Decomposing a symmetric matrix into a sum of products I just realized that if a single vector generates a matrix, say,
$A=aa^T$
so that 
$A=\begin{bmatrix} a_1^2&a_1a_2&a_1a_3\\a_1a_2&a_2^2&a_2a_3\\a_1a_3&a_2a_3&a_3^2\end{bmatrix}$
I can just look at it and say that it is going to have two zero eigenvalues and the third eigenvalue equal to the trace. That is so since I can always transform the single $(3\times1)$ vector to a basis where any two of its components are zero.
My question is whether any general symmetric $(3\times3)$ matrix with all three non zero eigenvalues can be written as a sum of products of the form,
$M=j_1aa^T+j_2bb^T+j_3cc^T+j_4(ab^T+ba^T)+j_5(bc^T+cb^T)+j_6(ca^T+ac^T)$,
where $j_i$ are scalars and a, b, c are $(3\times1)$ vectors?
Also, I intuitively used three vectors, but is it possible that such a matrix be written in this form using more than 3 vectors? 
And what are the implications of this on the eigenvalues of the matrix? Can we draw any conclusions from it? 
 A: A symmetric real matrix $A$ is diagonalizable and can be written as $U\Lambda U^T$, where $\Lambda=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$ is a matrix with $A$’s eigenvalues along the main diagonal and zeros elsewhere, and $U$ is an orthogonal matrix whose columns are all eigenvectors of $A$. $\Lambda$ can be expanded into $\sum_{i=1}^n\lambda_ie_i^Te_i$, where $e_i$ is the $i$th standard basis vector, so $A=\sum_{i=1}^n\lambda_iUe_ie_i^TU^T=\sum_{i=1}^n\lambda_i(Ue_i)(Ue_i)^T$. Now, $Ue_i$ is the $i$th column of $U$, which is just the eigenvector $u_i$, so we get $$A=\sum_{i=1}^n\lambda_iu_iu_i^T.$$ This decomposition is a weighted sum of projections onto orthonormal eigenvectors of $\mathbb R^n$. In fact, any diagonalizable matrix has a similar decomposition into a weighted sum of projections, although those projections might not be orthogonal in general.
A: If we take: 
$$a=\pmatrix{1\\0\\0}, b=\pmatrix{0\\1\\0}, c=\pmatrix{0\\0\\1}$$
your formula will naturally give:
$$M=\pmatrix{j_1&j_4&j_6\\j_4&j_2&j_5\\j_6&j_5&j_3}$$
which is indeed the general form of a symmetric matrix, without needing an assumption on eigenvalues.
