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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 $n$, can anyone show me how to prove $I+N\sim e^N$? Thank you~

Further, can this be generalized to exponential mapping or something like this?

Clarification: Terms may differ from place to place, but this is the definition of similarity here, which is not the same as equivalence. Those who wants to verify the similarity can use the following Mathematica snippet

NilpotentMatrix[n_] := Table[Boole[j == i + 1], {i, n}, {j, n}]; JordanDecomposition[MatrixExp[NilpotentMatrix[#]]][[2]]& [AnyOrderYouWant] //MatrixForm

to see whether the Jordan canonical form of $e^N$ is equal to $I+N$.

Update 2: $I$ subtracted.

Update: By using

JordanDecomposition[Table[Boole[i <= j <= i + 1] +10 Boole[j >= i + 2] (RandomReal[] - 0.5), {i, #}, {j, #}]][[2]] &[AnyOrderYouWant] //MatrixForm

I noticed a stronger proposition, that $A\sim N$, where $$ A=\begin{pmatrix}0&1&*&*\\&\ddots&\ddots&*\\&&0&1\\&&&0 \end{pmatrix}_{n\times n} $$ and $*$'s stand for an arbitrary triangle.

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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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