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Let $A= \begin{bmatrix} 1& -1 & 1\\ 0 & 1 & 1 \\ 0 & 0 & 1\\ \end{bmatrix}$. Compute $A^k$.

My attempt

I'm trying to compute $A^k$ using this approach as follows: $$ A=I+N= \begin{bmatrix} 1& 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}+ \begin{bmatrix} 0& -1 & 1\\ 0 & 0 & 1 \\ 0 & 0 & 0\\ \end{bmatrix} $$ with $$ N^2= \begin{bmatrix} 0& 0 & -1\\ 0 & 0 & 0 \\ 0 & 0 & 0\\ \end{bmatrix}, \, \text{and} \, \, N^3= \begin{bmatrix} 0& 0 & 0\\ 0 & 0 & 0 \\ 0 & 0 & 0\\ \end{bmatrix} $$

Then, $$ A^2=(I+N)^2=I+2N+N^2, \\ A^3=(I+N)^3=I+3N+3N^2, \\ A^4=(I+N)^4=I+4N+6N^2, \\ A^5=(I+N)^5=I+5N+10N^2, \\ A^6=(I+N)^5=I+6N+15N^2, $$

By induction, we can see $A^k=(I+N)^k=I+kN+f[k]N^2$. But, I couldn't figure out what $f[k]$ is. Any help?

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    $\begingroup$ binomial coefficient $k$ choose $2,$ which become $k(k-1)/2$ $\endgroup$ – Will Jagy Sep 24 '18 at 0:30
  • $\begingroup$ @WillJagy Exactly! It worked. Thank you. $\endgroup$ – Lod Sep 24 '18 at 0:46
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How about this. Take the exponential function $e^{tA}$, where $t$ is some parameter $$e^{tA}=\sum_{k=0}^\infty\frac{t^kA^k}{k!}=e^{t(I+N)}= e^{tI}e^{tN}=e^t\left[I+tN+\frac{(tN)^2}{2}\right]$$ where we used the matrix identity $e^{A+B}=e^Ae^B$ that is valid when matrices $A$ and $B$ commute. Since the functions $t^k$ are linearly independent, we obtain $$A^k=I+kN+\frac{k(k-1)}{2}N^2$$

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  • $\begingroup$ That's even a better compact approach. However, I did not get how $t^2$ became $k(k-1)$. What is it that you mean with "the functions $t^k$ are linearly independent"? $\endgroup$ – Lod Sep 24 '18 at 1:26
  • $\begingroup$ @Lod, you have to expand the exponential on the right side, the coefficients of $t^k$ on both sides of the equation have to be equal. The linear independence you can see from this: If you assume $\sum_{k=0}^\infty a_kt^k=0$ is valid for all $t$, this implies all $a_k=0$. To see this just take the $k$-th order derivative and then replace $t=0$. $\endgroup$ – minmax Sep 24 '18 at 1:37
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You have written $A=I+N$, and you know that $N^3$ (and hence all higher powers) are zero. If $X$ and $Y$ are two matrices that commute with each other, then you can still use the bionomial theorem: $$(X+Y)^n=\sum_{i+j=n}\binom{n}{i} X^i Y^j$$

Because $I$ commutes with $N$, and because all the higher powers of $N$ vanish, we can apply the formula to get

$$(I+N)^n=\sum_{i+j=n}\binom{n}{j} N^j=I+nN+\binom{n}{2}N^2$$

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