This proposition is stated in my linear algebra notes:
(Let $V$ be a inner product space) for any $x,y,z \in V$, if $〈x,y〉 = 〈x,z〉$ for any $x \in V$, then $y = z$. (#)
Now I try to apply this proposition to the following:
Let $T$ be a linear operator on a (finite-dimensional) inner product space $V$ over the field of real numbers.
Denote $T^*$ to be the adjoint of $T$, we know by definition that for any $x, y \in V$, $〈T(x),y〉 = 〈x,T^*(y)〉$.
Since we are working on a real inner product space, we can switch the order of the inner products directly without adding a conjugation.
Now for any $x \in V$, $〈T(x),x〉 = 〈x,T^*(x)〉= 〈T^*(x),x〉$.
This implies $〈x,T(x)〉=〈x,T^*(x)〉$ for any $x \in V$.
Then using (#), I can say $T(x) = T^*(x)$ for any $x \in V$.
This further implies $T = T^*$, or $T$ is self-adjoint.
which is strange. Does this mean every linear operator on a real inner product space is self-adjoint? Can anyone tell me where is the flaw in my "proof"? Thanks.
I do not think you have met the requirements of proposition (#).
(#) does provide a test of whether or not $T(x) = T^*(x)$ is true, for a fixed $x$.
You would further need to show $〈a,T(x)〉=〈a,T^*(x)〉$ for all $a$ in the space.