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If $$M=\begin{pmatrix} 1 & 1 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\\ \end{pmatrix}$$ is a matrix and $e^M=\sum_\limits{n=0}^{\infty}\frac{M^n}{n!}$ then find $\frac{1}{e}\sum_\limits{i=1}^{3}\sum_\limits{j=1}^{3} b_{ij}$ where $b_{ij}$ are entries of $e^M$.

I know the definition of $e^M$ and how to find $e^M$ for diagonalisable matrix, nilpotent matrix but I've no idea how to solve this please give some hints to solve this problem. Thanks in advance.

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    try using exponentials of the Jordan blocks one by one..in your case the matrix is itself a a Jordan block of $3×3$. – 2017-01-28

2 Answers 2

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We can write $M=I+N$ where

$$ I = \begin{pmatrix}1&0&0 \\ 0&1&0 \\0&0&1 \end{pmatrix} \qquad N = \begin{pmatrix}0&1&0 \\ 0&0&1 \\0&0&0 \end{pmatrix} $$

Since $I$ is the identity, $N$ and $I$ commute; hence $e^{I+N} = e^I e^N$ and $N$ is nilpotent with index $3$. So, $N^3=0$

$$ e^I e^N = eI\cdot(I+N+\frac{1}{2}N^2) = e\begin{pmatrix}1&1&\frac{1}{2} \\ 0&1&1 \\0&0&1 \end{pmatrix}$$

Thus the sum of the entries of the matrix $M$ is $\displaystyle 5.5e$.

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    Therefore the answer is 5.5. Thank you – 2017-01-28
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    Is this result is true for any matrix? – 2017-01-28
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    For any matrix can be expressed as M=D+N where D is diagonalisable and N is nilpotent matrix. – 2017-01-28
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    yes.you are right.i.e matrix $M$ in Jordan-Normal-Form. – 2017-01-28
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    Okay. Jordan normal form can be found by characteristic polynomial and minimal polnomial. So, if we don't about matrix and we have only ch(t) and m(t) for the above Jordan canonical form then we can find the same. – 2017-01-28
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    Oh! that's great result... Without knowing a matrix we can find expnential of matrix. Thanks. – 2017-01-28
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    Another way to solve :$ \lambda=1$ is the only eigenvalue of $M$, hence $(M-I)^3=0$. Then: $\frac{1}{e}e^M=e^{M-I}=M+\frac{1}{2}(M-I)^2$ – 2017-01-28
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HINT

You know how to diagonalize a matrix, that is, to find a diagonal matrix $D$ and an orthogonal matrix $P$ (which will be the matrix whose columns are the eigenvectors of $M$) such that $$ M = PDP^{-1} $$ ($D$ will have the eigenvalues of $M$ on the diagonal elements.)

Then for all $k$, $$ M^k = PD^kP^{-1} $$ since pairs of $P^{-1}P$ cancel out except at the left and right sides of the product.

Finally, notice how this lets you relate $e^M$ to $P e^D P^{-1}$ since you know how to exponentiate the diagonal matrix $D$.

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    This works for diagonalizable matrices. But here, the problem is that $M$ which visibly is a Jordan block, is **not** diagonalizable... – 2017-01-28