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$$A=\begin{bmatrix}1 & 0&0\\1 & 0&1\\0&1&0\end{bmatrix}$$
Find $A^{50}$ ?

Now from Cayley–Hamilton theorem, I get $A^3-A^2-A+I=0$ and $A^{50}=(A^4)^{12}A^2$ so I found $A^4$ which is $-2A-I$, then we have $A^{50}=B^{12}A^2$ where $B =A^4$ was calculated, now should I again use Cayley–Hamilton theorem to find $B^{12}$ or is there a better possibility?

ketan
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2 Answers2

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Let $f(x) = x^3-x^2-x-1 = (x^2-1)(x-1) = (x-1)^2(x+1)$.
Since $f(A) = 0$, we want to find polynomial $q(x), r(x)$ such that $$x^{50} = q(x)f(x) + r(x)\tag{*1}$$ with $\deg r(x) \le \deg f(x) - 1 = 2$. If we can figure out what is $r(x)$, then

$$A^{50} = q(A)f(A) + r(A) = r(A)$$

Write $r(x)$ as $a x^2 + b x + c$. To fix the coefficients, evaluate both side of $(*1)$ at $1$ and $-1$ and the derivative at the double root $1$, we get:

$$ \begin{cases} 1 &= a + b + c\\ 1 &= a - b + c\\ 50 &= 2a + b \end{cases} \quad\implies\quad \begin{cases} a &= 25\\ b &= 0\\ c &= 24\\ \end{cases} \quad\implies\quad r(x) = 25 x^2 - 24. $$ As a result,

$$A^{50} = 25 A^2 - 24 I = 25 \begin{pmatrix} 1 & 0 & 0\\ 1 & 1 & 0\\ 1 & 0 & 1 \end{pmatrix} - 24 \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0\\ 25 & 1 & 0\\ 25 & 0 & 1 \end{pmatrix} $$

achille hui
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    from where have you learnt this step:-> we want to find polynomial q(x),r(x) such that $x^{50} = q(x)f(x) + r(x)$.would you tell more regarding this step? plz add some basic explanation to first 2 steps. – ketan Mar 29 '14 at 18:36
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    @ketan, Cayley Hamilton theorem tell us we can reduce $A^{50}$ to a quadratic polynomial in $A$. To carry out the reduction, one need to subtract suitable multiple of $f(A)$ from $A^{50}$. If you think carefully, what we need is the properties of polynomials instead of anything that $A$ is an matrix. So we temporarily forget $A$ is a matrix and look at everything from the angle of polynomials. Once we are done, it will automatically works for the matrix $A$. Think of this as a general tactic to solve algebraic problem using correspondence between different types of algebraic structures. – achille hui Mar 29 '14 at 18:56
  • @ketan That is long division of polynomials. – OR. Mar 29 '14 at 18:57
  • $\deg r(x) \le \deg f(x) - 1 $ this is still mystery! how come this sir? also can this process be generalised? – ketan Mar 29 '14 at 19:13
  • @ketan if you have a $n \times n$ matrix $B$, Cayley Hamilton theorem tell us $B^n$ can be expressed in terms of $I,B,\ldots,B^{n-1}$. This implies for any $B^m$ with $m \ge n$, one can express $B^m$ in terms of $B^{m-n}, B^{m-n+1},\ldots,B^{m-1}$. If one repeat these steps, one find every $B^{m}$ with $m \ge n$ can be expressed in terms of $I,B,\ldots,B^{n-1}$. Translate this to polynomial, the degree of the remainder $r(x)$ is at most $n-1$. In our case of $A$, $n = 3$ and so degree of $r(x)$ is at most $2$. – achille hui Mar 29 '14 at 19:22
  • this seems a powerful way to deal with such probs. Thank you very much, sir. my good wishes to you. – ketan Mar 30 '14 at 03:04
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If you do a few calculations: \begin{equation} A^2= \begin{bmatrix} 1&0&0\\1&1&0\\1&0&1 \end{bmatrix} \end{equation}\begin{equation} A^4= \begin{bmatrix} 1&0&0\\2&1&0\\2&0&1 \end{bmatrix} \end{equation}\begin{equation} A^6= \begin{bmatrix} 1&0&0\\3&1&0\\3&0&1 \end{bmatrix} \end{equation} ...so then from here you can deduce the answer...

On a more rigorous note: note that $A^2=\begin{bmatrix} 1&0&0\\1&1&0\\1&0&1 \end{bmatrix} =(\begin{bmatrix} 0&0&0\\1&0&0\\1&0&0 \end{bmatrix}+I)$.

Now $\begin{bmatrix} 0&0&0\\1&0&0\\1&0&0 \end{bmatrix}^2 =0$.

So that $(\begin{bmatrix} 0&0&0\\1&0&0\\1&0&0 \end{bmatrix}+I)^{n}=\binom{n}{n-1} \begin{bmatrix} 0&0&0\\1&0&0\\1&0&0 \end{bmatrix}^1I^{n-1}+I^{n}=n\begin{bmatrix} 0&0&0\\1&0&0\\1&0&0 \end{bmatrix}I+I$.

(This last step is just using the binomial theorem combined with the result directly above it. You can write out the full binomial expansion and then just reduce to zero all higher powers of the nilpotent matrix.)

So then $A^{50}=(A^2)^{25}$, so that $n=25$ and the result follows.

Christiaan Hattingh
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