This is a linear algebra question that came up in my research, and I feel like there ought to be either a simple proof or a simple counterexample, but I have been unable to find either.

Assume $V$ and $W$ are real finite-dimensional inner product spaces, so there is an induced inner product on $\text{Hom}(V,W)$ defined via its canonical identification with $V^* \otimes W$. Suppose $\Phi : \text{Hom}(V,W) \to \text{Hom}(V,W)$ is a linear map with the property that

$\langle A , \Phi(A) \rangle > 0$

for every element $A \in \text{Hom}(V,W)$ with rank 1. Equivalently, using the identification $\text{Hom}(V,W) = V^* \otimes W$, this condition means

$\langle w , \Phi(v_\flat \otimes w) v \rangle > 0$

for all $v \in V$ and $w \in W$, where $v_\flat := \langle v,\cdot \rangle$.

Does it follow that $\Phi$ is invertible?

My instinct says no: for instance, if we pick bases of $V$ and $W$ and express $\Phi$ in the resulting basis of $\text{Hom}(V,W)$, then the condition says that all diagonal entries of the matrix for $\Phi$ are positive, bus that's certainly not enough to conclude that $\Phi$ is invertible (at least not without some information about the magnitude of the diagonal entries relative to the non-diagonal entries, which I don't have, as far as I know). But the hypothesis seems to say more than this since it is not tied to any specific bases of $V$ and $W$, e.g. the diagonal entries will be positive for any basis of $\text{Hom}(V,W)$ consisting of rank 1 elements. My intuition and algebraic knowledge are insufficient to say what that means, but it sounds like it means something.

Or is there a simple counterexample, e.g. where $V$ and $W$ are both 2-dimensional?


It does not follow that $\Phi$ is invertible.

Consider $V=W=\mathbb{R}^2$ together with the standard inner product. With respect to the standard basis we identify $\operatorname{Hom}(V,W)$ with the space of $2\times 2$ real matrices. We define the map $\Phi$ by \begin{align*} \operatorname{Mat}_{2,2}(\mathbb{R})&\to\operatorname{Mat}_{2,2}(\mathbb{R})\\ \left(\begin{matrix}a&b\\c&d\end{matrix}\right)&\mapsto\left(\begin{matrix}a-b&b-c\\c+d&d+a\end{matrix}\right). \end{align*}

Notice that $\Phi$ is not bijective since $\Phi(N)=0$ for \begin{align*} N=\left(\begin{matrix}1&1\\1&-1\end{matrix}\right). \end{align*}

On the other hand, for every $A\in\operatorname{Mat}_{2,2}(\mathbb{R})$ with entries $a,b,c,d$ we have \begin{align*} \langle A,\Phi(A)\rangle&=\operatorname{tr}\left(A\Phi(A)^T\right)=\operatorname{tr}\left(\left(\begin{matrix}a&b\\c&d\end{matrix}\right)\left(\begin{matrix}a-b&c+d\\b-c&d+a\end{matrix}\right)\right)\\ &=a^2-ab+b^2-bc+c^2+cd+d^2+da\\ &=\frac12\left((a-b)^2+(b-c)^2+(c+d)^2+(d+a)^2\right). \end{align*} This expression is positive unless $A$ is a scalar multiple of $N$. Note that $\operatorname{rk}(N)=2$. We can conclude that $\langle A,\Phi(A)\rangle$ positive whenever the matrix $A$ has rank $1$.

| cite | improve this answer | |
  • $\begingroup$ Thank you, this is exactly the kind of answer I was hoping for! $\endgroup$ – Chris Wendl Aug 3 '18 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.