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Let $T$ be a linear operator on an inner product space $V$. If $\langle T(x),y \rangle=0$ for all $x,y \in V$ then $T=T_0$ where it means zero transformation.
Prove this result if the equality holds for all $x,y$ in some basis for $V$.

The hint says use the theorem below.
[let $V$ and $W$ be vector spaces over $F$ and suppose that {$v_1,\dots,v_n$} is a bsis for $V$. For $w_1,\dots,w_n$ in $W$, there exists exactly one linear transformation $T:V \rightarrow W$ such that $T(v_i)=w_i$.]
But I don't know how to apply this.
Actually the true meaning of this question is still ambiguous to me.
It means that $x,y \in \beta$ for some basis $\beta$? And $\langle T(x),y \rangle=0$?

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  • $\begingroup$ How is it defined $T_0$? $\endgroup$ – Riccardo Jun 3 '13 at 16:13
  • $\begingroup$ I think the second sentence is supposed to be "If $\langle Tx,y\rangle = 0$ for all $x,y \in V$, then $T = 0$." $\endgroup$ – kahen Jun 3 '13 at 16:15
  • $\begingroup$ Yes, $T_0$ means zero transformation. $\endgroup$ – noname Jun 3 '13 at 16:15
  • $\begingroup$ And obviously no hypothesis on $\langle \cdot \rangle$? $\endgroup$ – Riccardo Jun 3 '13 at 16:21
  • $\begingroup$ @RicPed Yes, it's a normal inner product. Sine $\langle T(x),y \rangle=0$ for all $y \in V$, it means that $T(x)=0$ for all $x \in V$ so $T=T_0$. $\endgroup$ – noname Jun 3 '13 at 16:23
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Presumably what the question means is that if $v_k$ is a basis for $V$ and $w_k$ is a basis for $W$, then $\langle Tv_i, w_j \rangle = 0$ for all $i,j$.

Let $v_k$ be a basis for $V$ and $w_k$ be a basis for $W$. Then if $x \in V$ we can write $x = \sum_i \alpha_i v_i$ and similarly for any $y \in W$, we can write $y = \sum_j \beta_j w_j$. Then $\langle Tx, y \rangle = \sum_i \sum_j \alpha_i \beta_j \langle Tv_i, w_j \rangle = 0$. Hence $\langle Tx, y \rangle = 0$ for all $x,y$.

If $\langle Tx, y \rangle = 0$ for all $x,y$, then for each $x$, choose $y=Tx$, then you have $\langle Tx, Tx \rangle = \|Tx\|^2 = 0$. Hence $Tx = 0$ for all $x$, from which it follows that $T=0$.

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  • $\begingroup$ Thanks, but I know how to prove $T=0$. I want to know if that equality holds for all $x,y$ in some basis of $V$. $\endgroup$ – noname Jun 3 '13 at 16:25
  • $\begingroup$ I added a paragraph showing that if it holds for elements of the basis of $V$ and $W$, then it is true generally. $\endgroup$ – copper.hat Jun 3 '13 at 16:27
  • $\begingroup$ Just one more question. Why that theorem implys that $\langle T(v_i),w_j \rangle=0$? $\endgroup$ – noname Jun 3 '13 at 16:42
  • $\begingroup$ I presume the question was if the equality holds for $x$ in some basis of $V$ and $y$ in some basis of $W$, then show the result is true generally. So it is an assumption, because it is not true generally, of course (Just take $T$ to be the identity operator for example). $\endgroup$ – copper.hat Jun 3 '13 at 16:45
  • $\begingroup$ Oh, so I wonder why $\langle T(x),y \rangle = \sum_i\alpha_i\sum_j\beta_j \langle T(v_i),w_j \rangle=0$ means $\langle T(v_i),w_j \rangle=0$. Is that what you mean, right? $\endgroup$ – noname Jun 3 '13 at 16:59

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