Let $$A = \begin{pmatrix} 2 & 2\\ 2 & 5\end{pmatrix}$$ Show that there exists a matrix $B$ such that $A = \exp(B)$.
My thought is to use $\log(A)$, but first I need to make sure this is well-defined.
The eigenvalues of $A$ are $6$ and $1$, so
$$A=Q^{-1}\begin{pmatrix}6 & 0\\0 & 1\end{pmatrix}Q$$
for some invertible $Q$. My first attempt was to use the fact that
$$\log(x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^n}{n}$$
for $|x-1| < 1$. However, all the matrix norms I can think of give me $\|A-I\|>1$. So my second approach was to note that $1, 6 \in [1,6]$, which is compact, so Weierstrass approx. $\implies$ there exists polynomials $P_n:[1,6]\to\mathbb{R}$ s.t. $P_n\to log$ (uniformly). Then
$$P(Q^{-1}\begin{pmatrix}6 & 0\\0 & 1\end{pmatrix}Q)=Q^{-1}P(\begin{pmatrix}6 & 0\\0 & 1\end{pmatrix})Q=Q^{-1}\begin{pmatrix}P(6) & 0\\0 & P(1)\end{pmatrix}Q$$
for any polynomial $P$. Hence,
$$P_n(Q^{-1}\begin{pmatrix}6 & 0\\0 & 1\end{pmatrix}Q)\to Q^{-1}\begin{pmatrix}log(6) & 0\\0 & 0\end{pmatrix}Q$$
so
$$\log(A)=Q^{-1}\begin{pmatrix} \log(6) & 0\\ 0 & 0\end{pmatrix}Q$$
is well-defined. Does this look correct? If so, does anyone have a simpler solution? Thank you.
There's no reason to worry about whether the log is well defined.
Find any matrix $D$ such that $$ \exp(D) = \pmatrix{1&0\\0&6} $$ In particular, we can take $$ D = \pmatrix{\log(1) & 0\\0 & \log(6)} $$ (notably, $\log(1) = 0$ and $\log(6)$ is some positive number). Once you have such a matrix $D$, it immediately follows that $B = Q^{-1}DQ$ is a solution to your problem.
Is there a log function defined by some power series such that $D$ is the log of that diagonal matrix and $B$ is the log of $A$? Didn't we "guess" our solution by taking the "log" of the matrix? Maybe, but that doesn't really matter: we have a solution, so take the money and run.