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I need help to prove the following :

Let $V$ and $W$ be inner product spaces over the same field and $$T : V \rightarrow W$$ linear. Then $T$ preserves inner product iff $$\|T\alpha \| = \|\alpha \|$$

If $T$ preserves inner product then $T$ preserves norm .

For the converse, $\|T\alpha\|=\|\alpha \|$ implies ${\|T\alpha\|}^2 = {\|\alpha\|}^2$ We have to show , $$\langle T\alpha , T\beta \rangle = \langle \alpha , \beta \rangle $$ for any $\alpha , \beta \in V.$

Hint is to use the polarization identity $${\langle \alpha \mid \beta \rangle }={1\over 4} \|\alpha + \beta \|^2 - {1\over 4} \|\alpha - \beta \|^2$$ but I cannot figure out how to use this.

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    $\begingroup$ Hint: $T(\alpha\pm\beta)=T\alpha\pm T\beta$ $\endgroup$ – Hagen von Eitzen Jan 25 '16 at 18:38
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Using the linearity of $T$, you have

$$ \left< T\alpha|T\beta \right>_W = \frac{1}{4} \left( ||T\alpha + T\beta||_W^2 - ||T\alpha - T\beta||_W^2 \right) = \frac{1}{4} \left( ||T(\alpha + \beta)||_W^2 - ||T(\alpha - \beta)||_W^2 \right) = \frac{1}{4} \left( ||\alpha + \beta||_V^2 - ||\alpha - \beta||_V^2 \right) = \left< \alpha | \beta \right>_V.$$

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