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Given $G=\left\{A\in M_2(\mathbb{R})\mid A^\top XA = X\right\}$ where $X = \pmatrix{3&1\\1&1}$. Let a smooth path $A(t)$ in $G$ with $A(0)=I_2$.

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Consider the condition $A(t)^TXA(t)=X$. Differentiating, you obtain $$(A(t)^T)'XA(t)+A(t)^TX(A(t))'=0$$

since $X$ is a constant matrix and so differentiating gives zero. Now putting $t=0$ and replacing $(A(0))'$ by $Y$ and $A(0)$ by $I_2$ we obtain the result;

$$Y^TX+XY=0$$ as desired.

Note that $Y$ is an element of the Lie algebra, $g$ of the Lie group, $G$ since $A'(0)=Y$ and $A(t)\in G$. This is because a Lie algebra of a Lie group (a group and a manifold) is thought of as the tangent space to the Lie group at the identity, so the condition $Y=(A(0))'$ is enough. To convince yourself you can verify using the matrix exponential.

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$A(t)$ is a curve in $G$, which means $A(t)^t X A(t) = X$ for all $t$. Now we can differentiate this with respect to $t$ at $t=0$ (and observing that matrix multiplication satisfies the product rule) to get $A'(0)^t X A(0) + A(0)^t X A'(0) = X$. Since $A(0)$ is the identity, this proves the result.

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