# Proving that operator is linear if it is unitary and surjective

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: $$T(ax) = aT(x) \\ T(x+y) = T(x)+T(y)$$

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!

\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 $$T(\alpha x)=\alpha T(x)$$.

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

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.

• 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. Feb 3, 2021 at 13:06