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Let $F$ be a field and let $A=\begin{bmatrix}a&0&0\\1&a&0\\0&1&a\end{bmatrix}\in\mathscr{M}_{3\times 3}(F)$. Show that $$A^k=\begin{bmatrix}a^k&0&0\\ka^{k-1}&a^k&0\\\dfrac{1}{2}k(k-1)a^{k-2}&ka^{k-1}&a^k\end{bmatrix}$$ for all $k > 0$. (Exercise 797 from Golan, The Linear Algebra a Beginning Graduate Student Ought to Know.)

I know it can be proved by induction, but since the topic is about Krylov Spaces, eigenvalues and Jordan canonical form, I wonder if there is another way to solve this problem.

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    $\begingroup$ Inducción, José! $\endgroup$ – Pedro Tamaroff May 24 '15 at 20:59
  • $\begingroup$ it is spelled induction, and I'm asking for another way, thanks. $\endgroup$ – José May 24 '15 at 21:02
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    $\begingroup$ This is a trivial matter using induction, really. I doubt there is any gain going in any other route... $\endgroup$ – Mariano Suárez-Álvarez May 24 '15 at 21:04
  • $\begingroup$ Did you try Krylov spaces ? $\endgroup$ – Dietrich Burde May 24 '15 at 21:42
  • $\begingroup$ @Dietrich Burde: what is Krylov space? Any good online refs? I've never heard of this approach for such problems . . . $\endgroup$ – Robert Lewis May 24 '15 at 22:26
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There is extended reading giving formula for calculating functions such as polynomials on matrixes: http://en.wikipedia.org/wiki/Matrix_function#Jordan_decomposition

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For a route different to induction, express the matrix as $$A=aI_3+G$$ where $I_3$ is the identity matrix and $G$ is given by $$G=\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix}$$ A binomial expansion will yield $$\begin{align}A^k&=(aI_3+G)^k\\&={k \choose 0}a^kI_3+{k \choose 1}a^{k-1}I_3G+{k \choose 2}a^{k-2}I_3G^2+\sum_{m=3}^k{k \choose m}a^{k-m}I_3G^m\\&=a^kI_3+ka^{k-1}I_3G+\frac{k(k-1)}{2}a^{k-2}I_3G^2+\sum_{m=3}^k{k \choose m}a^{k-m}I_3G^m\end{align}$$ Now $I_3G$ is given by $$G=\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix}$$ while $I_3G^2$ is $$G=\begin{bmatrix}0&0&0\\0&0&0\\1&0&0\end{bmatrix}$$ and $I_3G^m$ for $m\geq3$ is the zero matrix.

Using these results, we arrive at the given expression for $A^k$.

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  • $\begingroup$ For the binomial expansion, would not explicitly (using matrix multiplication) working out $I_3G$, $I_3G^2$, and $I_3G^3$ suffice. Once we show that $I_3G^3$ is the zero matrix, then $I_3G^m$ for $m>0$ will also be zero matrices as well. $\endgroup$ – Alijah Ahmed May 24 '15 at 21:30
  • $\begingroup$ Obviously I like your methodology! Endorsed!!! $\endgroup$ – Robert Lewis May 24 '15 at 22:29
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Set

$N = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}; \tag{1}$

then a simple calculation reveals that

$N^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix} \tag{2}$

and

$N^3 = 0. \tag{3}$

Also,

$A = aI + N. \tag{4}$

Since $N$ and $I$ commute, we may apply the ordinary binomial theorem to (4), and find

$A^k = (aI + N)^k =$ $(aI)^k + k(aI)^{k - 1}N + \dfrac{k(k - 1)}{2}(aI)^{k - 2}N^2$ $ = a^kI^k + ka^{k - 1}I^{k - 1}N + \dfrac{k(k - 1)}{2}a^{k - 2}I^{k - 2}N^2$ $ = a^kI + ka^{k - 1}N + \dfrac{k(k - 1)}{2} a^{k - 2}N^2, \tag{5}$

since the expansion is trunated after the third term by virtue of $N^3 = 0$. When (5) is written out explicitly, we see that

$A^k = \begin{bmatrix} a^k & 0 & 0 \\ ka^{k - 1} & a^k & 0 \\ \dfrac{k(k - 1)}{2} a^{ k - 2} & ka^{k - 1} & a^k \end{bmatrix}, \tag{6}$

as per request.

Nota Bene: a few words on induction. It has been rightly noted that in deploying the binomial theorem we do not in fact escape induction. The only other option I can see is to prove this result by direct matrix mutiplication, again using induction. Here we hide the induction by invoking the binomial theorem, doing so since the problem statement asks for "another way"; so I assumed direct induction should be avoided. But in truth, this is a statement concerning the integers, so induction will almost certainly enter in at some point, being as it is an essential defining property of the integers. End of Note

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  • $\begingroup$ @Dietrich Burde: Yes, yes, yes . . . take 'er easy, Pal, still editing. And I'll say a few words about induction before I'm done! Cheers! $\endgroup$ – Robert Lewis May 24 '15 at 21:33
  • $\begingroup$ Sorry, everything is good. I just think, why not prove the (easy) matrix formula - by induction. $\endgroup$ – Dietrich Burde May 24 '15 at 21:35
  • $\begingroup$ @Dietrich Burde: me too. Cheers! $\endgroup$ – Robert Lewis May 24 '15 at 21:38

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