📂Matrix Algebra

Unipotent Matrix

Definition¹

For a square matrix $A$, if there exists a positive $k$ satisfying $(A - I)^{k} = O$, then $A$ is called unipotent. In this case $O$ is the zero matrix.

Explanation

Thus, if $A - I$ is a nilpotent matrix, then $A$ is unipotent. Since the definition of the matrix logarithm is as follows, if $A$ is unipotent then the value of $\log A$ exists.

$$ \log A = \sum\limits_{m=1}^{\infty} (-1)^{m+1} \dfrac{(A - I)^{m}}{m} \tag{1} $$

Properties

(a) If $A$ is unipotent, then $\log A$ is nilpotent.

　(a') If $A$ is unipotent, then $e^{\log A} = A$.

(b) If $X$ is nilpotent, then $e^{X}$ is unipotent.

　(b') If $X$ is nilpotent, then $\log( e^{X} ) = X$.

Proof

(a)

If $A$ is unipotent, there exists $k$ satisfying $(A - I)^{k} = O$. The definition of $\log A$ is the same as $(1)$, hence $\log A$ is nilpotent.

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(a')

Assume $A$ is unipotent. Then $A - I$ is nilpotent, so for some nilpotent matrix $N$ we have $A = I + N$. Let $A(t) = I + tN$. Then $\log A(t)$ is nilpotent and is a polynomial of finite degree in $t$. That is, there exists some $M$ such that

$$ \log A(t) = \sum\limits_{m=1}^{\infty} (-1)^{m+1} \dfrac{(tN)^{m}}{m} = \sum\limits_{m=1}^{M} (-1)^{m+1} \dfrac{t^{m}N^{m}}{m} $$

Also, for the same reason $e^{\log A(t)}$ is a polynomial of finite degree in $t$.

$$ e^{\log A(t)} = \sum\limits_{s=0}^{\infty} \dfrac{\left( \sum\limits_{m=1}^{M} (-1)^{m+1} \dfrac{t^{m}N^{m}}{m} \right)^{s}}{s!} = \sum\limits_{s=0}^{S} \dfrac{\left( \sum\limits_{m=1}^{M} (-1)^{m+1} \dfrac{t^{m}N^{m}}{m} \right)^{s}}{s!} $$

Now denote this by $g(t) = e ^{\log A(t)}$.

Properties of the matrix logarithm:

If $\| A - I \| \lt 1$, then $e^{\log A} = A$

Meanwhile, for sufficiently small $t$, $\| A(t) - I \| = \| tN \| \lt 1$ holds, so for infinitely many sufficiently small $t$ the following holds.

$$ g(t) = e^{\log A(t)} = A(t) $$

Therefore the two polynomials $g(t)$ and $A(t)$ are identical. Substituting $t = 1$ yields the following.

$$ e^{\log A} = A $$

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(b)

If $X$ is nilpotent, there exists $k$ satisfying $X^{k} = O$. Since $e^{X}$ is defined as below (see here), if $X$ is nilpotent then $e^{X}$ is unipotent.

$$ e^{X} = I + \sum\limits_{m = 1}^{\infty} \dfrac{X^{m}}{m!} $$

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(b')

Let $X(t) = Xt$ and $h(t) = \log e^{X(t)}$, and the proof proceeds by the same strategy as (a’).

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