# How to define the logarithm from positive definite matrices to symmetric matrices

Let $$S$$ be the set of positive definite matrices, and $$\mathfrak s$$ the set of symmetric matrices. The exponential map $$\exp: \mathfrak s \rightarrow S$$ is a smooth bijection. To argue that it is a diffeomorphism, I would like to define a smooth inverse $$\log: S \rightarrow \mathfrak s$$. For $$B \in S$$, the series

$$\log(B) \;\; =\;\; \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} (I - B)^k$$ only converges when the eigenvalues of $$B$$ are between $$0$$ and $$2$$. Perhaps a way we can define $$\log$$ in general is by setting $$\log(B) = P^{-1} \log(PBP^{-1})P$$, where $$P$$ is any orthogonal matrix such that $$PBP^{-1}$$ is diagonal, and for diagonal positive definite matrices we define the logarithm of $$\operatorname{diag}(\alpha_1, ... , \alpha_n)$$ to be $$\operatorname{diag}(\log(\alpha_1), ... , \log(\alpha_n))$$.

How can we show that this map is well defined and smooth?

• By the way, I guess you can also use the series. Just take $\log(B)=\log\left(\frac{1}{\|B\|}B\right)+\log(\|B\|)I$, where $I$ is the identity matrix and $\|B\|$ is the norm of $B$, that is, the largest eigenvalue. Apr 15 '20 at 19:00
• Thanks. That seems like a pretty good way to do it. Is it obvious that $B \mapsto ||B||$ is smooth on $\operatorname{GL}_n(\mathbb R)$?
– D_S
Apr 15 '20 at 20:33
• Good point, I guess it is not smooth. But anyway, you can always move a neighborhood of any symmetric matrix to the desired interval $(0,2)$. It's probably better to use some constant, so $\log(B)=\log\left(\frac{1}{\alpha}B\right)+\log\alpha\,I$. Then you have to show that this is well defined. Apr 16 '20 at 6:10

On finite dimension, this process is very simple. Take any normal matrix $$A$$ and any continuous function $$f\colon\sigma(A)\to\mathbb{C}$$, where $$\sigma(A)$$ is the spectrum of $$A$$. (In case of positive operators, you may take functions $$f\colon \mathbb{R}^+\to\mathbb{C}$$.)
Now, what you are suggesting is essentially to choose a basis of eigenvectors $$x_i$$ corresponding to eigenvalues $$\lambda_i$$ (so $$Ax_i=\lambda_i x_i$$) and define a new operator $$B$$ by $$Bx_i=f(\lambda_i)x_i$$. This is surely well defined and we can denote $$B=:f(A)$$
Slightly "more standard" formulation is the following. You can always do so-called spectral decomposition of the operator $$A$$. This means to write it in the form $$A=\sum_i \lambda_i P_i$$, where $$\lambda_i$$ are the eigenvalues of $$A$$ and $$P_i$$ are orthogonal projections to the corresponding eigenspaces. Then you can define $$f(A):=\sum f(\lambda_i)P_i.$$
Now you can try to prove all the stuff you need. Namely that the mapping of matrices induced by $$f$$ is again continuous, how does the derivative look like and so on.