Problem: Let $n$ be a positive integer and $V = M(n, \mathbb{C} )$ be the vector space of all $n \times n$ complex matrices. The dimension of $V$ is $n^2$. Fix $A \in V$ and consider the set
\[
\mathcal{G} _A = \left\{ I = A^0, A, A^2, A^3,\dots, A^{n^2-1} \right\} .
\]
Prove that $\mathcal{G} _A$ can not be a basis of $V$.
Solution: We will show that $\mathcal{G} _A$ is linearly dependent, which will conclude that it can not be basis. Since $A \in M(n, \mathbb{C} )$, the characteristic polynomial $m(t) = \det (A - tI)$ is a degree $n$ polynomial. We write it as
\[
m(t) = a_0 + a_1 t + a_2 t^2 + \dots + t^n.
\]
Then according to the Cayley-Hamilton theorem, the matrix $A$ must satisfy the polynomial $m(t)$, that is
\[
m(A) = a_0 I + a_1 A + a_2 A^2 + \dots + A^{n} = \mathbf{0}
\]
is the zero matrix. Since the coefficient of $A^n$ is $1$, the set
\[
S_A = \left\{ I, A, A^2,\dots, A^n \right\}
\]
is linearly dependent. As $S_A \subseteq \mathcal{G} _A$, the set $\mathcal{G} _A$ is also linearly dependent. Hence, $\mathcal{G} _A$ is not a basis for $V$.