I want to show that if $\lambda$ is a real eigenvalue of a symplectic matrix $A$ then its char poly is of the form $\det(A-\mu id) = (\lambda-\mu)(\frac{1}{\lambda}-\mu) \det(\hat{A}- \mu id) $ where $\hat{A}$ is again a symplectic matrix.

My proof goes like this: First, we take the eigenvector $e_1$ to the eigenvalue $\lambda$. We can extend this vector to a symplectic basis $e_1,f_1,e_2,f_2,\ldots,f_n$ satisfying the canonical conditions on such a basis with symplectic form $\omega.$

With respect to this basis, the first column of $A$ is $(\lambda,0,\ldots,0)^T$ by the eigenvector property.

The entry $(2,2)$ of the matrix is given by $\omega(e_1,f_1) = \frac{1}{\lambda} \omega(e_1,Af_1) = \frac{1}{\lambda} \omega(e_1,f_1) = \frac{1}{\lambda}.$

Moreover, all other entries in the second row are zero( besides the entry $(2,2)$) as for any $k \in \operatorname{span}\{{f_1}\}^{\perp}.$

$$\omega(e_1,Ak) = \frac{1}{\lambda} \omega(Ae_1,Ak) = \frac{1}{\lambda} \omega(e_1,k) = 0.$$

Thus, we can calculate the determinant of $A-\mu id$ in this basis by first expanding with respect to the first column. Since the only non zero entry is $(\lambda-\mu)$ we only get one term. Then we calculate the determinant of the submatrix with entries $(2,\ldots,n)\times(2,\ldots,n).$ We expand this determinant with respect to the second row. Here, the only non-zero entry is $(2,2).$ This one is $\frac{1}{\lambda-\mu}.$ So we indeed end up with something of the form $\det(A-\mu id) = (\lambda-\mu)(\frac{1}{\lambda}-\mu) \det(\hat{A}- \mu id) .$

But now the big question is: Why is the submatrix $\hat{A}$ corresponding to the entries $(3,\ldots,n)\times(3,\ldots,n)$ of $A$ again symplectic?


Perhaps you want to prove that if $\lambda\in \mathbb{C}$ is an eigenvalue of $A$ with multiplicity $\alpha$, then the same is true for $1/\lambda$.

Since $A$ satisfies $A^{-1}=\Omega^{-1}A^T\Omega$, $A^{-1}$ is similar to $A^T$; consequently $A^{-1}$ is similar to $A$ and we are done.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.