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

$\displaystyle\left(B^*B\right)^*=B^*\left(B^*\right)^*=B^*B$

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.

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

2 Comments »

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
[…] 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

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This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

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