Quick question:
Let $V$ be a finite dimensional inner product space and $T$,$U$ linear operators.
If $\left<Ta,b\right> = \left<Ua,b\right> \;\forall a,b \in V$ then $T = U$?
I am not sure if this reasoning is correct:
If that statement is true for all $a,b \in V$ then in particular for $b = 0$ we have:
$$\left<(T-U)a,0\right> = 0 \quad\forall a \in V \Rightarrow (T-U)a = 0 \; \forall a \in V$$
and then $U = T$
Thanks in advance!
On
It's certainly true that
$\langle Ta, b \rangle = \langle Ua, b \rangle \Longrightarrow \langle (T - U)a, b \rangle = 0 \tag 1$
for all $a, b \in V$, so the line of reasoning is correct up to this point. It is, however, false that
$\langle (T - U)a, 0 \rangle \Longrightarrow (T - U)a = 0; \tag 2$
if fact,
$\langle (T - U)a, 0 \rangle = 0, \tag 3$
always, no matter what $(T - U)a$ might be. This follows from the standard linearity properties of and inner product $\langle \cdot, \cdot \rangle$; that is,
$\langle x, y + z \rangle = \langle x, y \rangle + \langle x, z \rangle; \tag 4$
whence
$\langle x, 0 \rangle = \langle x, 0 + 0 \rangle = \langle x, 0 \rangle + \langle x, 0 \rangle \Longrightarrow \langle x, 0 \rangle = 0. \tag 5$
Since
$\langle (T - U)a, b \rangle = 0 \tag 6$
holds for all $a, b \in V$ by hypothesis, we are free to set
$b = (T - U)a \tag 7$
in (6) and then we obtain
$\Vert (T - U)a \Vert^2 = \langle (T - U)a, (T - U)a \rangle = 0, \tag 8$
from which we conclude
$(T - U)a = 0 \tag 8$
for all $a \in V$; Thus
$Ta = Ua. \tag 9$
In arguing (6)-(9) I have used the proposition that
$\langle x, x \rangle = 0 \Longleftrightarrow x = 0, \tag{10}$
which is generally taken as axiomatic for inner products $\langle \cdot, \cdot \rangle$.
No, $\langle (T - U) a, 0 \rangle = 0$ holds for any $T$ and $U$ whatsoever since we always have $\langle x, 0 \rangle = 0$. So, you cannot conclude anything from this.
To get a correct proof, you would need to apply the hypothesis with $b := (T - U) a$.