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I am new here so please let me know if I must resentence the exercice. I considered it too short not to include it in the title too.

Let $ A= \begin{pmatrix} a & b & c \\ 0 & a & b \\ 0 & 0 & a \\ \end{pmatrix}$

Find $A^n$

I tried finding the matrix's first 4 or 5 powers and it looks like a pettern can be noticed but I can't seem to find a form for the element on line 3 row 3. I haven't tried using a binomial expansion though, writing the matrix as $$ \begin{pmatrix} 0 & b & c \\ 0 & 0 & b \\ 0 & 0 & 0 \\ \end{pmatrix}+ a \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix} $$

the second one being the identity matrix times $a.$

Michael Hardy
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Dareal1 62
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  • Why is it $A(x)$ and not just $A$? How does $x$ come into play? – lisyarus Oct 31 '20 at 19:30
  • Sorry, i was still thinking about my previous exercice – Dareal1 62 Oct 31 '20 at 19:34
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    You should begin by noticing that $$\begin{bmatrix} 0 & b & c \ 0 & 0 & b \ 0 & 0 & 0 \end{bmatrix}^{\large3} = \text{the $3\times3$ zero matrix}$$ and $$ \left( a \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \right)^{\large n} = a^n \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} $$ – Michael Hardy Oct 31 '20 at 19:44

2 Answers2

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Hint

If $$N=\begin{pmatrix}0&b&c\\0&0&b\\0&0&0\end{pmatrix},$$ then $$A=aI+N$$

(where $I$ is the identity matrix), and $N^3=0$. Applying Binomial theorem yields $$A^n=(aI+N)^n=\sum_{k=0}^n\binom{n}{k} a^{n-k}N^k=\cdots$$

Surb
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  • I mentioned taking this method into consideration at the very end but thanks for letting me know it's the right one – Dareal1 62 Oct 31 '20 at 19:32
  • @Dareal162: Well, you can still compute $A^n$ brute force if you wish, but this will be very long... and if you wish an other method, as far as I know, there are none... – Surb Oct 31 '20 at 19:38
  • Sorry if i'm being annoying but $N^4=0$, not $N^3$ – Dareal1 62 Oct 31 '20 at 19:44
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    You are obviously not annoying (and I can make mistake, so it's a good thing to mention them). So, I agree that $N^k=0$ for all $k\geq 3$, and thus, as you mentionned, $N^4=0$. But $N^3=0$ as well. In general, if $N\in \mathcal M_{n\times n}(\mathbb R)$ is nilpotent, then $N^n$ must holds.@Dareal162 – Surb Oct 31 '20 at 19:46
  • Oh yeah, now i see... damn you calculus... – Dareal1 62 Oct 31 '20 at 19:48
  • @Surb : You seem to be quite active here so it should be worth the trouble to notice this typographical difference: $$ \begin{align} & \sum_{k=0}^n \binom n o a^{n-k} N^k = ... \ {} \ \text{versus } & \sum_{k=0}^n \binom n o a^{n-k} N^k = \ldots \ {} \ \text{versus } & \sum_{k=0}^n \binom n o a^{n-k} N^k = \cdots \end{align} $$ The first one is akin to a spelling error; that's one of the reasons why \ldots and \cdots exist. Another is that in LaTeX (as opposed to MathJax, which is used here if you write $a+b=...$ then you see [continued below] $\qquad$ – Michael Hardy Oct 31 '20 at 19:56
  • $\qquad$something like $a+b=\text{...},$ i.e. you see neither $a+b=...$ nor $a+b=\ldots$ nor $a+b=\cdots. \qquad$ – Michael Hardy Oct 31 '20 at 19:58
  • . . . and it's worth being aware of the following: $$ \begin{align} & a+b+\text{...}+z \ {} \ \text{versus } & a+b +...+z \ {} \ \text{versus } & a+b+\ldots+z \ {} \ \text{versus } & a+b+\cdots+z \end{align} $$ The first two above are incorrect usages by most standards. The first one is similar to what you get in full-fledged LaTeX if your code says a+b+...+z. $\qquad$ – Michael Hardy Oct 31 '20 at 20:00
  • @MichaelHardy: Thanks for the remark. On latex I always use \hdots, which doesn"t work here, and I didn't knew there were an other way to write it. So, thanks for that. – Surb Oct 31 '20 at 20:04
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Let $X=aI,Y=bJ$ and $Z=cJ^2$, where $J$ denotes the $3\times3$ nilpotent Jordan block. Then $X,Y,Z$ commute. Since $J^3=0$, among all degree-$n$ monomials of them, only $X^n,\,X^{n-1}Y,\,X^{n-1}Z$ and $X^{n-2}Y^2$ are nonzero. It follows that \begin{aligned} A^n &=(X+Y+Z)^n\\ &=X^n+nX^{n-1}Y+nX^{n-1}Z+\binom{n}{2}X^{n-2}Y^2\\ &=\pmatrix{a^n&na^{n-1}b&na^{n-1}c+\binom{n}{2}a^{n-2}b^2\\ 0&a^n&na^{n-1}b\\ 0&0&a^n}. \end{aligned}

user1551
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