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I'm trying to prove the following:

Let $V$ be a vector space over $\mathbb{C}$ and let $T$ be a linear operator on $V$. If $\langle T v,v\rangle=\langle v,v\rangle$ for all $v\in V$ satisfying $\langle v,v\rangle=1$, then $T$ is the identity operator.

Can anyone point me to a proof of a similar result or give one here? Thanks!

Batominovski
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J-J
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2 Answers2

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$$\langle Tv,v\rangle = \langle v,v\rangle =1$$ for any $v$ such that $\langle v,v\rangle=1$, so for any $x\in V$ $$\langle (T-I)x,x\rangle=\|x\|^2\langle (T-I)\frac{x}{\|x\|},\frac{x}{\|x\|}\rangle = \|x\|^2\left( \langle Tv,v\rangle - \langle v,v\rangle\right)=0$$ where $v=\frac{x}{\|x\|}$ so that $\langle v,v\rangle=1$

why does it follow that $T-I=0$? But if that's true, then $T=I$.

To see that $T-I=0$ see Relation between $T=0$ and $(Tx,x)=0$

Squirtle
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    This is perfect. I know the theorem which implies $T-I=0$, your initial argument suffices. – J-J Oct 17 '18 at 16:28
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    Thank you. glad to help! – Squirtle Oct 17 '18 at 16:31
  • I don't follow the last part: the expansion of $\langle Ax, x \rangle$ should include terms of the form $\overline{b_k} b_j \langle A e_k, e_j \rangle$. What I'd say would be: an arbitrary nonzero $v$ is equal to $\lambda v'$ for $\lambda = \lVert v \rVert$ and $v' = v / \lVert v \rVert$ where $\lVert v' \rVert = 1$, so $\langle Tv, v \rangle = |\lambda|^2 \langle Tv', v' \rangle = |\lambda|^2 \langle v', v' \rangle = \langle v, v \rangle$; and also for $v=0$, $\langle Tv, v \rangle = \langle v, v \rangle = 0$ is trivial. – Daniel Schepler Oct 17 '18 at 18:34
  • @DanielSchepler thank you for catching these issues! – Squirtle Oct 18 '18 at 02:27
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Lemma: Let $V$ be a vector space (of possibly an infinite dimension) over $\mathbb{C}$ with a nondegenerate sesquilinear (not necessarily Hermitian) form $\langle\_,\_\rangle$. If $F:V\to V$ is a linear transformation such that $\langle Fv,v\rangle =0$ for every $v\in V$, then $F$ is the zero map.

Proof: Fix a basis $B$ of $V$. Now, for each $a\in B$, we get $\langle Fa,a\rangle=0$. For two distinct $a,b\in B$, we have $$|p|^2\langle Fa,a\rangle +p\bar{q}\langle Fa,b\rangle +\bar{p}q\langle Fb,a\rangle +|q|^2\langle Fb,b\rangle =\big\langle F(pa+qb),pa+qb\big\rangle =0$$ for all $p,q\in\mathbb{C}$. Since $\langle Fa,a\rangle=\langle Fb,b\rangle =0$, we see that $$p\bar{q}\langle Fa,b\rangle +\bar{p}q\langle Fb,a\rangle=0.$$ Taking $p=q=1$, we have $\langle Fa,b\rangle +\langle Fb,a\rangle=0$. Taking $p=1$ and $q=i$, we have $-i\langle Fa,b\rangle +i\langle Fb,a\rangle =0$. That is, $\langle Fa,b\rangle=0$ for all $a,b\in B$.

Fix $a\in B$, and vary $b\in B$. We get that $\langle Fa,b\rangle=0$ for all $b\in B$ implies that $\langle Fa,v\rangle =0$ for every $v\in V$. As $\langle\_,\_\rangle$ is nondegenerate, $Fa=0$ for each $a\in B$. This proves that $F$ is identically $0$.

Corollary: Let $V$ be a vector space (of possibly an infinite dimension) over $\mathbb{C}$ with a positive-definite Hermitian form $\langle\_,\_\rangle$. If $F:V\to V$ is a linear transformation such that $\langle Fv,v\rangle =0$ for every $v\in V$ such that $\langle v,v\rangle=1$, then $F$ is the zero map.

Proof: Let $S$ denote the set of $v\in V$ such that $\langle v,v\rangle =1$. If $\langle Tv,v\rangle =0$ for all $v\in S$, then $\langle Fv,v\rangle=0$ for all $v\in V$ because for every $v\in V$ with $v\neq 0$, $\lambda v\in S$ for some $\lambda >0$. By the lemma, $F=0$.

Remark: If $V$ is a complex Hilbert space, the corollary also holds, with similar proof.

From the problem statement, we get $\langle (T-\text{id})v,v\rangle=0$ for all $v\in V$. From the lemma, we get $T-\text{id}=0$, or $T=\text{id}$. So, this is a proof for an arbitrary dimension, for arbitrary nondegenerate sesquilinear form which may or may not be an inner product, where the usual Gram-Schmidt won't work. I forgot the condition $\langle v,v\rangle =1$ here.

But anyhow, from the problem statement, we get $\langle (T-\text{id})v,v\rangle=0$ for all $v\in V$ with $\langle v,v\rangle=1$. From the corollary, we get $T-\text{id}=0$, or $T=\text{id}$.

The problem statement must require that $\langle\_,\_\rangle$ is a positive-definite Hermitian form (which the OP unfortunately didn't specify). If the form is not necessarily positive-definite, consider a negative-definite Hermitian form $\langle\_,\_\rangle$. Then, $\langle v,v\rangle=1$ is never satisfied. And if the form is not required to be Hermitian, you can construct a sesquilinear form for which $\langle v,v\rangle$ is never a real number except when $v=0$. In these cases, any $T$ will work.

Interestingly, the lemma does not hold if $\langle\_,\_\rangle$ is a nondegenerate (not necessarily symmetric) bilinear form. An example is given by $V=\mathbb{C}^2=\mathbb{C}e_1\oplus\mathbb{C}e_2$ with the bilinear form $$\langle \alpha e_1+\beta e_2,\gamma e_1+\delta e_2\rangle =\alpha\beta+\gamma\delta.$$ Take $F:V\to V$ to be $$F(\alpha e_1+\beta e_2)=\beta e_1-\alpha e_2.$$ Then, $F\neq 0$, but $\langle Fv,v\rangle=0$ for all $v\in V$. So, with $T=\text{id}+F$, the OP's question also fails for bilinear forms.

  • I don't see how you've satisfied the conditions of the hypothesis in your final setting. Could you please clarify this further? It seems you are on to something (though it's not $this$ exact question). – Squirtle Oct 17 '18 at 16:30