The Unapologetic Mathematician

Mathematics for the interested outsider

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 B is nondegenerate if for every nonzero ket vector \lvert v\rangle there is some bra vector \langle w\rvert so that \langle w\rvert B\lvert v\rangle\neq0. But before we go looking for such a bra vector, the transformation B has turned the ket vector \lvert v\rangle into a new ket vector B\lvert v\rangle=\lvert B(v)\rangle. If we find that B(v)=0, then there can be no suitable vector w with which to pair it. So, at the very least, we must have B(v)\neq0 for every v\neq0. That is, the kernel of B is trivial. Since B is a transformation from the vector space V to itself, the rank-nullity theorem tells us that the image of B is all of V. That is, B must be an invertible transformation.

On the other hand, if B is invertible, then every nonzero ket vector \lvert v\rangle becomes another nonzero ket vector \lvert w\rangle=B\lvert v\rangle. Then we find that

\displaystyle\langle w\rvert B\lvert v\rangle=\langle w\vert w\rangle>0

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 B to be invertible.

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

\displaystyle\langle w\vert w\rangle=\langle v\rvert B^*B\lvert v\rangle

Since the form defined by the bra-ket pairing is invertible, so is the form defined by B^*B. And this is a sensible concept, since B^*B 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.

About these ads

July 17, 2009 - Posted by | Algebra, Linear Algebra


  1. 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 | Reply

  2. [...] 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 [...]

    Pingback by The Determinant of Unitary and Orthogonal Transformations « The Unapologetic Mathematician | July 31, 2009 | Reply

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


Get every new post delivered to your Inbox.

Join 366 other followers

%d bloggers like this: