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 be the self-adjoint transformation corresponding to our (conjugate) symmetric form, we might have a nonzero vector such that . However, the vector need not be completely trivial as far as the form is concerned. There may be another vector so that .
Let us work out a very concrete example. For our vector space, we take with the standard basis, and we’ll write the ket vectors as columns, so:
Then we will write the bra vectors as rows — the transposes of ket vectors:
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:
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 . Then the basis vector is just such a one of these vectors “orthogonal” to itself (with respect to our new bilinear form). Indeed, we can calculate
However, this vector is not totally trivial with respect to the form . For we can calculate
Now, all this is prologue to a definition. We say that a form (symmetric or not) is “degenerate” if there is some non-zero ket vector so that for every bra vector we find
And, conversely, we say that a form is “nondegenerate” if for every ket vector there exists some bra vector so that