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I know that this has a proof at Relation between $T=0$ and $(Tx,x)=0$ (and some other questions) but my question is specifically about how to prove this by contradiction rather than directly. (I thought it would be easier this way, but I'm really struggling to come up with a proof).

The exact statement I'm trying to prove is this: let $V$ be a vector space, and $T : V \to V$ a linear transformation. Assume that thist vector space is over the field $\mathbb{C}$, then prove that if $\forall x\Big(\langle x, T(x)\rangle = 0 \Big)$, then $T = 0$.

I started like this

  • Assume that the statement is false, then it must be that $\langle x, T(x) \rangle = 0$ and $T \ne T_0$.
  • Therefore, there exists $a, b \in V$ such that $b = T(a)$ and $a \ne 0$ and $b \ne 0$. By assumption we have that $\langle a, b \rangle = 0$. From here I think I should try to show that $b=0$, and the fact that this is over a complex field must be important (as the statement is false if we are working over a real field). Unfortunately I got really stuck here. I thought about trying to express $a$ in terms of something like $x + iy$ (for some vectors $x, y$ both with real coordinates) and then to write $b = T(x + iy)$, so that we have

$$ \begin{align} \langle x + iy, T(x + iy) \rangle &= \langle x, T(x) \rangle + \overline{i} \langle x, T(y) \rangle + i \langle y, T(x) \rangle + i \overline{i} \langle y, T(y) \rangle \\ &= 0 + \overline{i} \langle x, T(y) \rangle + i \langle y, T(x) \rangle + 0 \\ &= i\left(- \langle x, T(y) \rangle + \langle y, T(x) \rangle \right) \end{align} $$

  • And this must be equal to zero, therefore $- \langle x, T(y) \rangle + \langle y, T(x) \rangle = 0$, that is $\langle x, T(y) \rangle = \langle y, T(x) \rangle$ but I have no idea what to do with this.

Any help would be much appreciated (sorry for the mess).

peritremic
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  • The title is missing $=0$? – Ted Shifrin Nov 18 '22 at 18:19
  • What does $(a,b)\in T$ mean? Do you mean, vectors $a$ and $b$ such that $T(a)=b$? Not the most felicitous notation, given that ordered pairs often show up in linear algebra playing other roles. – Arturo Magidin Nov 18 '22 at 18:23
  • Yes, because functions are sets of ordered pairs. Sorry, it's not good notation. – peritremic Nov 18 '22 at 18:23
  • Then perhaps just write "there exist nonzero vectors $a$ and $b$ with $T(a)=b$ and $\langle a,b\rangle = 0$"... – Arturo Magidin Nov 18 '22 at 18:24
  • You should specify that $x$ and $y$ are vectors with real coordinates, as that makes them uniquely determined by $a$ and $b$. – Arturo Magidin Nov 18 '22 at 18:24
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    You aren't really doing a proof by contradiction. You are doing a "fake proof by contradiction". You are assuming $b\neq 0$; but you are trying to do a direct computation from which you will deduce that $b=0$, without every using the fact that $b\neq 0$. This is just a direct proof with the line "assume $b\neq 0$" added at the top, and the line "which contradicts our assumption that $b\neq 0$" at the bottom. It isn't really a proof by contradiction, even though it is presented as such. – Arturo Magidin Nov 18 '22 at 18:27
  • Thanks Ted Shifrin and Arturo Magidin, I have updated the question. – peritremic Nov 18 '22 at 18:28

1 Answers1

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Your idea is good, but contradiction is not necessary.

I usually have inner products semilinear in the first variable and linear in the second one, fix the argument for your convention, if needed.

First we “polarize”: since $\langle x+y,T(x+y)\rangle=0$ for every $x,y\in V$, we can expand and obtain $$ \langle x,T(y)\rangle+\langle y,T(x)\rangle=0 $$ (write down the steps). This holds for every $x$ and every $y$, so it remains valid when we replace $x$ by $ix$ and we obtain $$ -i\langle x,T(y)\rangle+i\langle y,T(x)\rangle=0 $$ Removing the factor $i$ and summing the two relations, we obtain $$ 2\langle y,T(x)\rangle=0 $$ for every $x,y\in V$. Now you should be able to finish.

egreg
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  • Thanks! We just set $y = T(x)$, from which it follows that $T(x) = 0$ right? If you don't mind me asking how did you think to first consider $x + y$ and then relate this to $v$, rather than starting with $v$ and then trying to manipulate the specific $v$ - is it just a standard technique? – peritremic Nov 18 '22 at 18:44
  • @peritremic Yes, it's a standard technique when you have a “quadratic form” like $x\mapsto \langle x,T(x)$. Well this is not really a quadratic form, but it works… – egreg Nov 18 '22 at 18:46