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In the above definition, when it says $ (T \alpha|T \beta)=(\alpha|\beta) $, the inner product $(|)$ acting on $W$ is the same as acting on $V$? Or not necessarily? The book uses the same notation, so I had this question.

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    $\begingroup$ Not necessarily the same. $\endgroup$ – mechanodroid Feb 17 at 23:31
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    $\begingroup$ What would that even mean were $V$ and $W$ not the same? $\endgroup$ – Alex Provost Feb 18 at 0:00
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The inner product defined on $V$ may not be the same as the inner product defined on $W$, but given your $\alpha$ and $\beta$ in $V$, $(\alpha|\beta)$ is a just a number. $T$ preserves inner products if evaluating $(T\alpha|T\beta)$ gets you the same number as before.

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It sounds like you could an example. Take $V = \Bbb{R}^2$, with $(x \mid y)_V = x \cdot y$, the dot product, and $W = P_1(\Bbb{R})$, the space of real polynomials of degree at most $1$, equipped with the inner product $$(p \mid q)_W = \int_{-1}^1 f(x)g(x) \; \mathrm{d}x.$$ Define $T : V \to W$ by $$T(a, b) = x \mapsto \frac{a}{\sqrt{2}} + \frac{\sqrt{3}b}{\sqrt{2}} x.$$ Then \begin{align*} (T(a, b) \mid T(c, d))_W &= \left(\left.\frac{a}{\sqrt{2}} + \frac{\sqrt{3}b}{\sqrt{2}} x \; \right| \; \frac{c}{\sqrt{2}} + \frac{\sqrt{3}d}{\sqrt{2}} x\right) \\ &= \int_{-1}^1 \left(\frac{a}{\sqrt{2}} + \frac{\sqrt{3}b}{\sqrt{2}} x\right)\left(\frac{c}{\sqrt{2}} + \frac{\sqrt{3}d}{\sqrt{2}} x\right) \; \mathrm{d}x \\ &= \left[\frac{ac}{2}x + \frac{1}{2}\left(\frac{\sqrt{3}b}{2} + \frac{\sqrt{3}d}{2}\right)x^2 + \frac{1}{2}bdx^3\right]_{x=-1}^{x=1} \\ &= ac + bd = ((a, b) \mid (c, d))_V \end{align*} That is, $T$ preserves inner products.


Of course, in the above example, I've really side-stepped your issue by differentiating between the two inner products $(\cdot \mid \cdot)_V$ and $( \cdot \mid \cdot)_W$. It is unfortunately common for different inner products on different spaces to have the same notation. You'd be expected to know which inner product to use from context.

For example, if I write (given the above example) $((1, 1) \mid (0, 2))$, I'm clearly talking about the dot product, and hence the number $2$. On the other hand, if I write $(1 + x, x - 4)$, I'm obviously using the inner product on $W$, which results in $-\frac{22}{3}$.

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