If $P$ is the standard matrix an orthogonal projection, prove that so is $P^k$. 
Suppose $P$ is the standard matrix for an orthogonal projection of $\mathbb R^n$ onto a subspace of $\mathbb  R^n$. Prove that $P^k$ is the standard matrix of an orthogonal projection of $\mathbb R^n$ onto some subspace (for any positive integer $k$).

This is my attempt:
We have to prove symmetric and  idempotent. First, to prove symmetric $(P^k)^T=(P^T)^k$ since $P^T=P$ then $(P^k)^T=P^k$.
Second idempotent $(P^k)=P^{k-2} P^2=P^{k-2}.P=P^{k-4} P^2 P=\dots=P^4=P^2 P^2 =P.P=P^2=P$
but I am not sure whether what I've done is correct.
 A: I'm not sure myself what our OP user 146264 means by "standard" projection, but I think it's pretty clear from the context and what is stated that $P$ is a matrix satisfying $P^2 = P$ and $P = P^T$.  And that what is wanted is to show that $P^k$, $k \ge 1$, also has these properties, i.e., that $(P^k)^2 = P^k$ and that $(P^k)^T = P^k$.  Both these assertions follow readily from the corresponding properties of $P$ and elementary properties of the transpose operation:
I.)  For any square matrices $A$ and $B$ it is well-known that $(AB)^T = B^TA^T$; from this we have $(A^2)^T = (AA)^T = A^TA^T = (A^T)^2$; thus,
$(A^3)^T = (AA^2)^T = (A^2)^TA^T = (A^T)^2A^T = (A^T)^3; \tag{1}$
now a simple induction may be performed, viz.
$(A^j)^T = (A^T)^j \Rightarrow (A^{j + 1})^T = ((A^j)A)^T = A^T(A^j)^T = A^T(A^T)^j = (A^T)^{j + 1}, \tag{2}$
whence we conclude $(A^k)^T = (A^T)^k$ for all positive integers $k$.  So, replacing $A$ with $P$, we see that $(P^k)^T = P^k$; $P^k$ is in fact symmetric as was desired to show.
It seems to me user14264 argued this correctly, though I have given a somewhat more expansive and detailed presentation.
II.)  To show $(P^k)$ is idempotent, that is, $(P^k)^2 = P^k$, it is probably easiest to note that $P^2 = P \Rightarrow P^k = P$ for $k \ge 1$; indeed, we have $P^3 = P^2P = PP = P^2 = P$ and thus again we may perform a basic inductive step:
$P^j = P \Rightarrow P^{j + 1} = P^jP = PP = P^2 = P, \tag{3}$
whence $P^k = P$ for all $k \ge 1$.  Now it is easy to see that $P^k$ is idempotent, since
$P^{2k} = P = P^k$.
It appears that our OP was pretty close to getting this one, right on top of it in fact, but didn't quite finish tying the knot.  The argument he presented is a little awkward and needs refinement, but has the seed of truth in it.  My take on it, in any event.
Observe:  since $P^k = P$, the range of $P^k$ is the range of $P$, so they project onto the same subspace of $\Bbb R^n$, and they have the same properties, so if $P$ is a "standard orthogonal projection" or whatever, so is $P^k$.
Moral of the Story:  Taking powers of idempotents doesn't get you very far!
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
