Logarithm of a Jordan block

Question

I don't understand how to find a logarithm of Jordan block. Wikipedia says, that we can use Mercator series, but I don't know why it works for matrices - I know only the proof for real numbers. In some other books I've only found explanation on the level "by contruction we have that $ e^B=A $" and the elements of matrix $ B $. Can anybody explain formally how to find this logarithm?

Edit. By finding the logarithm I mean not just deriving the formula for it, but rigorous proof that this formula is correct. If $ A $ is a Jordan block of the form $ \lambda I + N $, where $ N $ is nilpotent and $ I $ is identity matrix, I know that

$$ B=\log{A}=I\log{\lambda}+\sum_{j=1}^{n-1}\frac{(-1)^{j+1}}{j\lambda^j}N^j. $$

Unfortunately, I don't understand how to prove that $ e^B=A. $ I tried to do it by the definition of exponential, but I wasn't successful.

Answer 1

A Jordan block is in the form $A=\lambda I_n+J$ where $J$ is the nilpotent Jordan block of dimension $n$ and $\lambda\in\mathbb{C}\setminus \{0\}$.

When $\lambda\notin (-\infty,0]$, the principal logarithm of $A=\lambda(I+u J)$ (where $u=\dfrac{1}{\lambda}$) is

$\log(A)=\log(\lambda)I+uJ-\dfrac{1}{2}(uJ)^2+\dfrac{1}{3}(uJ)^3+\cdots$

with the definition: $\log(\lambda)=\log(r)+i\theta$ when $\lambda=r\exp(i\theta), \theta\in(-\pi,\pi)$.

Note that the above sum is finite because $J^n=0$ and that we use the standard series of $\log(1+v)$ when $|v|<1$.

EDIT.

Assume that, on the complex numbers, $f(y)=\sum_i a_iy^i,|y|<u$ and $y=g(x)=\sum_{j\geq 1} b_jx^j,|x|<v$ is s.t. $b_1\not= 0$ and for every $|x|<v$, one has $|g(x)|<u$. Let $\phi_k(x)=\tau_k(\sum_{i=0}^k a_i(\sum_{j=1}^k b_jx^j)^i)$ where $\tau_k$ keeps only the monomials of degree $\leq k$..

Then $(f\circ g)(x)=\lim_{k\rightarrow\infty}\phi_k(x)$ (uniformly locally convergent series with radius $\geq v$).

One has a similar result for the matrices (same proof. -that is your business- the key being $||A^p||\leq ||A||^p$): Let $f(B)=\sum_i a_iB^i$ and $B=g(A)=\sum_{j\geq 1} b_jA^j$;

$(f\circ g)(A)=\lim_{k\rightarrow\infty}\phi_k(A)$ for $||A||<u$ and even for $\rho(A)=\max((|\lambda_i|)_i<u$.

Now, if $(f\circ g)(x)=x$, then $A=\lim_{k\rightarrow\infty}\phi_k(A)$.

In particular, if $A=uJ$, then $uJ=\phi_{n-1}(uJ)$.

Answer 2

An informal "proof":

For any two polynomials $p(x),q(x)$, the composition $p\circ q$ is another polynomial. For instance: $$ p(x) = 1 + x^2, \quad q(x) = x - x^3\\ p(q(x)) = 1 + (x - x^3)^2 = 1 + x^2 - 2x^4 + x^6. $$ The same is true for polynomials on matrices. For instance, $$ p(X) = I + X^2, \quad q(X) = X - X^3\\ p(q(X)) = I + (X - X^3)^2 = I + X^2 - 2X^4 + X^6. $$ Composing "matrix polynomials" in this way always gives you the same result as composing scalar polynomials.

Likewise, we can consider the power series, noting that the composition of two power series is another power series. $$ p(x) = \exp(x) = \sum_{k=0}^\infty \frac{x^k}{k!}, \quad q(x) = \log(x) = \sum_{k=1}^\infty -\frac{(1-x)^k}{k},\\ p(q(x)) = x. $$ Because $\exp(\log(x)) = x$ applies for scalar power series, it must be the case that for any matrix $A$, we have $p(q(A)) = A$ (as long as $q(A)$ is defined). That is, $\exp(\log(A)) = A$, where $\log$ was defined by the Mercator series.

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If I have understood your question correctly, the primary concern is this: given some expression for the logarithm $\log(A)$ of a matrix $A$, how do we know that this matrix will really satisfy $\exp(\log(A)) = A$, where $\exp$ is the matrix-exponential?

One argument goes as follows.

1. Show that $\exp(\log(A)) = A$ holds whenever $A$ is diagonalizable
2. Show that the functions $X \mapsto \exp(X)$, $X \mapsto \log(X)$ are continuous (since it is the uniform limit of continuous functions over any closed and bounded subset of its domain)
3. Show that for any non-diagonalizable matrix $A$ (a Jordan block for instance), there exist a sequence of diagonalizable matrices $(A_n)_{n=1}^\infty$ such that $A_n \to A$
4. By continuity, $$ A = \lim_{n \to \infty}A_n = \lim_{n \to \infty} \exp(\log(A_n)) = \exp(\log(A)) $$ as desired.