# Invertibility of Symplectic Matrices

1. Jan 25, 2012

### WWGD

Hi, All:

http://en.wikipedia.org/wiki/Symplectic_vector_space ,

claims that symplectic matrices are invertible

, i.e., skew-symmetric nxn-

matrix with entries w(b_i,b_j) , satisfying the properties:

i)w(b_i,b_i)=0

ii)w(b_i,b_j)=-w(b_j,b_i)

iii)w(b_i,.)=0 , i.e., w(b_i,b_j)=0 for all b_j

are invertible.

Even for small n , calculating the determinant seems to get out of hand;

Is there an easy way of seeing this?

TIA

Thanks.

iii)w(bi,.)=0 , then bi=0

2. Jan 25, 2012

### morphism

Call the matrix of your symplectic form J. Then a matrix A is symplectic (wrt your form) iff A^t J A = J. From this you get (detA)^2 = 1, since det J != 0 (which is because J is coming from nondegenerate bilinear form). In particular, detA=+/-1 is nonzero.

In fact, you can show that detA=1, but this much more difficult...

3. Jan 25, 2012

### WWGD

A couple of followup questions:

i)AFAIK, symplectic forms and symmetric/quadratic forms represent a generalization of

, I think it was inner-product and maybe the norm or length, maybe even a metric;

anyone know?

TIA.

4. Jan 25, 2012

### WWGD

Thanks, morphism, but I was actually trying to show that J itself is nonsingular; I don't see

how the nondegeneracy condition helps show invertibility. I guess one could choose a basis

{b1,..,bn}, and then set up something like:

w(bi,bj)=biTJbj , where T is the matrix (w_ij:=w(bi,bj)) , for the above choice

of basis {b1,b2,..,bn} and then w(bi,bj)=0 for all bj only if bi=0 . Maybe we can

also set up a symplectic basis {b1',b2',..,bn'} , i.e., a basis so that w(bi',bj')=

δi'j' . But then the result may be basis-dependent.

Any Ideas?

5. Jan 25, 2012

### morphism

A symplectic form is a nondegenerate skew-symmetric bilinear form.
A (real) inner product is a nondegenerate symmetric bilinear form. (Well, it's not just nondegenerate - it's positive-definite.)

The invertibility is basis-independent, because there's an invertible change of basis matrix involved. I.e. if you write A for the matrix of your nondegenerate bilinear form in one basis, and write B for it in another basis, then there will be an invertible matrix S such that A=SBS^{-1}.

Last edited: Jan 25, 2012