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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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    I think$I$understand, thank you, @zyx .2012-04-19