# When is $x\otimes y^\ast$ positive definite?

As an exercise, I was told to determine when the linear map $x\otimes y^\ast:v\mapsto \left\langle v,y \right\rangle x$ is self adjoint, positive definite, and positive semi-definite, when the inner product space is over $\mathbb C$. I found that $(x\otimes y^\ast)^\ast=y\otimes x^\ast$ and from there that $x\otimes y^\ast$ is self adjoint iff $x$ is a real multiple of $y$ (or they're both zero).

For positive definite, I'm not sure about the definition... Should the general definition asks that $\left\langle u ,Au \right\rangle >0$? Is this what I'm meant to check? Because I know positive definiteness is defined for bilinear forms and I'm not sure what the bilinear form corresponding to an operator $A$ should be...

If this is indeed the case, is the following correct?

In the complex case, positive (semi)definite implies self adjoint so we can sub in $x=\alpha y$ for $\alpha\in \mathbb R$. Then I find $x\otimes y^\ast$ is positive semidefinite whenever $\alpha \geq 0$, and positive definite if $\alpha >0$.

• Yes, the bilinear form corresponding to $A$ is $\langle u,v \rangle_A = \langle u, Av \rangle$ Sep 21, 2016 at 13:34

Yes, your reasoning and your answer are correct: the map is positive semidefinite when $\alpha \geq 0$, and positive definite when $\alpha > 0$.
Everything in your answer is correct, with one exception. The map $x\otimes x^*$ is never positive definite. Because, for any $v$ with $\langle v,x\rangle=0$ (which always exists as long as your space has at least dimension two), $$(x\otimes x^*)v=\langle v,x\rangle x=0.$$ In the particular case of finite dimension, positive definite implies invertible. The operator $x\otimes x^*$, on the other hand, has rank one so it is never invertible unless your space is one-dimensional.