A way of Matrix decomposition

Question

if $M$ is $3\times 3$ sysmmetric matrix, $\det(M)=0$, it seems that we can decompose $M$ by this way:$M^{ij}=a^ib^j+b^ia^j$. Is it correct? How to prove it?

Answer 1

I assume that the entries of $M$ are real numbers.

$M$ can be written in the form $M^{ij} = a^i b^j + b^j a^j$ if and only if for the column-vectors $a = (a^1,\dots,a^n)$ and $b = (b^1,\dots,b^n)$, we have $$ M = ab^T + ba^T = \pmatrix{a &b}\pmatrix{0&1\\1&0}\pmatrix{a &b}^T. $$ As a consequence of Sylvester's law of inertia, there exist vectors $a,b$ such that this holds if and only if $M$ is symmetric and one of the following conditions hold:

1. $M$ has rank $2$ with one positive eigenvalue and one negative eigenvalue,
2. $M$ has rank $1$,
3. $M = 0$.