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Let $ A=\left(\matrix{a&b&0&0 \\ 0&a&b&0 \\ 0&0&a&b \\ 0&0&0&a }\right) $

$ \boxed{1} $ Decompose $ A $ under the form $ aI + bJ $, where $I$ and $J$ are two matrices to be determined.

$\boxed{2}$ Prove that $\exists k\in \mathbb{N}$ such that $J^k = 0.$

$\boxed{3}$ Deduce $A^n$ for all $n\in \mathbb{N} $

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5  
You probably mean $aI + bJ$. Have you tried something? –  Patrick Da Silva Jul 2 '12 at 16:22
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It says "to be determined"...but it isn't really difficult to determine what the natural choices for $I,J$ are. –  Cameron Buie Jul 2 '12 at 16:56
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Welcome to the site, by the way! It is helpful if you let us know what you've already tried and discovered--that way, we can avoid telling you something you already know, and we can give you an answer more likely to be useful to you. This looks like it may be a homework problem. If so, please add the homework tag. Don't worry, people will still help you out! –  Cameron Buie Jul 2 '12 at 17:03
    
I downvoted. OP showed no effort or attempt. –  user2468 Jul 2 '12 at 18:32
    
Hints: for property 2, check this page about nilpotent matrices. You should be able to compute the characteristic polynomial very easily in this case. –  user2468 Jul 2 '12 at 18:35

1 Answer 1

up vote 1 down vote accepted

For the first part, observe that

$\begin{eqnarray*} A & = & \left(\begin{array}{cccc}a & b & 0 & 0\\0 & a & b & 0\\0 & 0 & a & b\\0 & 0 & 0 & a\end{array}\right)\\ & = & \left(\begin{array}{cccc}a & 0 & 0 & 0\\0 & a & 0 & 0\\0 & 0 & a & 0\\0 & 0 & 0 & a\end{array}\right)+\left(\begin{array}{cccc}0 & b & 0 & 0\\0 & 0 & b & 0\\0 & 0 & 0 & b\\0 & 0 & 0 & 0\end{array}\right)\\ & = & a\left(\begin{array}{cccc}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\end{array}\right)+b\left(\begin{array}{cccc}0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\\0 & 0 & 0 & 0\end{array}\right), \end{eqnarray*}$

so (as mentioned in the comments) there is a fairly natural choice for $I$ and $J$.


For the second part, the simplest thing to do in such situations is generally to multiply $J$ by itself a few times and see what happens. (In fact, looking ahead to part $3$, you'll need to know what the various powers of $J$ are, anyway, so this is pretty much all you can do.) You shouldn't have to do this very many times, so if you don't end up with the zero matrix fairly quickly, something has gone wrong.


For the third part, it is very useful to realize that for any $4\times 4$ matrix $B$, we have $IB=B=BI$. This means in particular that multiplication by $I$ is like leaving the matrix alone, and that $I$ commutes multiplicatively with any matrix. Usually, if $B,C$ are arbitrary same-sized square matrices, we have (for example) that $(B+C)^2=B^2+BC+CB+C^2\neq B^2+2BC+C^2$, since we won't generally have $BC=CB$. However, if we happen to know that $B$ and $C$ commute, then we don't have to worry about that sort of thing, and we will have $(B+C)^2=B^2+2BC+C^2$.

Further note that if $k$ is any non-negative integer, then $(aI)^k=a^kI^k$ and $(bJ)^k=b^kJ^k$. This is nice, because it means that $(bJ)^k=0$ for sufficiently large $k$ (refer to part $2$, and note that multiplying a $4\times 4$ matrix by the $4\times 4$ zero matrix will yield the zero matrix), and that $(aI)^k=a^kI$ for all $k$ (easy induction since $IB=B=BI$ for any $4\times 4$ matrix $B$)

Finally, since everything involved will commute, we may use the binomial expansion formula $$(x+y)^n=\sum_{k=0}^n\binom n k y^kx^{n-k},$$ where $$\binom n k=\frac{n!}{k!(n-k)!}.$$ In particular, we'll use this with $x=aI$ and $y=bJ$, so that the expression ends up being much nicer (as higher powers of $y=bJ$ will vanish).


Hopefully, that will be enough to get you going. Let me know if you get stuck somewhere, and I'll see about getting you unstuck.

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