Diagonalizing Linear Transformations on Finite-Dimensional Real Vector Spaces

[Treadstone 71]

"Let T be a linear transformation on a finite dimensional real vector space V. Show that T is diagonalisable if and only if there exists an inner product on V relative to which T is self-adjoint."

The backward direction is easy. As for the forward direction, I don't understand how given an arbitrary vector space, you can go about defining an inner product without knowing something more about it.

 

[AKG]

Although it's not really relevant, your vector space isn't arbitrary, it's isomorphic to Rⁿ for some n. Anyways, you don't have to know anything about V, you just have to be able to define an inner product on it, and an inner product is just a function $\langle .,.\rangle\, :\, V\times V \to \mathbb{R}$ that is symmetric, positive definite, and bilinear (there may be another condition or two, you can look it up). So just define a function that has these properties, and is also such that T is self-adjoint with respect to it.

Now the problem is, how to find an inner product such that T is self-adjoint relative to it. Well what things do you know about T, given that it's diagonalizable? Second, you're not going to write out what <v,w> is for each individual v and w in V. Given that inner products are multilinear, it suffices to define an inner product on a ______. But you should know that if T is diagonalizable, there is a ______ with some strong relation to T. Fill in the blank, figure out what that "strong relation" is, and use it to prove that T is self-adjoint w.r.t. your inner product.

 

[Hurkyl]

As for the forward direction, I don't understand how given an arbitrary vector space, you can go about defining an inner product without knowing something more about it.

The only thing there is to know about a (finite-dimensional real) vector space is its dimension.

Anyways, I think the backward direction gives you a strong hint about how to proceed with the forward direction.

 

[Treadstone 71]

Got it. Thanks.