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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.

Update:

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.2012-04-15

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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 $N$"?2012-04-15
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    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.2012-04-15
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    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 such that $A^k=0$?2012-04-15
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    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.2012-04-18
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    I think I understand, thank you, @zyx .2012-04-19