Suppose the characteristic polynomial of $A=\pmatrix{a&b\\c &d} $ is $(\lambda-1)^2$. Compute $A^{2011}-2011A$.
I think this is about diagonalization of $A$. What I got is only $ad-bc=1$ and $a+d=2$. I tried to find its eigenvector to diagonalize $A$. But the expression it too complex.
On
Since the characteristic polynomial of $A$ is $(\lambda-1)^2$, $A$ is similar to$$B=\begin{bmatrix}1&1\\0&1\end{bmatrix}.$$Therefore, $A^{2\,011}-2\,011A$ is similar to$$B^{2\,011}-2\,011B=\begin{bmatrix}-2\,010&0\\0&-2\,010\end{bmatrix}.$$Since this is just $2\,010\operatorname{Id}$,$$A^{2\,011}-2\,011A=\begin{bmatrix}-2\,010&0\\0&-2\,010\end{bmatrix}.$$
On
Hint: Write $\lambda^{2011}-2011\lambda=q(\lambda)(\lambda-1)^2+a\lambda+b$. Find $a,b$ by setting $\lambda=1$ in this equation and in its derivative.
On
characteristic equation is $(\lambda-1 )^2=0$ or $(\lambda^2-2\lambda+1)=0$.
So $A^2-2A+I=0\implies A^2-A=A-I$
$A^2=2A-I$
$A^3=2A^2-A=A^2+A-I=3A-2I$
$A^4=3A^2-2I=A^2+2(A-I)=4A-3I$
$.\ .\ .$
$A^n=nA-(n-1)I$
for $n=2011,\ A^{2011}-2011A=-2010I$
On
If $A = I$ then $A^{2011}-2011A = -2010I$.
Otherwise, $A$ has Jordan form $J = \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}$.
For any entire function $f : \mathbb{C} \to \mathbb{C}$ we have
$$f(J) = \begin{bmatrix} f(1) & f'(1) \\ 0 & f(1) \end{bmatrix}$$
In particular for $f(z) = z^{2011} - 2011z$ we have $f'(z) = 2011z^{2010}-2011$ so
$$f(J) = \begin{bmatrix} 1^{2011} - 2011\cdot 1 & 2011\cdot 1^{2010} - 2011 \\ 0 & 1^{2011} - 2011\cdot 1 \end{bmatrix} = \begin{bmatrix} -2010 & 0 \\ 0 & -2010\end{bmatrix} = -2010I$$
so again $f(A) = -2010I$.
If you look at the Jordan form, there are two possibilities. Either $J=I$, or $$J=\begin{bmatrix} 1&1\\0&1\end{bmatrix}.$$ In the first case, $A=I$, so $$A^{2011}-2011A=-2010 I.$$In the other case, $$ J^n=\begin{bmatrix}1&n\\ 0&1\end{bmatrix}, $$ so $$ J^{2011}-2011 J=\begin{bmatrix} -2010&0\\0&-2010\end{bmatrix}=-2010 I. $$ So in both cases we get $$A^{2011}-2011 A=-2010 I.$$