Linear independence of matrices $I, A, A^2$ I want to prove that $I,A,A^2\:$matrices $\in M_{2\times 2}$ are $\textit {linearly independent}$. 
I consider the following matrices and their "corresponding" vectors:
$I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\to (1,0,0,1)$
$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}\to (a,b,c,d)$
$A^2=\begin{pmatrix} a^2+ b c & ab+db \\ ac+dc & d^2+bc\end{pmatrix}\to (a^2+bc, ab+db, ac+dc,d^2+bc)$
What I am trying to do, is to prove that these vectors are $\textit{linearly independent}$. 
For that direction, I want to prove that (at least) one $3\times 3$ determinant is $\neq 0$. 
It seems that all $3\times 3$ determinants $\left( e.g. \begin{vmatrix} 1 & 0 & 0\\ a & b & c\\ a^2+bc & ab+db & ac+dc\end{vmatrix} \right)$ are equal to zero! What am I really missing? 
 A: The thing you're trying to prove is not true. Consider the case $A = I$. 
More generally, if $A$ is any $2 \times 2$ matrix, then you can compute its characteristic polynomial, $c(x)$, the determinant of $A - xI$. That will be some quadratic polynomial in $x$, like
$$
3x^2 - 22 x^1 + 8x^0.
$$
It turns out that if you plug in $A$ for $x$ , and treat the $A^0$ as $I$, you'll get a $2 \times 2$ matrix...that's the zero matrix! That's called the Cayley-Hamilton theorem. 
As an example: 
$$
A = \begin{bmatrix}1 & 3 \\ -1 & 1 \end{bmatrix} \\
A- xI = \begin{bmatrix}1-x & 3 \\ -1 & 1-x \end{bmatrix} \\
c(x) = det(A - xI) = (1-x)^2 + 3 = x^2 -2x + 4
$$
Now plug $A$ into $c$ to get
$$
c(A) = A^2 - 2A + 4I \\
 = \begin{bmatrix} -2 & 6 \\ -2 & -2 \end{bmatrix} -  \begin{bmatrix}2 & 6 \\ -2 & 2 \end{bmatrix}  +  \begin{bmatrix}4 & 0 \\ 0 & 4 \end{bmatrix} \\
= \begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix} 
$$
A: First of all your matrix is a root of its characteristic polynomial . The bad news is that its square is linear dependant from its lower order terms. The good news is that the algebra generated by this matrix (if its characteristic polynomial is irreducible which it's mostly is) is the splitting field of its roots. 
