What's wrong with this derivation about sympletic matrices?

Question

Let $M$ be a $2\times 2$ matrix, the definition of sympletic that I have is that $M$ is sympletic if

$$MJM^T = J,$$

being $J$ the matrix

$$J = \begin{pmatrix}0 & -1 \\1 & 0\end{pmatrix}.$$

Now, consider a path of sympletic matrices $M : (-\epsilon,\epsilon)\to GL_2(\mathbb{R})$ with $M(0)=I$, the identity matrix. In the book I'm reading the author claims that if we define $m = M'(0)$ then we have

$$Jm + m^T J = 0,$$

but that is not what I'm getting. Indeed, taking the derivative on the defining condition of a sympletic matrix, since $J$ is constant we have

$$M'(0)JM^T(0)+M(0)J(M^T)'(0)=0,$$

but now $M(0)=M^T(0)=I$ and also $M'(0)=m$ while $(M^T)'(0)=(M')^T(0)=m^T$, hence we get

$$mJ+Jm^T = 0,$$

which is a little different from the author's claim. Indeed if I take the transpose there we get

$$(mJ+Jm^T)^T = (mJ)^T+(Jm^T)^T = J^Tm^T+mJ^T = 0$$

and since $J^T = -J$ we get

$$Jm^T+mJ=0,$$

which is also different from the author's claim.

So what is wrong here? Am I missing something or perhaps there's a typo in the book? How do we get the relation the author talks about?

EDIT: I was able to get the result using a trick: multiplying the defining relation by $J$ and using $J^2 = -I$ we have

$$MJM^TJ = -I,$$

thus differentiating we fint that $m = Jm^T J$ and hence multiplying by $J$ on both sides on the left we have $Jm + m^T J = 0$. This works and gives the expected result. But why just differentiating the defining relation directly gives the wrong result? What I've missed in what I did?

Answer 1

You could take the equation you derived, $mJ+Jm^\top=0$, and multiply it on the left by $J$ and on the right by $-J$ to get, since $J^2=-I$, $Jm+m^\top J=0$.

Answer 2

POST-COMMENT addition: This argument really only applies to orthogonal matrices, not to ones that are invertible. I've left it here because of the discussion in the comments.

For a path $s \mapsto M(s)$ through the identity, $M(s) \in GL_2$, you have $M(s) M^t(s) = I$. Differentiate in $s$ to get $$ M'(s) M^t(s) + M(s) M'(s)^t = 0 $$ Plug in $M(0) = I$ to get $$ M'(0) I + I M'(0)^t = 0 $$ so $$ m + m^t = 0 $$ i.e., $m^t = -m$.

Plug this into $$ Jm^T+mJ=0, $$ to get $$ -Jm - m^t J = 0 $$ which you can negate to get $$ Jm + m^t J = 0. $$