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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?

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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 $1$.

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

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