# Proof negative-definiteness of a nonsymmetric and rank-deficient matrix

Consider the vectors $$\mathbf{a} \in \mathbb{R}^N$$ and $$\mathbf{b} \in \mathbb{R}^N$$ with $$N>1$$ and $$\mathbf{a} \neq \mathbf{b}$$. The product $$\mathbf{C}=\mathbf{a} \mathbf{b}^T \in \mathbb{R}^{N \times N}$$ will yield a square, non-symmetric matrix $$\mathbf{C}$$ with rank=1.

How do I formulate an optimization problem on $$\mathbf{a}$$, given a specific $$\mathbf{b}$$, such that $$\mathbf{C} = \mathbf{a} \mathbf{b}^T$$ is negative definite? How do I specify this constraint?

$$\min_{\mathbf{a}} \text{a user-specified objective} \\ \text{subject to} \ldots ?$$

I know, a square, nonsymmetric matrix $$\mathbf{D}$$ is negative definite if and only if its symmetric part $$(\mathbf{D} + \mathbf{D}^T)/2$$ is negative definite. However, in my case, the symmetric part $$(\mathbf{C} + \mathbf{C}^T)/2$$ will have a maximum rank of 2, and it may have zero eigenvalues... See also this wiki-link.

• As you noted, if $N > 2$ then your matrix $C$ is never negative definite (e.g., if $\mathbf{v}$ is orthogonal to $\mathbf{a}$ and $\mathbf{b}$ then $\mathbf{v}^TC\mathbf{v} = 0$). Is it not enough for your $C$ to be negative semidefinite? If semidefiniteness is OK, then this type of problem can be solved via semidefinite programming solvers (SeDuMi etc, or frameworks like CVX for MATLAB). Feb 3 at 15:31
• ok, thanks, I thought so. But what do I need to do for $N=2$? Feb 4 at 21:30
• For $N = 2$, you can basically type the problem as-is into CVX in MATLAB. Just tell it what your objective function is and then specify the constraint $\mathbf{a}\mathbf{b}^T + \mathbf{b}\mathbf{a}^T \leq 0$, where $\leq 0$ here means negative semidefinite. Feb 5 at 0:36
• thanks, didn't know about CVX yet! Seems powerful Feb 5 at 9:44