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Show that the matrix $A=\begin{bmatrix} 1 & 0 \\ 2 & 1 \\ \end{bmatrix}$ can be decomposed as a sum of a unit and nilpotent matrix.

Hence evaluate the matrix $A^{2007}$.

I read about nilpotent matrices the other day. It said "A square matrix such that is the zero matrix for some positive integer matrix power" is nilpotent.

But how to use this in this problem is beyond me.

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    $\begingroup$ If by "unit" you mean invertible (as in ring theory), the matrix itself is already a unit. If you mean the identity, then you should say so (and there is no real question, since the only decomposition possible is $A=I+N$ where $N=A-I$; indeed you find $N^2=0$ here). $\endgroup$ Feb 5, 2015 at 9:12

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$A=\begin{bmatrix} 1 & 0 \\ 2 & 1 \\ \end{bmatrix} = I + N = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 2 & 0 \end{bmatrix}$

Observe that the second matrix is nilpotent for $k = 2$.

Now apply binomial theorem:

$A^{2007} = (I + N) ^ {2007} = \sum_{i = 0}^{2007}(\binom{2007}{i} * I^{i} * N^{2007 - i})$ (Holds since the matrices commute)

and note that all the terms except for the following 2 become 0 (because $N$ is nilpotent):

$A ^{2007} = (I + N)^{2007} = \binom{2007}{2007} $ * $I^{2007}$ * $N^0$ + $\binom{2007}{1} * I^{2006} * N $

Which gives

$ A ^ {2007} = \begin{bmatrix} 1 & 0 \\ 4014 & 1 \end{bmatrix}$

Edit:

As pointed out in the comments by Prof. Marc, things work out here since the matrices commute $(A * I = I * A \ \forall A$ where $I$ is the identity matrix) . Otherwise, the binomial theorem does not hold.

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    $\begingroup$ I think one essential point is not mentioned here. The binomial formula is in general not valid for matrices. It is so only in the special case where the two matrices involved commute. Here that is OK, since one of them is the identity. But it should be mentioned. $\endgroup$ Feb 5, 2015 at 9:15
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$A=\begin{bmatrix} 1 & 0 \\ 2 & 1 \\ \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}+\begin{bmatrix} 0 & 0 \\ 2 & 0 \\ \end{bmatrix}=E+A_1$

$A_1=\begin{bmatrix} 0 & 0 \\ 2 & 0 \\ \end{bmatrix}$ is nilpotent matrix because $A_1^2=\begin{bmatrix} 0 & 0 \\ 2 & 0 \\ \end{bmatrix} \cdot\begin{bmatrix} 0 & 0 \\ 2 & 0 \\ \end{bmatrix}=\begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}$

And what about the second question. Take into account that $$A^2=\begin{bmatrix} 1 & 0 \\ 2 & 1 \\ \end{bmatrix}\cdot \begin{bmatrix} 1 & 0 \\ 2 & 1 \\ \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 4 & 1 \\ \end{bmatrix}$$ $$A^3=\begin{bmatrix} 1 & 0 \\ 4 & 1 \\ \end{bmatrix}\cdot \begin{bmatrix} 1 & 0 \\ 2 & 1 \\ \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 6 & 1 \\ \end{bmatrix}$$ $$\ldots$$ $$A^n=\begin{bmatrix} 1 & 0 \\ 2n & 1 \\ \end{bmatrix}$$

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To more easily compute a high power of a square matrix$~A$, a standard approach is to try to diagonalise it: if $A=PDP^{-1}$ for some invertible matrix$~P$ and a diagonal matrix$~D$, then you can write $A^n=PD^nP^{-1}$, and the power $D^n$ is computed by just taking the $n$-th power of each diagonal entry.

Your given matrix $A$ is not diagonalisable though, even if you allow complex coefficients (in $P$ and $D$). A more general approach, which in princilple always works if you allow complex coefficients, is to write $A$ and the sum of a diagonalisable matrix $S$ and a nilpotent matrix $N$ that commutes with it: $SN=NS$. Such a decomposition exists and is unique, and can be found after finding a Jordan normal form for$~A$. Here that is of no concern, since the decomposition is obvious: $S=I$ (which commutes with everything) and $N=\binom{0~0}{2~0}$.

Given such a decomposition, one can apply the binomial theorem to $A^n=(S+N)^n$, exactly because $S$ and $N$ commute with each other (if this were not the case, the binomial theorem would not apply to $(S+N)^n$): $$ A^n=(S+N)^n = \sum_{k\geq0}\binom nkS^{n-k}N^k. $$ The sum over $k$ can be limited to those values for which $N^k\neq0$, which are finite in number because $N$ is nilpotent. In the example $N^2=0$, so it suffices to take $k=0,1$. Also $N^0=I$ and for $S=I$ the powers $S^n,S^{n-1}$ are both the identity, and can be removed from the formula. The relevant binomial coefficients are $\binom n0=1$ and $\binom n1=n$. What remains is $A^n=I+nN$.

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You know what is a unit matrix. Then \begin{equation} A= \begin{bmatrix} 1& 0\\ 0&1\\ \end{bmatrix}+ \begin{bmatrix} 0& 0\\ 2&0\\ \end{bmatrix} \end{equation} Check what is \begin{equation} \begin{bmatrix} 0& 0\\ 2&0\\ \end{bmatrix} * \begin{bmatrix} 0& 0\\ 2&0\\ \end{bmatrix} \end{equation} and you will get the answer.

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