Decomposing a symmetric matrix into a sum of products

Question

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 keyword for google-oriented searches in this domain is "rank one matrices". — Jean Marie, Dec 11 '16 at 07:46

score 7 · Answer 1 · edited Oct 12 '21 at 17:55

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.

...and a further generalization is afforded by SVD. – J. M. ain't a mathematician Dec 11 '16 at 09:56 — J. M. ain't a mathematician, Dec 11 '16 at 09:56

score 1 · Answer 2 · answered Dec 11 '16 at 08:58

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.

Decomposing a symmetric matrix into a sum of products

2 Answers2