If A=[3 -4;1 -1],then A^n=....??

Question

anonymous · Answer

You have to find the diagonlization: $A = PDP^{-1}$ where $P$ is some matrix and $D$ is a diagonal matrix. $$ \large
\begin{split}
A^n& = \underbrace{AAA...A}_{n} \
&=  \underbrace{PDP^{-1}PDP^{-1}PDP^{-1}...PDP^{-1}}_{n} \
&= \underbrace{PDIDIDI...IDP^{-1}}_{n} \
&= PD^nP^{-1}
\end{split}
$$
Since $D$ is a diagonal matrix, it's $n$th power is trivial.

anonymous · Answer

To find $P$ and $D$, you need to eigen vectors and eigen values of the matrix $A$.  $P$ is just the eigen vectors and $D$ is just their respective eigen values in the diagonal positions.  Finding $P^{-1}$ just a matter of finding the inverse.

anonymous · Answer

for this matrix eigen values are same,its 1.So i cant find linearly independent eigen vectors to form P.And hence P is singular and cant find its inverse.