Is matrix with entries $a_{ij} = x_ix_j$ positive semidefinite?

Question

I wonder if I can proof that the square matrix with entries $a_{ij} = x_ix_j$ for a given vector $x$ is positive semidefinite.

I hope it is because this matrix is somehow related to this question where I asked if mean square error is convex function in linear regression. I actually calculated the Hessian matrix and obtained the one I give you (multiplied by 2). The problem is that I don't know how to proof that it is positive semidefinite so that I can show that mean square error is convex. Please give a proof or a counterexample.

What do you think the sign of $\sum\limits_{i,j}a_{ij}u_iu_j$ could be, for every $(u_i)$? — Did, Oct 20 '16 at 17:22

score 4 · Accepted Answer · answered Oct 20 '16 at 17:25

4

You can easily check that $A=xx^T$. Then, for any $y$, $$ y^TAy=y^Txx^Ty=(x^Ty)^Tx^Ty\geq0. $$

answered Oct 20 '16 at 17:25

Martin Argerami

205,756

Or prove that the eigenvalues are $|x|^2$ (Euclidean norm) and $0$. ;-) – egreg Oct 20 '16 at 17:37
Yes, after first proving that $A$ is symmetric :) – Martin Argerami Oct 20 '16 at 17:40

score 2 · Answer 2 · answered Oct 20 '16 at 17:25

2

If $x=(x_1,x_2,\dots,x_n)$ is the row vector, then $A=x^Tx$. So give a column vector $y$, you have that $y^TAy=(y^Tx^T)(xy)=(xy)^T(xy)$. What is that never negative?

answered Oct 20 '16 at 17:25

Thomas Andrews

177,126

Is matrix with entries $a_{ij} = x_ix_j$ positive semidefinite?

2 Answers2