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Let $$ N=\begin{pmatrix}0&1&&\\&\ddots&\ddots&\\&&0&1\\&&&0 \end{pmatrix}_{n\times n} $$ and $I$ is the identity matrix of order $n$. How to prove $I+N\sim e^N$?

Clarification: this is the definition of similarity, which is not the same as equivalence.


I noticed a stronger relation, that $A\sim N$, if $$ A=\begin{pmatrix}0&1&*&*\\&\ddots&\ddots&*\\&&0&1\\&&&0 \end{pmatrix}_{n\times n} $$ and $*$'s are arbitrary numbers.

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Now you can answer your question. This is explicitly encouraged. –  Norbert Apr 15 '12 at 0:34

1 Answer 1

up vote 1 down vote accepted

By subtracting $I$ this is equivalent to asking about the similarity class of a nilpotent square matrix of size $N$. The similarity type of $N$ is determined by the dimensions of the kernels of powers of $N$. In the upper triangular case the list of dimensions of $\ker N^i$ is $1,2,3,4,...,n$ for both of the matrices you consider. Hence they are similar.

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Can you give more words on "The similarity type of $N$ is determined by the dimensions of the kernels of powers of $N$"? –  ziyuang Apr 15 '12 at 1:09
Some subspace is killed by $N$, but a possibly larger space is killed by $N^2$, and a larger one by $N^3$. Continue until you get the null space of $N^n$ which is the vector space $V$ on which $I$ and $N$ act. If you choose (compatible) bases for all these kernels you will see that it gives a basis for $V$ and the same exercise for another operator $N'$ with the same diagram of kernel dimensions leads to another basis. Changing from one basis to the other is the similarity. It is easier to draw a picture than describe in words. –  zyx Apr 15 '12 at 1:14
Why won't the "chain" of $A^i$ terminate before $i$ reaches $n$ in the triangle case? I mean, why is it impossible that there exist a $k<n$ such that $A^k=0$? –  ziyuang Apr 15 '12 at 12:15
For both of the matrices you consider, the elements immediately above the diagonal are nonzero. You can find an ordered basis where the action is $Nv_i = a_i v_{i-1} + \ldots$ and all $a_i$ are nonzero. –  zyx Apr 18 '12 at 20:51
I think I understand, thank you, @zyx . –  ziyuang Apr 19 '12 at 0:01

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