Tukukkk

up vote 3 down vote favorite 1 If ${u_1, cdots, u_n}$ is an orthonormal basis for $mathbb{R}^n$ , and if $A$ can be expressed as $$A = c_1u_1u_1^T + cdots + c_nu_nu_n^T$$ then $A$ is symmetric and has eigenvalues $c_1, cdots, c_n$ . I'm trying to prove this. Here's what I have so far. I figure I need to show: $A$ is symmetric. I can achieve this by showing that $A$ has an orthonormal set of $n$ eigenvectors (or equivalently, that $A$ is orthogonally diagonalizable). If $P$ orthogonally diagonalizes $A$ then $D = P^TAP equiv A = PDP^T$ . $(PDP^T) = (PDP^T)^T$ by trivial manipulations, knowing that $D$ is diagonal and thus $D^T = D$ . $c_1, cdots, c_n$ are the eigenvalues of $A$ . I think both of these would be satisfied if I could show that $c_1u_1u_1^T + cdots + c_nu_nu_n^T$ was equivalent...