\mathrm{nullspace}(P)$, we write $x = Px + (x-Px)$; so $P$ is the projection onto the first component, i.e., $P$ is the projection onto its columnspace. Since the nullspace is orthogonal to the image, $P$ is an orthogonal projection.
