Prove that the sum of (symmetric) projection matrices is the identity matrix 1- How can i prove that the sum of projection matrices(symmetric) is identity matrix, P1+P2+P3+...+Pk=I
2- How can i prove that each column in projection matrix is the multiple of the eigenvectors ? 
 A: 1)
Let $v$ be a vector and $v_i = P_k i$. I am assuming that the projections are all different and the dimension is $k$. Then clearly $v = \sum_i v_i$. Hence the result (note if $P_1 = P_2 \ldots $ then your result is not valid.
2)
Let $v$ be a vector, and $v_i = P_i v$. Then since $P_i^2=P_i$, we have 
$$
P_i v_i = v_i$$
So either $v_i =0$ or it is an eigenvector of $P_i$ with eigenvalue = 1.
Now pick $v = (1,0,0,...)^T$. Then $v_i$ is the first column of $P_i$. If $v=(0,1,0...)^T$
then $v_i$ is the second column and so on
Clarification: The assumption is the projections provide a disjoint sum of the original space. 
A: If $A$ is symmetric on a real space, or Hermitian on a complex space (finite-dimensional spaces of dimension $n$ assumed), then $A$ has an orthonormal basis $\{ e_{j} \}_{j=1}^{n}$ of eigenvectors. Equivalently, there exist finite-dimensional symmetric (Hermitian) projections $\{ P_{j}\}_{j=1}^{k}$ such that $\sum_{j} P_{j} = I$, $P_{j}P_{j'}=0$ for $j \ne j'$, $AP_{j}=P_{j}A$ and
$$
           A = \sum_{j=1}^{k}\lambda_{j}P_{j}.
$$
This decomposition is unique if one assumes that $\{ \lambda_{j}\}_{j=1}^{k}$ is the set of distinct eigenvalues of $A$. This way of stating that $A$ has an orthonormal basis of eigenvectors is the Spectral Theorem for Hermitian matrices. This form is coordinate free, but it definitely depends on the particular choice of inner-product.
The projection $P_{j}$ satisfies $AP_{j}=\lambda_{j}P_{j}$, and the range of $P_{j}$ consists of the subspace spanned by all eigenvectors of $A$ with the common eigenvalue $\lambda_{j}$; in particular, if $P_{j}$ is represented in a matrix form with respect to some basis, then the columns of $[P_{j}]$ will always be eigenvectors of matrix representation $[A]$ of $A$, with eigenvalues $\lambda_{j}$; of course some of these column vectors may be $[0]$.
