4
$\begingroup$

Let $X$ be unitary space. Show that surjective operator $T: X\to X$ such that $\langle Tx, Ty\rangle = \langle x, y \rangle \forall x, y \in X$ is linear.


The only thing I was able to conclude (trivial stuff) is that $$ \langle Tx, Ty\rangle = \langle x, y \rangle \implies \langle Tx, Tx\rangle = \langle x, x \rangle \implies \sqrt{\langle Tx, Tx\rangle} = \sqrt{\langle x, x \rangle} \\ \implies \| Tx\| = \|x\| $$ so $T$ is isometry.

To show that it is linear I want to have: $$ T(ax) = aT(x) \\ T(x+y) = T(x)+T(y) $$

$\endgroup$

3 Answers 3

3
$\begingroup$

Here is an interpretation of the answers so far given, breaking down the argument, and highlighting an important intermediate result.

First fact. Let $T$ be a surjective isometry. Then $T$ admits an adjoint operator. In other words, for every $y$ in $X$, there exists a (necessarily unique) $z$ such that $$ \langle T(x), y\rangle = \langle x, z\rangle , \quad\forall x\in X. $$

Proof (@Udalricus.S) Given $y$, find $z$ such that $T(z)=y$. Then $$ \langle T(x), y\rangle = \langle T(x), T(z)\rangle = \langle x, z\rangle . \tag*{$\square$} $$

Incidentally, the vector $z$ referred to above is often denoted by $T^*(y)$, so the highlited expression in the statement becomes the familiar one: $$ \langle T(x), y\rangle = \langle x, T^*(y)\rangle . $$

Second fact. Let $T:X\to X$ be a function (not necessarily linear, unitary or surjective) and suppose that there exists another function $T^*:X\to X$, such that $$ \langle T(x), y\rangle = \langle x, T^*(y)\rangle \quad\forall x, y\in X. $$ Then $T$ is linear.

Proof. (@Chrystomath) Given $x_1$ and $x_2$ in $X$, we have for all $y$ that $$ \langle T(x_1+x_2), y\rangle = \langle x_1+x_2, T^*(y)\rangle = $$$$ = \langle x_1, T^*(y)\rangle + \langle x_2, T^*(y)\rangle = \langle T(x_1), y\rangle + \langle T(x_2), y\rangle = $$$$ = \langle T(x_1) + T(x_2), y\rangle . $$ Since $y$ is arbitrary, we deduce that $T(x_1+x_2)=T(x_1) + T(x_2)$. Similarly one proves that $T(\lambda x)=\lambda T(x)$. QED

As an added bonus, should $X$ be a Hilbert space, one can show by means of the closed graph Theorem that the map $T$ of the second fact above is moreover bounded!

$\endgroup$
2
$\begingroup$

\begin{align} \langle T(x+y),T(z)\rangle&=\langle x+y,z\rangle\\ &=\langle x,z\rangle +\langle y,z\rangle\\ &=\langle T(x),T(z)\rangle+\langle T(y),T(z)\rangle\\ &=\langle T(x)+T(y),T(z)\rangle \end{align} Since $T$ is onto, $T(z)$ can be any vector, hence $T(x+y)=T(x)+T(y)$.
Same proof can be modified for $T(\alpha x)=\alpha T(x)$.

$\endgroup$
1
  • $\begingroup$ Did you mean $\langle T(x),T(z)\rangle + \langle T(y),T(z)\rangle$ in the end? $\endgroup$
    – blahblah
    Feb 3, 2021 at 13:13
2
$\begingroup$

For example, use the following: Let $x,y\in X$. By surjectivity, there is a $z\in X$ such that $y=Tz$. Then

$$\langle T(\alpha x),y\rangle = \langle T(\alpha x), Tz\rangle =\langle \alpha x, z\rangle=\alpha\langle x, z\rangle = \alpha\langle Tx, y\rangle$$

Then we have for all $y$

$$\langle (T(\alpha x)-\alpha Tx),y\rangle=0$$

By non-degeneracy, it follows that $T(\alpha x)-\alpha Tx=0$. I think the second property can be shown in a similar way.

$\endgroup$
1
  • 1
    $\begingroup$ Indeed, $\langle T(x+y), z\rangle=\langle T(x+y), Tw\rangle= \langle x+y,w\rangle=\langle x,w\rangle+\langle y,w\rangle$ and now you reverse everything and you're done. Here, again, $w\in T^{-1} (\{z\})$ is arbitrary. $\endgroup$ Feb 3, 2021 at 13:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .