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
 A: Observe the following computation:
$$
A^2 \;\; =\;\; \left [ \begin{array}{cc}
1 & -1 \\
2 & -2\\
\end{array} \right ]\left [ \begin{array}{cc}
1 & -1 \\
2 & -2\\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{cc}
-1 & 1 \\
-2 & 2\\
\end{array} \right ].
$$
Similarly if we compute $A^3$ we find
$$
A^3 \;\; =\;\; \left [ \begin{array}{cc}
-1 & 1 \\
-2 & 2\\
\end{array} \right ]  \left [ \begin{array}{cc}
1 & -1 \\
2 & -2\\
\end{array} \right ] \;\; =\;\;  \left [ \begin{array}{cc}
1 & -1 \\
2 & -2\\
\end{array} \right ]. 
$$
Therefore we have that $A^{2n} = -A$ and $A^{2n+1} = A$ for all $n \in \mathbb{N}$.  We conclude that 
$$
A^{100} + A^5 \;\; =\;\; -A + A \;\; =\;\; \textbf{0}_{2\times 2}.
$$
A: 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 $A^{100} + A^5 = -A + A = 0$.
A: If you know that $A^2+A=0$, which follows from the value you computed for the characteristic polynomial and is easily checked explicitly, then you can compute the value of any polynomial $P[A]$ in$~A$ by computing the remainder $R$ of $P$ by $X^2+X$, and then computing $R[A]$. In the current case you are suspecting that $P[A]=0$, which amounts to saying $R=0$ (since $R$ is of degree at most $1$, and $c_0I+c_1A=0$ only holds for the given $A$ if $c_0=c_1=0$). You could check explicitly that $X^2+X$ divides $X^{100}+X^5$.
Since in this $X^2+X$ has the easy factorisation $X(X+1)$ with easy simple roots, there is a standard trick to find the remainder $R$ without doing the whole long division: since $P=Q(X^2+X)+R$ for some quotient$~Q$, you get $P[\lambda]=R[\lambda]$ for each root$~\lambda$ of $X^2+X$, that is for $\lambda\in\{0,-1\}$. You compute $P[0]=0=P[-1]$ easily for $P=X^{100}+X^5$, so you must have $R[0]=0=R[-1]$. With $R$ of degree at most$~1$ this forces $R=0$. So indeed you can be sure that $P[A]=0$.
Of course your argument with an explicit diagonal matrix $D$ similar to $A$ is also perfectly valid.
