Prove that if $\langle Tx,x\rangle =0$ for all $x \in X$, then $T = 0$

Solution 1:

Hint: The fact that $X$ is a complex inner product space is significant; this fails to hold for real inner product spaces.

You've done well to consider $\langle T(u + av), (u + av) \rangle$. Now, note the following: since your equality is true for all $a$, it is true for any single value of $a$. In particular, we have

$\langle Tu,v \rangle + \langle Tv,u \rangle = 0$ (set $a = 1$)
$\langle Tu,v \rangle - \langle Tv,u \rangle = 0$ (set $a = i$)

Isomorphism between quotient modules [closed]

Bijecting a countably infinite set $S$ and its cartesian product $S \times S$

Non-integer powers of negative numbers

Checking my understanding: $1 - 1 + 1 - 1 + 1 - ... = \frac{1}{2}$

If Monty Hall doesn't know where the prize is, should the contestant still switch doors, after Monty opens one door and unveiss a goat?

Is the $n$-th prime smaller than $n(\log n + \log\log n-1+\frac{\log\log n}{\log n})$?

Dirichlet's theorem on primes in arithmetic progression

What is the greatest integer that divides $p^4-1$ for every prime number $p$ greater than $5$?

A finite group which has a unique subgroup of order $d$ for each $d\mid n$.

Number of solutions of $x^2_1+\dots+x^2_n=0,$ $x_i\in \Bbb{F}_q.$

Axiom Systems and Formal Systems

Maximum Likelihood Estimator for $\theta$ when $X_1,\dots, X_n \sim U(-\theta,\theta)$

Prove that if $\langle Tx,x\rangle =0$ for all $x \in X$, then $T = 0$

Solution 1:

Related

Recent Posts