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I have $A=\begin{bmatrix} 4&2\\-2&0\end{bmatrix}$ and I had to show that $A=N+D$ where $N$ is nilpotent and $D$ is diagonalizable.

I then found $N=\begin{bmatrix} 2&2\\-2&-2\end{bmatrix}$ and $D=\begin{bmatrix} 2&0\\0&2\end{bmatrix}$ which is diagonal (and diagonalizable).

I now need to give an expression for $A^{n}$ and have no clue where what to do. Can I get a hint?

McLovin
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    $N$ and $D$ commute because $D = 2 I$. So you can use the usual binomial expansion and that $N^2 = 0$ to find $A^n = (N + D)^n$. Note that $D^n$ is quite easy to find because $D$ is diagonal! – M. Wang May 29 '20 at 15:09

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Observe that diagonal matrix $\;D\;$ is way more than merely diagonal: it is a scalar matrix, since $\;D=2I\;$ , and thus it commutes with all square matrices $\;2\times2\;$, and thus you can apply Newton's binomial theorem:

$$A^n=\left(N+D\right)^n=\sum_{k=0}^n\binom nk N^nD^{n-k}$$

...and now you could use what yuou know about $\;N,\,D\;$ ...

DonAntonio
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  • Might be a dumb question but do we have to know that two matrices $A,B$ commute in order to apply Newton's binomial to $(A+B)^{n}$? – McLovin May 29 '20 at 15:16
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    @Pilpel Of course, ottherwise it fails spectacularly. For example, $;A+B)^2;$ becomes $$(A+B)^2=(A+B)(A+B)=A^2+AB+BA+B^2...$$ Remember that the formula of Newton applies because we can gather together similar terms, which are all the produycts no matter their order .Here, that order matters if the matrices don't commute. – DonAntonio May 29 '20 at 15:19
  • @Pilpel In this case one of the matrices is a multiple of the identity, so they do commute. – amd May 29 '20 at 23:35