# 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:

1. 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.)
2. 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.

-
Hint for the hint: What is $P(x-Px)$? – Mr. F Apr 3 '12 at 0:24
I have no clue.... that was a homework question. All I know are the points under "what I know". :) – Kyra Apr 3 '12 at 1:35
I bet you know a lot more than what you've listed under "what I know". I bet you know that if $A,B,C$ are matrices then $A(B+C)=AB+AC$. Now look at EMS' hint again. – Gerry Myerson Apr 3 '12 at 1:42

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.

-