Prove that $\det A=0$ if $I \notin \operatorname{span} \{A,A^2,\dots,A^n\}$ Why if $A \in \textsf{M}_{n\times n}(F)$ and 
$$I \notin \operatorname{span} \{A,A^2,\dots,A^n\}$$ 
so $\det (A)=0$?
I understand that $A$ is not invertible if $I \notin \operatorname{span} \{A,A^2,\dots,A^n\}$, but can't explain why.
 A: If$$\operatorname{Id}=\alpha_1 A+\alpha_2 A^2+\cdots+\alpha_n A^n=A.(\alpha_1\operatorname{Id}+\alpha_2 A+\cdots+\alpha_n A^{n-1}),$$then $A$ is invertible (and its inverse is $\alpha_1\operatorname{Id}+\alpha_2 A+\cdots+\alpha_nA^{n-1}$). So, $\det A\neq0$.
On the other hand, if $\det A\neq0$, then the constant term of the characteristic polynomial of $A$ is not $0$. If this characteristic polynomial is $c_0+c_1\lambda+\cdots+(-1)^n\lambda^n$, then, by the Cayley-Hamilton theorem, $c_0\operatorname{Id}+c_1A+\cdots+(-1)^nA^n=0$, and therefore $\operatorname{Id}$ is a linear combination of $A,\ldots,A^n$.
A: Cayley Hamilton says that $A$ satisfies its own char. poly. Suppose that 
$$
\det(A - xI)= c_0 + c_1 x + \dots + c_n x^n.
$$
Then we know by CH that
$$
c_0 I + c_1 A + \dots + c_n A^n = 0
$$
If $c_0 \ne 0$, then 
\begin{align}
I 
&=  -\frac{c_1}{c_0} A - \dots - \frac{c_n}{c_0} A^n  \\
&=  A(-\frac{c_1}{c_0} I - \dots - \frac{c_n}{c_0} A^{n-1})  \\
\end{align}
so that $A$ is invertible, which contradicts the assumption. 
Hence we conclude that $c_0 = 0$. But that makes $\lambda = 0$ a root of $c$, so $A$ has a nontrivial nullspace, hence is not invertible. 
