Consider $I, t \in \mathbb{R}^d$ and $g$ is some element in a group of transformations (for example like the affine group in $\mathbb{R}^2$).

I was wondering when the inner product $ \langle gI, t \rangle$ is equal to $ \langle I, g^{-1} t\rangle $ i.e.:

$$ \langle gI, t \rangle = \langle I, g^{-1} t\rangle$$

What kind of space does it have to be? What kind of group (action) does $g$ has to have? What type of conditions does the inner product have to have? Any conditions for that to hold...or maybe that equation can never hold?

These are some of my thoughts:

To get familiar with the problem I considered the dot product (for the inner product) and examined both sides of the equation:

$$ \langle gI, t \rangle = \langle I, g^{-1} t\rangle $$

$$ \sum^{d}_{i=1} (gI)_i t_i = \sum^{d}_{i=1} I_i (g^{-1} t)_i $$

however, doing that didn't reveal that much to me. My intuition says that maybe cyclic groups might be import, though not sure if they are.

Some other idea I have, maybe leaving it in terms of dot products rather than expanding the summation leads to:

$$(gI)^T t = I^T g^{-1}t$$

$$I^T g^T t = I^T g^{-1}t$$

so $g^{T} = g^{-1}$? Seems strange to me.

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    $\begingroup$ Wouldn't it just be when the group elements are unitary? $\endgroup$ – Cameron Williams May 23 '15 at 0:44

Suppose that $V$ is an inner product space. Let $T:V \to V$ be a linear transformation. What is known as the adjoint of $T$, denoted by $T^*$, has the defining characteristic $$ \langle T(\textbf{v}),\textbf{w}\rangle=\langle\textbf{v},T^*(\textbf{w})\rangle , \mbox{ for all }\textbf{v},\textbf{w}\in V.$$ Furthermore, $T$ is referred to as unitary precisely when $TT^*=T^*T=id_V$. That is, when $T^* = T^{-1}$. More information on adjoints can be found here.

So, it is clear that unitary transformations in $\mathbb{R}^d$ will satisfy your relation. It should be noted that when we restrict $V$ to be $\mathbb{R}^d$, we tend to call the transformations orthogonal instead of unitary.

Furthermore, when we restrict $V$ to be finite-dimensional, the notion of a unitary linear transformation is equivalent to an isometric linear transformation. More information on (linear) isometries can be found here.

  • $\begingroup$ I should have added in my original question to explain they why, but I will ask it here. Why is that true for unitary transformations do the trick? Is it suppose to be obvious? What I have (for the special case of the dot product) is that $(gI)^T t = I^T g^{-1}t$, so $I^T g^T t = I^T g^{-1}t$, so $g^T = g^{-1}$, is that correct? That doesn't seem to be what a unitary transformation suppose to look like... $\endgroup$ – Charlie Parker May 23 '15 at 1:13
  • $\begingroup$ The concept of the transpose of a linear transformation is not well-defined, and I would avoid it. However, if we are working in a finite-dimensional inner product space, then I can find a matrix $A$ which is representative of the linear transformation $T$. The adjoint of A is then the conjugate-transpose of A. Clearly, when we are working in a real inner product space, the adjoint of A is just the transpose of A. So, unitary matrices do indeed "look like that". $\endgroup$ – eloiprime May 23 '15 at 1:23
  • $\begingroup$ I see, obviously when $g$ is real and we are in the "linear algebra"context, then it just says that $g$ needs to be orthogonal (special case of unitary for reals). Ok, this special case makes sense to me now. But for the general linear transformation, what is the correct answer? why is $\langle T(v) , w \rangle = \langle v , T^{*}(w) \rangle$ ? $\endgroup$ – Charlie Parker May 23 '15 at 1:29
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    $\begingroup$ That is actually the definition of $T^*$, and it turns out that the choice of $T^*$ is unique to each $T$. In general, the linear transformations that satisfy your relation are unitary. Note, that in the finite-dimensional case, all unitary transformations are referred to as isometries. The definition of an isometry $S:V\to V$ is that the norm of $S(v)$ is equal to the norm of $v$ for all $v \in V$. $\endgroup$ – eloiprime May 23 '15 at 1:47
  • $\begingroup$ Thanks eloiPrime! Maybe you should add some comments noting that. Maybe this link could go there as well: en.wikipedia.org/wiki/Hermitian_adjoint $\endgroup$ – Charlie Parker May 23 '15 at 1:49

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