Exploring Inner Product Spaces: Answers to Your Questions

[samkolb]

I have some questions about inner product sapces.

1. If V is a vector space over R and ( , ):VxV-->R is an inner product on V, then for v,w in V, is the value of (v,w) independent of my choice of basis for V used to compute (v,w)?

2. If V is an arbitrary n dimensional vector space over R, are there some standard ways to define an inner product on V?

3. If V is an arbitrary n dimensional vector space over R, and v and w are linearly independent vectors in V, is it always possible to define an inner product on V such that (v,w) is not zero?

 

[turin]

3. If V is an arbitrary n dimensional vector space over R, and v and w are linearly independent vectors in V, is it always possible to define an inner product on V such that (v,w) is not zero?

I believe so.

Case 1: v and w are not orthogonal ... done.

Case 2: v and w are orthogonal according to "the naive" inner product. You can either skew the space or shuffle the dual space.

Example 1, skewing the space
$$v\rightarrow\left(\begin{array}{c}1\\0\end{array}\right) \qquad w\rightarrow\left(\begin{array}{c}0\\1\end{array}\right)$$
inner product:
$$(,)\rightarrow\left(\begin{array}{cc}1&1\\0&1\end{array}\right)$$
i.e.
$$(v,w) \rightarrow \left(\begin{array}{cc}1&0\end{array}\right)\left(\begin{array}{cc}1&1\\0&1\end{array}\right)\left(\begin{array}{c}0\\1\end{array}\right) = \left(\begin{array}{cc}1&0\end{array}\right)\left(\begin{array}{c}1\\1\end{array}\right) \textrm{ or } \left(\begin{array}{cc}1&1\end{array}\right)\left(\begin{array}{c}0\\1\end{array}\right) = 1$$
This sort of inner product actually occurs in, for example, crystalography.

Example 2, shuffling the dual space:
$$v\rightarrow\left(\begin{array}{c}1\\0\end{array}\right) \qquad w\rightarrow\left(\begin{array}{c}0\\1\end{array}\right)$$
inner product:
$$(,)\rightarrow\left(\begin{array}{cc}0&1\\-1&0\end{array}\right)$$
i.e.
$$(v,w) \rightarrow \left(\begin{array}{cc}1&0\end{array}\right)\left(\begin{array}{cc}0&1\\-1&0\end{array}\right)\left(\begin{array}{c}0\\1\end{array}\right) = \left(\begin{array}{cc}1&0\end{array}\right)\left(\begin{array}{c}1\\0\end{array}\right) \textrm{ or } \left(\begin{array}{cc}0&1\end{array}\right)\left(\begin{array}{c}0\\1\end{array}\right) = 1$$
This sort of inner product actually occurs in, for example, the Majoranna Lagrangian. Actually, I'm not sure if this satisfies your definition of an inner product, because (v,v) and (w,w) vanish in this case.

 

[samkolb]

Turin.

I know it's been a while since you replied to my post, but I have a question about it. I think the first two by two matrix you used is a matrix representation of a particular bilinear form, but I don't think it satisfies the conditions for an inner product since it is not symmetric.

 

[turin]

OK, yes, the official definition of inner product requires

$$\langle{}v,w\rangle=\langle{}w,v\rangle^*$$

So, based on this requirement, neither one of the two suggestions that I gave is valid. However, if you restrict the field to R, then I believe that you can get away with

$$\langle{}v,w\rangle \equiv \left(\begin{array}{cc}v_1&v_2\end{array}\right) \left(\begin{array}{cc}1&a\\a&1\end{array}\right) \left(\begin{array}{c}w_1\\w_2\end{array}\right)$$

if

$$-1<a<1$$

This is obviously symmetric, and the condition on a guaruntees that no vector with nonzero real-valued components can have a vanishing inner product. Then

$$\left(\begin{array}{cc}1&0\end{array}\right) \left(\begin{array}{cc}1&a\\a&1\end{array}\right) \left(\begin{array}{c}0\\1\end{array}\right) = a$$

 

[samkolb]

That works! It took me some time to see why the condition -1<a<1 guarantees that the inner product of a vector with itself vanishes if and only if the vector is the zero vector, but I get it now. Thanks alot.

Sam