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I was doing an exercise of a past exam in which one of the things I had to do was calculating the $n$th power of a Jordan matrix $$J=\begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}.$$ I started calculating until the 5th power but I couldn't guess the expression for the $a_{1,2}$, $a_{1,3}$ and $a_{2,3}$ components. When I looked it up in an online calculator it showed that:

$$J^n=\begin{pmatrix} 2^n & \frac{2^n·n}{2} & \frac{2^n·(n^2-n)}{8} \\ 0 & 2^n & \frac{2^n·n}{2} \\ 0 & 0 & 2^n \end{pmatrix}.$$

My question is: when the relationships are as difficult as these　(especially the $a_{1,3}$ component), are there any tricks for figuring out the$n$th power component? Because these kind of relationships are difficult to think in the middle of an exam.

Xander Henderson
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Alberto De Celis Romero
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3 Answers3

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Hint : $$\begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix} = 2 I+\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$$

TheSilverDoe
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  • Actually I tried calculating it this way, but I couldn't guess the relation :/ I calculated the power as $J^n=(2I+M)^n$, where $M$ is the second matrix you wrote. – Alberto De Celis Romero Sep 04 '20 at 14:15
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    @AlbertoDeCelisRomero Well, you don't have to "guess" the relation : just expand with the Newton's binomial formula (because $2I$ and $M$ commute), and see that because $M^3=0$, just three terms remain. – TheSilverDoe Sep 04 '20 at 15:07
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$J=2I+A$, where $A=\begin{pmatrix} 0&1&0 \\0&0&1 \\ 0&0&0\end{pmatrix}$. As $A$ and $I$ commute, we can apply the binomial formula.

Observe that $A^2=\begin{pmatrix} 0&0&1 \\0&0&0 \\ 0&0&0\end{pmatrix}$, $A^k=0$ for $k\ge 3$, so the $n$-th power is $$J^n=2^n I+n2^{n-1}A+\frac{n(n-1) 2^{n-2} }2A^2=2^n I+n2^{n-1}A+n(n-1) 2^{n-3}A^2. $$

Bernard
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  • Oh sorry didn't see the answer. Thank you very much for your help! It's very useful! So you can only do this when the matrices commute, right? – Alberto De Celis Romero Sep 05 '20 at 02:24
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    @AlbertoDeCelisRomero: yes. More generally, in a ring, you can apply the binomial formula if the elements commute. – Bernard Sep 05 '20 at 07:44
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Hint:

Expand

$$\begin{pmatrix} a_{n+1} & b_{n+1} & c_{n+1} \\ 0 & d_{n+1} & e_{n+1} \\ 0 & 0 & f_{n+1} \end{pmatrix}=\begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix}\begin{pmatrix} a_n & b_n & c_n \\ 0 & d_n & e_n \\ 0 & 0 & f_n \end{pmatrix}$$

and contemplate the six linear recurrence relations you obtain. ($a,d,f$ are immediate.)

Additional hint:

The characteristic polynomial is $(2-\lambda)^3$ so the matrix has a triple Eigenvalue $2$, and you can expect the expressions of the elements of the powers to be in the form of $2^n$ times a quadratic polynomial in $n$.

  • I haven't heard about the additional hint you have said before. Is there any theorem I can look for to see when is this true? It can help a lot in future exercises!

    From the relation $a_{n+1}=2a_n$ I suppose that $a_n=d_n=f_n=2^n$ right? Nevertheless for $b_{n+1}$ the relation is $b_{n+1}=2b_n+d_n=2(b_n+2^{n-1})$, which is only true for $n=4$. For $n=3$ we would have $12=2(4+4)$ which is not true, and for $n=5$ we have $80=2(32+2^4)$, which is also false. Am I having a mistake here?

     – Alberto De Celis Romero Sep 04 '20 at 14:13
  • Ah, I just noticed a mistake in it XD, it's true for all $n$ – Alberto De Celis Romero Sep 04 '20 at 14:23
  • But still I don't understand well how to get the expression $2^{n-1}n$ for $b_n$ :/ – Alberto De Celis Romero Sep 04 '20 at 14:24