The Unapologetic Mathematician

Nondegenerate Forms I

The notion of a positive semidefinite form opens up the possibility that, in a sense, a vector may be “orthogonal to itself”. That is, if we let $H$ be the self-adjoint transformation corresponding to our (conjugate) symmetric form, we might have a nonzero vector $v$ such that $\langle v\rvert H\lvert v\rangle=0$. However, the vector need not be completely trivial as far as the form is concerned. There may be another vector $w$ so that $\langle w\rvert H\lvert v\rangle\neq0$.

Let us work out a very concrete example. For our vector space, we take $\mathbb{R}^2$ with the standard basis, and we’ll write the ket vectors as columns, so:

\displaystyle\begin{aligned}\lvert1\rangle&=\begin{pmatrix}1\\{0}\end{pmatrix}\\\lvert2\rangle&=\begin{pmatrix}0\\1\end{pmatrix}\end{aligned}

Then we will write the bra vectors as rows — the transposes of ket vectors:

\displaystyle\begin{aligned}\langle1\rvert&=\begin{pmatrix}1&0\end{pmatrix}\\\langle2\rvert&=\begin{pmatrix}0&1\end{pmatrix}\end{aligned}

If we were working over a complex vector space we’d take conjugate transposes instead, of course. Now it will hopefully make the bra-ket and matrix connection clear if we note that the bra-ket pairing now becomes multiplication of the corresponding matrices. For example:

\displaystyle\begin{aligned}\langle1\vert1\rangle&=\begin{pmatrix}1&0\end{pmatrix}\begin{pmatrix}1\\{0}\end{pmatrix}=\begin{pmatrix}1\end{pmatrix}\\\langle1\vert2\rangle&=\begin{pmatrix}1&0\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}0\end{pmatrix}\end{aligned}

The bra-ket pairing is exactly the inner product we get by declaring our basis to be orthonormal.

Now let’s insert a transformation between the bra and ket to make a form. Specifically, we’ll use the one with the matrix $S=\begin{pmatrix}0&1\\1&0\end{pmatrix}$. Then the basis vector $\lvert1\rangle$ is just such a one of these vectors “orthogonal” to itself (with respect to our new bilinear form). Indeed, we can calculate

$\displaystyle\langle1\rvert S\lvert1\rangle=\begin{pmatrix}1&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1\\{0}\end{pmatrix}=\begin{pmatrix}1&0\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}0\end{pmatrix}$

However, this vector is not totally trivial with respect to the form $S$. For we can calculate

$\displaystyle\langle2\rvert S\lvert1\rangle=\begin{pmatrix}0&1\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}1\\{0}\end{pmatrix}=\begin{pmatrix}0&1\end{pmatrix}\begin{pmatrix}0\\1\end{pmatrix}=\begin{pmatrix}1\end{pmatrix}$

Now, all this is prologue to a definition. We say that a form $B$ (symmetric or not) is “degenerate” if there is some non-zero ket vector $\lvert v\rangle$ so that for every bra vector $\langle w\rvert$ we find

$\displaystyle\langle w\rvert B\lvert v\rangle=0$

And, conversely, we say that a form is “nondegenerate” if for every ket vector $\lvert v\rangle$ there exists some bra vector $\langle w\rvert$ so that

$\displaystyle\langle w\rvert B\lvert v\rangle\neq0$

July 15, 2009 - Posted by | Algebra, Linear Algebra

1. “We say that a form (symmetric or not) is “degenerate” if there is some ket vector ”

Do you want to say non-zero ket vector?

Comment by Johan Richter | July 19, 2009 | Reply

2. Yes, sorry. I caught this in the next post, but didn’t here.

Comment by John Armstrong | July 19, 2009 | Reply

3. [...] the orthogonal groups. This covers orthogonality with respect to general (nondegenerate) forms on an inner product space , the special case of orthogonality with respect to the underlying [...]

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