Exponential function and matrices

Solution 1:

Note that every real symmetric matrix can be orthogonally diagonalized. Using this fact, it is easy to prove that the exponential function is a surjective function from $S_n$ to $\Sigma_n$, and I will leave the proof to you. The part that needs more work is to show that the exponential function is injective. That is, given $\exp(A)=\exp(B)$ for two real symmetric matrices $A$ and $B$, we want to show that $A=B$.

Let $A$ and $B$ be orthogonally diagonalized as $A=Q_1D_1Q_1^T$ and $A=Q_2D_2Q_2^T$. As $Q_1\exp(D_1)Q_1^T=\exp(A)=\exp(B)=Q_2\exp(D_2)Q_2^T$, $D_1$ and $D_2$ have identical eigenvalues. It follows that there exist permutation matrices $P_1,P_2$ such that $$P_1D_1P_1^T = P_2D_2P_2^T = D := \lambda_1I_{m_1} \oplus \cdots \oplus \lambda_kI_{m_k}$$ for some distinct real numbers $\lambda_1,\ldots,\lambda_{k}$ and some sizes $m_1,\ldots,m_k$ that sum up to $n$. So, $\exp(A)=\exp(B)$ implies that $Q\exp(D) = \exp(D)Q$, where $Q = P_2^TQ_2^TQ_1P_1$. Therefore $Q$ must be a direct sum of some smaller real orthogonal submatrices, or more specifically, $Q = Q_{m_1} \oplus \cdots \oplus Q_{m_k}$, where $Q_{m_i}Q_{m_i}^T=I_{m_i}$. Now it can be verified that $QD = DQ$ and hence $A = Q_1 D_1 Q_1^T = Q_2 D_2 Q_2^T = B$.