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How do we prove that the exponential function is a bijection between the set $S_n$ of real symmetric matrices with size $n$ and the set $\Sigma_n$ of real symmetric positive definite matrices with size $n$? Thanks.

Rod Carvalho
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1 Answers1

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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 $B=Q_2D_2Q_2^T$. 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 $A = Q_1 D_1 Q_1^T = Q_2 D_2 Q_2^T = B$.

Maman
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user1551
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  • Where in the proof, using the fact that $=⨁{=1}^k _{}$ is useful ? Is this to prove that the orthogonal matrix $Q$ is also diagonalizable, hence there will a be a co-diagonalization criterion with $QD = DQ$ but it will not help to conclude ? Thanks – Maman Jun 09 '22 at 11:57
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    @Maman It is used to establish the inference that *“if $Q$ commutes $\exp(D)$ then $Q$ commutes with $D$”. One can prove this statement without considering the block structure of $Q$ (e.g. by showing that $D$ is a polynomial in $\exp(D)$), but for some reason, I opted to use this block structure when I wrote my answer. – user1551 Jun 09 '22 at 12:06
  • Yes ok thank you for clarifying this ! Indeed for the last equality $QD=DQ$, I used your second argument ($D$ is polynomial in $\exp(D)$ with Lagrange interpolation, notice that the converse is also true). – Maman Jun 09 '22 at 13:26