How to finds the smallest $n$ such that $I=A^0, A^1, A^2, \dots, A^n$ are linearly dependent? 
Let $A$ be the following $2\times 2$ matrix:
$$A=\begin{pmatrix}1&2\\3&4\end{pmatrix}$$
Find the smallest value of $n$ such that matrices $I=A^0, A^1, A^2, \dots, A^n$ are linearly dependent.

I don't quite know how to begin to answer this question since so far we have only defined linear independance for vectors, and these are matrices...

Calculate the powers of $A$ upto $A^2$ and write them in a row:
$$\underbrace{\begin{pmatrix} \fbox{1} & 0 \\ 0 & 1 \end{pmatrix}}_{I}, \underbrace{\begin{pmatrix} \fbox{1} & 2 \\ 3 & 4 \end{pmatrix}}_{A},\underbrace{\begin{pmatrix} \fbox{7} & 10 \\ 15 & 22 \end{pmatrix}}_{A^2}$$
Use the coefficient at the position $(1,1)$ in the matrix $I$ to annihilate the coefficients at $(1,1)$ in $A$ and $A^2$:
$$\underbrace{\begin{pmatrix} 0 & 2 \\ 3 & 3 \end{pmatrix}}_{A-I},\underbrace{\begin{pmatrix} 0 & 10 \\ 15 & 15 \end{pmatrix}}_{A^2-7I}$$
Now notice that $A^2-7I = 5(A-I)$. Therefore:
$$0 = A^2-7I - 5(A-I) = A^2-5A-2I$$
Since $A$ is obviously not a multiple of $I$, we conclude that the smallest $n$ such that $\{I, A, \ldots, A^n\}$ is lineraly dependent is $n = 2$.
