Inner product, projection matrices and intersection between spaces

Question

Let $\Bbb C^n$ be the inner product of the complex numbers and $U_1 , U_2$ two subspaces also given $p_1 ,P_2$ are projection matrices on $U_1 , U_2$

1. Prove that if $P_1 +P_2$ is a projection matrix then $U_1 \cap U_2 = \{0\}$
2. Prove that if $U_1 \perp U_2$ then $P_1 +P_2 $ is a projection matrix , on which space does it project?

---

For the first part:

First of this is what I know about projection I do not know how many of these are useful in this case

1.$P_1$ and $P_2$ project on their column spaces so $U_1 = col(P_1)$ $U_2 = col(P_2)$

2.assume that $p$ is the projection of $x$ on $U$ then $x=p+q$ while $p \in U$ and $q \in U^\perp$

3.if $v \in U$ we get $Pv=v$ and if $v \in U^\perp$ we get $Pv=0$

4.we also know that projection matrix has eigenvalues $1$ and $0$

I tried using points 3 and 4 to show that we have a different eigenvalue but I am not sure if it is correct

assume that $v \in U_1 \cap U_2$ then $P_1v=v$ and $P_2v=v$ so $(P_1+P_2)v=v+v=2v $ which means we have eigenvalue $2$ and that contradicts $v \in U_1 \cap U_2$ so the it is $U_1 \cap U_2 = \{0\}$

The second part:

$U_1= span\{u_1 \dots u_k \},U_2=span\{w_1 \dots w_k \}$ while the vectors are orthogonal basis for $U_1$ and $U_2$

$P_1=\frac{ <x,u_1>}{||u_1||^2}u_1 + \dots + \frac{ <x,u_k>}{||u_k||^2}u_k$

$P_2=\frac{ <x,v_1>}{||v_1||^2}v_1 + \dots + \frac{ <x,v_k>}{||v_k||^2}v_k$

and if the basis is orthogonal it can be orthonormal (not sure of this) so

$P_1=<x,u_1>u_1 + \dots + <x,u_k>u_k$

$P_2=<x,v_1>v_1 + \dots + <x,v_k>v_k$

since the inner product is in complex we have $<v,u>=u^*v$

$P_1= u_1^*xu_1 + \dots u_k^* xu_k$

$P_2= v_1^*xv_1 + \dots v_k^* xv_k$

But I got stuck from here and could not continue.. I assume it projects on the column space of each of them but again I am not sure

Sorry if the translations are not correct hopefully it can be understandable

Thanks for any help and tips!

Answer 1

The first part you have essentially solved. Here is one way of presenting the same argument contrapositively: Suppose $U_1 \cap U_2 \neq 0$ and choose a non-zero element $x \in U_1 \cap U_2$. Then both $P_1$ and $P_2$ fix $x$ and so $(P_1 + P_2)x = P_1x + P_2x = 2x$. However, this shows that $P_1 + P_2$ is not a projection since then $(P_1 + P_2)^2x = 4x \neq 2x$ since $x \neq 0$. This proves the first part contrapositively.

For the second part suppose $U_1 \perp U_2$. Then one readily checks that $P_1P_2 = P_2P_1 = 0$ and so $$(P_1+P_2)^2 = P_1^2 + P_1P_2 + P_2 P_1 + P_1^2 = P_1 + P_2,$$ which shows that $P_1 + P_2$ is a projection. It is straightforward to check that $P_1 + P_2$ is the projection onto $U_1 + U_2$.

Answer 2

If $U_1 \bigcap U_2 \not =0$ let $x$ be in both spaces then $(p_1+p_2)(x)=2x$, so thta $p_1+p_2= q$ is not a projector $q\circ q (x)=4x\not =2x=q(x)$

if $U_1 \perp U_2$, then $U_2=im(p_2)\subset U_1^{\perp}= ker (p_1)$ so that $p_1\circ p_2=0$ and similarly $p_2\circ p_1=0$. Then $(p_1+p_2)(p_1+p_2)= p_1^2+p_1.p_2+p_2 p_1+p_2^2=p_1+p_2$ is a projection.

It image is $U_1+U_2$ as it is contained in this space and as $(p_1+p_2)(x_1+x_2)= p_1(x_1)+p_2(x_1)+p_2(x_1)+p_2(x_2)=x_1+0+0+x_2$ with obvious notations.