Let $k<n$ and $v_{1},v_{2},...,v_{k} \in \mathbb{R}^{n}$ be non-zero vectors, orthogonal with respect to the standard inner product in $\mathbb{R}^{n}$, where the vectors of $\mathbb{R}^{n}$ represent column vectors. Let $\lambda_{1},... \lambda_{k} \in \mathbb{R}$ and suppose that $A= \lambda_{1} v_{1} v_{1}^{T} + \lambda_{2} v_{2} v_{2}^{T} +...+ \lambda_{k} v_{k} v_{k}^{T} \in \mathbb{R}^{n \times n}$. Prove that $A$ is diagonalizable and give an orthogonal base of eigenvectors of A.
I started by showing that $v_{1},v_{2},...,v_{k}$ are eigenvectors of $A$: Take $v_{i} \in \lbrace v_{1},v_{2},...,v_{k} \rbrace$. Then $A v_{i}= \lambda_{1} v_{1} v_{1}^{T} v_{i} +...+ \lambda_{i} v_{i} v_{i}^{T} v_{i} +...+ \lambda_{k} v_{k} v_{k}^{T} v_{i}= \lambda_{1} v_{1} \langle v_{1},v_{i} \rangle +...+ \lambda_{i} v_{i} \langle v_{i}, v_{i} \rangle +...+ \lambda_{k} v_{k} \langle v_{k}, v_{i} \rangle = \lambda_{i} v_{i} \langle v_{i}, v_{i} \rangle = \lambda_{i} \Vert v_{i} \Vert^{2} v_{i}$. This is true because $v_{i}$ is orthogonal with all the other vectors $\lbrace v_{1},...,v_{i-1},v_{i+1},...,v_{k} \rbrace$.
I then showed that $dim(E_{0}) \geq n-k$: Take $v \in \lbrace v_{1},...,v_{k} \rbrace^{\perp}$. Then $Av=\lambda_{1} v_{1} v_{1}^{T}v +...+ \lambda_{k}v_{k}v_{k}^{T}v=\lambda_{1}v_{1} \langle v_{1},v \rangle +...+\lambda_{k}v_{k} \langle v_{k},v \rangle=0=0v$. Thus all vectors in $vct \{ v_{1},...,v_{k} \}^{\perp}$ have eigenvalue 0. Thus $vct \{ v_{1},...,v_{k} \}^{\perp} \subset E_{0}$, which means that $n-k \leq dim(E_{0})$.
Could someone tell me how to go on from here?