# Isomorphism between vector spaces, and preserving structure

Say $T$ is an isomorphism between inner product spaces, $T:X \to X^1$. Both spaces are defined over same field. By definition, $T$ is bijective and linear operator. It is claimed that $$<Tx, Ty>=<x,y>, \forall x,y \in X$$

I have been trying to prove this claim from bijective and linearity properties of the operator. I have thought on it for a while. But I have no clue on where to start on the proof.

Any clue would be greatly appreciated!

• An isomorphism of inner-product spaces preserves the inner-product by definition. It's not a claim but a definition. In general isomorphisms of vector spaces between inner-product spaces need not preserve the inner-product. Commented May 11, 2017 at 7:29
• Ok... I understood the definition in wrong way... thanks everyone! Commented May 11, 2017 at 7:39

You didn't find a proof for this because this property doesn't follow just from bijectivity and linearity. Consider for example the operator $$T: \mathbb{R}^2 \to \mathbb{R}^2, \quad T\begin{pmatrix} x \\ y\end{pmatrix} = 2\begin{pmatrix} x \\ y\end{pmatrix}.$$ This operator is clearly bijective and linear, but $$\langle T\begin{pmatrix} x_1 \\ y_1\end{pmatrix}, T\begin{pmatrix} x_2 \\ y_2\end{pmatrix} \rangle = 4\langle \begin{pmatrix} x_1 \\ y_1\end{pmatrix}, \begin{pmatrix} x_2 \\ y_2\end{pmatrix} \rangle \neq \langle \begin{pmatrix} x_1 \\ y_1\end{pmatrix}, \begin{pmatrix} x_2 \\ y_2\end{pmatrix}\rangle.$$ Linear operators $T$ from an inner product space $X$ to an inner product space $X_1$ which fulfill $$\langle Tx, Ty\rangle_1 = \langle x, y\rangle$$ for all $x, y \in X$ are called isometric. A reasonable question to ask would be: prove that every linear isometric operator is injective.