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Let $A$ be a real, symmetric and positive definite matrix and suppose $B$ is a real symmetric matrix such that $\exp(B) = A$.

Is $B$ unique?

The solution of my homework sheet says that $B$ is uniquely determined by $A$ since, for every eigenvalue $t \in \mathbb{R}$ of $B$, the $t$-eigenspace of $B$ equals the $e^t$-eigenspace of $A$.

So far I was only able to show one inclusion of this equality. How do I prove that if $r>0$ is an eigenvalue of $A$ and $v$ a vector such that $Av = rv$, then $Bv = \log(r)v$?

Is this even true?

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    $\begingroup$ Use a dimensional argument. $B$ is real symmetric, so it has a full basis of eigenvectors. So all of $\exp(B)$'s eigenvectors are accounted for, and (since this is in finite dimension) $A$ can't have any "extra" eigenvectors that didn't arise from $B$ (note that we're using the injectiveness of $e^x$ here: the analogous argument doesn't hold for $B^2 = A$). $\endgroup$
    – Erick Wong
    May 22, 2016 at 19:20

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Thanks for the hint, Erick.

I now came up with the following argument. Suppose $A$ is a real positive definite symmetric Matrix and that $B$ is a real symmetric logarithm of $A$, i.e. $B$ is real and symmetric and $\exp(B) = A$.

We wish to show that $B$ is uniquely determined by $A$. Since $B$ is diagonalizable, it suffices to show that $\sigma(B)$, the set of eigenvalues of $B$, is uniquely determined by $A$, and that every eigenspace $E_t(B) = \lbrace v \in \mathbb{R}^n \, : \, Bv = tv \rbrace$, $t \in \sigma(B)$ is uniquely determined by $A$.

Since $B$ is symmetric, $B$ is similar to a diagonal matrix $D$, which has entries from $\sigma(B)$. Hence $\exp(B) $ is similar to $\exp(D)$. It follows that

$$\sigma(\exp(B)) = \lbrace e^t \, : \, t \in \sigma(B) \rbrace \Leftrightarrow \sigma(B ) = \lbrace \log(s) \,: \, s \in \sigma(A) \rbrace$$ Since $t \mapsto e^t$ is a bijection from $\mathbb{R}$ to $(0, \infty)$.

We next show that for every eigenvalue $t$ of $B$ we have

$$E_{t}(B) = E_{e^t}(\exp(B))$$

Let $v \in E_{t}(B)$. Then

$$\exp(B)v = \sum_{k\geq 0}(\frac{B^k}{k!})v = \sum_{k\geq 0}(\frac{B^kv}{k!}) = \sum_{k\geq 0}(\frac{t^kv}{k!}) = e^t v$$

Hence $E_t(B) \subset E_{e^t}(\exp(B))$. In order to prove equality, we prove that the dimensions of both spaces are equal. Since both $B$ and $\exp(B)$ are diagonalizable, we have the direct sum decompositions

$$ \bigoplus_{t \in \sigma(B)} {E_t(B)} = \mathbb{R}^n = \bigoplus_{t \in \sigma(B)} {E_{e^t}(\exp(B))}$$

Hence

$$\sum_{t \in \sigma(B)} {\dim(E_t(B))} = n = \sum_{t \in \sigma(B)} {\dim(E_{e^t}(\exp(B)))}$$

This implies $\dim(E_t(B)) = \dim(E_{e^t}(\exp(B)))$ for every $t \in \sigma(B)$, as desired.

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  • $\begingroup$ If the argument is correct, this gives an alternative for the proof given here $\endgroup$
    – m.s
    May 23, 2016 at 19:51

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