# Kernel decomposition of a finite rank integral opeartor

Given a self-adjoint finite rank integral operator P on $$L_2[0,1]$$, it has the eigen-decomposition $$P=\sum_{i=1}^k\lambda_i \langle u_i,\cdot\rangle u_i$$ where $$u_i$$ are eigenfunctions and $$\lambda_i$$ are eigenvalues for P.

Let $$p(x,y)$$ be the kernel for $$P$$, i.e. $$Pf(x)=\int_0^1 p(x,y)f(y)dy$$, what can we say about $$p(x,y)$$? Is it always true that there are finite number of $$f_k$$ which are orthogonal and we can write $$p(x,y)=\sum_{i,i'} \langle f_i,Pf_i'\rangle f_i(x)f_i'(x)$$ ? If such expression exists, what are the relationship between $$f_i$$ and the eigenfunctions of P, namely $$u_j$$ ?

We can assume $$0 for some constant C.

Thanks!

From the fact that $$P$$ is selfadjoint, you can deduce that $$u_1,\ldots,u_k$$ are orthonormal: for $$j\ne m$$, $$\lambda_j\langle u_j,u_m\rangle=\langle Pu_j,u_m\rangle=\langle u_j,Pu_m\rangle=\lambda_m\langle u_j,e_m\rangle.$$ So $$\langle u_j,u_m\rangle=0$$. And then $$Pu_j=\lambda_j\|u_j\|^2\,u_j$$, so $$\|u_j\|=1$$.
Then you can take $$p(x,y)=\sum_{j=1}^k\lambda_j u_j(x)u_j(y).$$ You have $$\int_0^1\sum_{j=1}^k\lambda_j u_j(x)u_j(y)\,f(y)\,dy=\sum_{j=1}^k\lambda_j u_j(x)\int_0^1 u_j(y)\,f(y)\,dy=\sum_{j=1}^k\lambda_j \langle u_j,f\rangle\,u_j(x)=(Pf)(x).$$