Let $T$ be a linear operator on an inner product vector space $V$. I'd like to prove that if $$(Tx|x)=0 \quad \forall x \in V$$ then $T$ is the null operator.
I can't figure out this proof using contradiction, do I have to choose an appropriate vector?
Asked
Active
Viewed 1,168 times
2
Jendrik Stelzner
- 17,081
omega-stable
- 1,245
- 12
- 21
-
@Crostul That would be a linear functional. I mean a linear map from $V$ to $V$ – omega-stable Nov 28 '16 at 01:37
-
8The statement is false: $T:\Bbb{R}^2 \to \Bbb{R}^2$ defined by $T(x,y)=(-y,x)$ is such an example. – Crostul Nov 28 '16 at 01:40
-
Its only false under the standard inner product, or am I wrong? Anyway, I get your point, Thanks! – omega-stable Nov 28 '16 at 01:41
-
@I.Padilla This is actually false for "any" inner product (say in finite dimension or a Hilbert space, as long as $\dim V\geq 2$). – Luiz Cordeiro Nov 28 '16 at 02:57
1 Answers
5
You can prove this for a $\Bbb C$omplex inner-product space $V$.
But if $\,V\,$ is an inner-product space over $\mathbb R$, then the implication doesn't hold.
Crostul gives in a comment the concrete example
$$T:\Bbb{R}^2\to\Bbb R^2, (x_1,x_2)\mapsto(-x_2,x_1)$$
of a non-zero operator satisfying the initial condition. $T$ rotates by $+90°$, hence $\,x\perp Tx\,$ for all $\,x\in\Bbb R^2$.
Let $\,V\,$ be an inner-product space over $\mathbb C$.
Every sesquilinear form on $V$ satisfies the polarisation identity. For the sesquilinear form $(T(\,\cdot\,)|\,\cdot\,)$ it reads
$$(Tx|y)\:=\:\frac14\sum_{k=0}^3\:i^k\left(\,T(x+i^ky)\,\big|\, x+i^ky\,\right).$$
Thus by assumption, $(Tx|y)=0\,$ for all $\,x,y\in V$, hence $T$ must be zero.
Hanno
- 6,776