If $A =\left[ \begin{smallmatrix} 3 & -4 \\ 1 & -1\end{smallmatrix}\right]$, then find $A^n$ 
If $$ A = \begin{bmatrix}
3 &  -4 \\
1 & -1 
\end{bmatrix}$$
Then find $A^n$

I have tried solving it using diagonalization $PDP^{-1}$ but I am getting only one independent eigenvector i.e $$ \begin{bmatrix}
2\\
1
\end{bmatrix}$$
Please tell me the correct method to solve it.
 A: You can split the matrix $A$ into $B+ I$ where $B=  \begin{bmatrix}
2 &  -4 \\
1 & -2
\end{bmatrix}$
Now it is interesting to observe that $B^2=O$, the null matrix
So using binomial
$A^n=(I+B)^n= C^n_0 I^n + C^n_1 I^{n-1} B + C^n_2 I^{n-2} B^2 +\cdots+ C^n_n B^n$
Which simplifies to:
$A^n= I + nB$ :)
Note: we are able to use binomial this way  because, one of the matrix is identity matrix and the product becomes commutative.
A: This matrix is not diagonalizable. Start multiplying and look for patterns. E.g.,
\begin{align*}
\begin{bmatrix} 3 & -4 \\ 1 & -1\end{bmatrix}^2 &= \begin{bmatrix} 5 & -8 \\ 2 & -3\end{bmatrix} \\
\begin{bmatrix} 3 & -4 \\ 1 & -1\end{bmatrix}^3 &= \begin{bmatrix} 7 & -12 \\ 3 & -5\end{bmatrix} \\
&\vdots
\end{align*}
Does this give you any ideas?
A: $$|A-\lambda I| = (3-\lambda)(-1-\lambda) - 1\times(-4)\\= 4-3-2\lambda+\lambda^2 \\= 1-2\lambda+\lambda^2=\\(1-\lambda)^2$$
so $\lambda =1$ double root.
Which eigenspace?
$$\text{null}(A-\lambda I)\\\begin{bmatrix}2&-4\\1&-2\end{bmatrix}$$
We see $[2,1]^T$ is eigenvector but also there can be only one because of degeneracy.
Therefore $A$ is not diagonalizable. You have to try find some other type of matrix than diagonal matrix for $D$ to be if you want to put $A$ on form $PDP^{-1} =A$. For example you can try 


*

*Jordan Canonical Form or 

*Smith's form
A: Since you have a defective eigenvalue, to proceed with your method you’ll need to find a generalized eigenvector and put the matrix into Jordan normal form. However, there’s no need to go to all that trouble since powers of the matrix can be computed without working out the full decomposition.  
Using the same method as the one in my answer to another of your questions (in fact, the two problems are congruent) we write $A^n=aI+bA$ and solve the system of equations generated by substituting $A$’s eigenvalues for $A$: $$a+b=1 \\ b=n,$$ (the second equation is obtained by differentiating $a+b\lambda=\lambda^n$) therefore, as in your other question, $A^n=nA-(n-1)I$.  
More generally, if the repeated eigenvalue is $\lambda$ and $A$ is not a multiple of the identity, we get the system of equations $$a+b=\lambda^n \\ b=n\lambda^{n-1}$$ so $$A^n=(1-n)\lambda^n I+n\lambda^{n-1} A = \lambda^n I+n\lambda^{n-1}(A-\lambda I).$$ From Cayley-Hamilton, we know that $(A-\lambda I)^2=0$, so this result can also be obtained by writing $A$ as the sum of the identity and the nilpotent matrix $N=A-I$ and then applying the Binomial Theorem, as has been suggested in other answers. This is much less work that computing generalized eigenvectors and then performing tedious matrix multiplications to find $A^n$.
A: Indeed, $A$ has one and only one eigenvalue: $1$. And $(2,1)$ is an eigenvector corresponding to that eigenvalue. Now, consider the vector $(1,0)$. Then$$A.(1,0)=(3,1)=(2,1)+(1,0).$$So, if $M=\left[\begin{smallmatrix}2&1\\1&0\end{smallmatrix}\right]$, then$$M^{-1}.A.M=\begin{bmatrix}1&1\\0&1\end{bmatrix},$$or$$A=M.\begin{bmatrix}1&1\\0&1\end{bmatrix}\cdot M^{-1}.$$So,\begin{align}A^n&=M.\begin{bmatrix}1&1\\0&1\end{bmatrix}^n.M^{-1}\\&=\begin{bmatrix}2&1\\1&0\end{bmatrix}.\begin{bmatrix}1&n\\0&1\end{bmatrix}.\begin{bmatrix}0&1\\1&-2\end{bmatrix}\\&=\begin{bmatrix}2n+1&-4n\\n&1-2n\end{bmatrix}.\end{align}
Note that the equality $\left[\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right]^n=\left[\begin{smallmatrix}1&n\\0&1\end{smallmatrix}\right]$ has to do with the fact that$$(\forall x,y\in\mathbb R):\begin{bmatrix}1&x\\0&1\end{bmatrix}.\begin{bmatrix}1&y\\0&1\end{bmatrix}=\begin{bmatrix}1&x+y\\0&1\end{bmatrix}.$$
