## Nondegenerate Forms II

Okay, we know what a nondegenerate form is, but what does this mean for the transformation that represents the form?

Remember that the form represented by the transformation is nondegenerate if for every nonzero ket vector there is some bra vector so that . But before we go looking for such a bra vector, the transformation has turned the ket vector into a new ket vector . If we find that , then there can be no suitable vector with which to pair it. So, at the very least, we must have for every . That is, the kernel of is trivial. Since is a transformation from the vector space to itself, the rank-nullity theorem tells us that the image of is all of . That is, must be an invertible transformation.

On the other hand, if is invertible, then every nonzero ket vector becomes another nonzero ket vector . Then we find that

where this last inequality holds because the bra-ket pairing is an inner product, and is thus positive-definite. Indeed, a positive-definite (or negative-definite) form *must* be nondegenerate. Thus it is sufficient for to be invertible.

Incidentally, this approach gives us a good way of constructing a lot of positive-definite transformations. Given an invertible transformation , we expand

Since the form defined by the bra-ket pairing is invertible, so is the form defined by . And this is a sensible concept, since is self-adjoint. Indeed, we take its adjoint to find

This extends our analogy with the complex numbers. An invertible transformation composed with its adjoint is a self-adjoint, positive-definite transformation, just as a nonzero complex number multiplied by its conjugate is a real, positive number.

Just to say that I really enjoy the math exposition in your blog. Recent posts cleared up a lot of confusion in my mind about linear algebra, especially about how different ways of describing things fit together (eg matrix / tensor / braket notations). Please continue!

Comment by Piotr | July 18, 2009 |

[…] we can use the fact that . We can also divide out by , since we know that is invertible, and so its determinant is nonzero. We’re left with the observation […]

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