Inner product, projection matrices and intersection between spaces

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 $$U_2$$

$$P_1=\frac{ }{||u_1||^2}u_1 + \dots + \frac{ }{||u_k||^2}u_k$$

$$P_2=\frac{ }{||v_1||^2}v_1 + \dots + \frac{ }{||v_k||^2}v_k$$

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

$$P_1=u_1 + \dots + u_k$$

$$P_2=v_1 + \dots + v_k$$

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

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

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

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!

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 $$U_1 + U_2$$.
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.