<s>[INST] Find $A^{100}+A^5$ Let $A= \begin{pmatrix} 1&-1 \\ 2&-2 \\  \end{pmatrix}$. We are to find $A^{100} +A^5$
My attempt : I found characteristic equation , $x^2+x=0$, so eigenvalues are $0$ and $-1$ hence its diagnol matrix is $$D=  \begin{pmatrix} 0&0 \\ 0&-1 \\  \end{pmatrix}$$  So $D^{100}+D^5$ is null matrix.
But I am getting something else if I use method given here To Find $A^{50}$ by achille hui.
Please help in this matter. Thanks
 [/INST] If its characteristic equation is $x^2 + x = 0$, then that means that $x^2 = -x$, ie, $A^2 = -A$.
Thus, $A^{100} = (A^2)^{50} = (-A)^{50} = A^{50} = (A^2)^{25} = (-A)^{25} = -A^{25}$
At this point you've got an odd power, so simply factor out $-A$, so
$$\cdots = -A\cdot A^{24} = -A\cdot (A^2)^{12} = -A\cdot (-A)^{12} = -A\cdot A^{12} = -A\cdot (A^2)^6 = -A\cdot(-A)^6 = -A\cdot A^6$$
$$\cdots = -A\cdot (A^2)^3 = -A\cdot (-A)^3 = (-A)^4 = A^4 = (A^2)^2 = (-A)^2 = A^2 = -A$$
Thus, this shows that $A^{100} = -A$.
Doing the same thing with $A^5$ gives $A^5 = A\cdot (A^2)^2 = A\cdot (-A)^2 = A\cdot A^2 = A\cdot (-A) = -A^2 = -(-A) = A$.
Thus, indeed $A^{100} + A^5 = 0$.
This was the dumb way, but it's a method that's easily generalizable.
In this case, it's easier if you consider the powers of $A$:
$A^2 = -A$
$A^3 = A\cdot A^2 = A\cdot -A = -A^2 = A$
$A^4 = A\cdot A^3 = A\cdot A = A^2 = -A$
at which point you'll notice that $A^k = A$ if $k$ is odd, and otherwise $A^k = -A$ if $k$ is even. Thus, $A^{100} = -A$ and $A^5 = A$, so $A^{100} + A^5 = -A + A = 0$.
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