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"Any unipotent matrix is similar to an upper-triangular matrix with 1's on the diagonal"...

This is usually alleged, but I have no idea how to demonstrate that, starting with the definition : $A$ is unipotent if and only if there is $k\in \mathbb{N}$ so that $(A-I_n)^k=0$.

And I browsed Internet for hints but found nothing useful. I am not looking here for a ready-made solution, but I would like to understand what is the procedure, what are the steps one has to make, in order to proceed from definition to the result I stated above.

Thanks in advance, if someone is able to detail the path to do it.

Andrew
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2 Answers2

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Here I work over the complex field $\Bbb C$.

The main steps are (1.): show that $1$ is the only possible eigevalue of $A$; (2.) cast $A$ into Jordan form. To wit:

First, look at what the condition

$(A - I_n)^k = 0 \tag 1$

reveals about the eigenvalues of $A$: that they must all be $1$, for if

$A \vec x = \lambda \vec x = \lambda I_n \vec x, \; \vec x \ne 0, \tag 2$

then

$(A - I_n) \vec x = (\lambda I_n - I_n) \vec x = (\lambda - 1)I_n \vec x = (\lambda - 1) \vec x, \tag 3$

from which we find

$(\lambda - 1)^k \vec x = (A - I_n)^k \vec x = 0; \tag 4$

now since $\vec x \ne 0$ we infer that

$(\lambda - 1)^k = 0 \Longrightarrow \lambda = 1. \tag 5$

Now, we may cast $A$ into Jordan normal form; that is, we may find a nonsingular matrix $P$ such that

$PAP^{-1} = D + N, \tag 6$

where $D$ is a strictly diagonal matrix whose diagonal entries are the eigenvalues of $A$, and $N$ is strictly upper triangular; since the only eigenvalue of $A$ is $1$, we have

$PAP^{-1} = I + N, \tag 7$

which is the requisite result. $OE\Delta$.

Robert Lewis
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You have already given us that a matrix $A$ is unipotent $\Longleftrightarrow$ $(A-I)^k=0$ for some $k \in \mathbb N$. Notice that this is an equivalent statement to saying that $A-I$ is nilpotent.

A useful characteristic of a nilpotent matrix is that if a matrix $X$ is nilpotent, then $\text{trace}(X)=0$, so it is easy to see that any matrix with 0s on the diagonal is nilpotent.

By Schur's Theorem, any $A \in \mathbb C ^{n\text{x}n}$ admits the decomposition $A=UTU^*$ where U is a unitary matrix and T is a triangular matrix.

We can manipulate this to show,

$$A-I=UTU^*-I=UTU^*-U^*IU=U(T-I)U^*,$$

$$(A-I)^k=U(T-I)^kU^*=0.$$

Thus $(A-I)$ is nilpotent when $T-I$ is nilpotent. Using our characteristic of Nilpotent Matrices we have, $$\text{trace}(T-I)=0,$$ which holds for when $t_{ii}$ entries are equal to 1.

Finally, back to Schur's Theorem we have that $A=UTU^*$, so A is (unitarily) similar to a triangular matrix with 1s on the diagonal.

Joel Biffin
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  • Thanks Joel. It looks fine, although quite different from Robert's approach. I am embarrassed by the fact that you say : when T-I is nilpotent, then A-I is nilpotent... OK, but the hypothesis is : A-I is nilpotent... Don't we need to assume the reverse : if A-I is nilpotent, then T-I is nilpotent ? – Andrew Jan 22 '19 at 10:01
  • @Andrew sorry about the wait on the response. Yes, Schur's Theorem is equality rather than an implication. In other words, if you do the substitution $V=U^$ then you have the reverse similarity transformation $T=VAV^$. Hence the (unitariy) similarity argument holds for both directions. A is similar to T if the decomposition exists and equivalently T is similar to A. – Joel Biffin Mar 07 '19 at 23:57