# Does element-wise exponentiation of a matrix preserve its positive semi-definiteness?

Given a real symmetric positive semi-definite matrix $\mathbf{A}$, is $\mathbf{A}^{\cdot k},k\in\mathbb{N}$ also positive semi-definite?

Matrix $\mathbf{A}\in\mathbb{R}^{n\times n}$ is positive semi-definite if $\mathbf{x}^T\mathbf{A}\mathbf{x}\geq0,\forall\mathbf{x}\in\mathbb{R}^n$.

The elements of $\mathbf{A}^{\cdot k}$ are $\{\mathbf{A}^{\cdot k}\}_{ij}=A_{ij}^k,1\leq i,j \leq n$.

• Oop, misread the exponentiation used :) – rschwieb Feb 7 '14 at 15:23
• Yeah, I know that it's true for normal exponentiation. – sk1ll3r Feb 7 '14 at 15:26

We prove that if $A$ and $B$ are positive semi-definite, then $A\circ B$ (the component-wise multiplication) is also positive semi-definite.
Note that the eigenvalues of then Kronecker product $A\otimes B$ are product of eigenvalues of $A$ and $B$, so $A\otimes B$ is positive semi-definite. Now note that $A\circ B$ is a principle sub-matrix of $A\otimes B$.