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

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

Like this:

Posted by John Armstrong | Algebra, Linear Algebra

3 Comments	»

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 | 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 [...]

Pingback by | July 31, 2009 | Reply

« Previous | Next »

About this weblog

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”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 21, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9150304794311523, "perplexity_flag": "middle"}