Finding a 2x2 Matrix raised to the power of 1000 Let $A= \pmatrix{1&4\\ 3&2}$. Find $A^{1000}$.
Does this problem have to do with eigenvalues or is there another formula that is specific to 2x2 matrices?

Two additional methods that you can use once you know the eigenvalues $\lambda_1$ and $\lambda_2$:


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*Decompose $A$ into $\lambda_1P_1+\lambda_2P_2$, where $P_1 = {A-\lambda_2I\over\lambda_2-\lambda_1}$ and $P_2={A-\lambda_1I\over\lambda_1-\lambda_2}$ are projections onto the corresponding eigenspaces with $P_1P_2=P_2P_1=0$. If you expand $A^{1000}$ using the binomial theorem, you’ll find that all but two terms vanish, giving $\lambda_1^{1000}P_1+\lambda_2^{1000}P_2$.

*Use the Cayley-Hamilton theorem to write $A^{1000}=aI+bA$ for some undetermined coefficients $a$ and $b$, then use the fact that this equation is also satisfied by the eigenvalues, which gives you the system of linear equations $a+b\lambda_i=\lambda_i^{100}$ to solve for $a$ and $b$.  


When $A$ has repeated eigenvalues, you’ll need to modify the above methods a bit, but the underlying ideas are still the same.
