Linearly independence of the set $\{I_n, A, A^2, \dots, A^{n^2} \}$, where $A$ is a square matrix

Question

I'm trying to check whether or not this set is linearly independent for all $n$, where $A$ is $n \times n$ and $A, A^2, \dots, A^{n^2}$ are distinct matrices and $I_n$ is the identity matrix.

Clearly, if we take $n = 2$, and $A = 3 I_n$ then the set $\{I, A, A^2, A^3, A^4 \}$ is not linearly independent. Is this good enough?

Answer 1

The vector space of $n \times n$ -matrices has dimension $n^2$, the set $\{I_n, A, A^2, \cdots, A^{n^2} \}$ has $n^2+1$ elements.

Conclusion ?

Answer 2

By the way, you can show something (much) stronger:

Given any $n \times n$ matrix $A$, the set $\{I_n, A, A^2, \dots, A^n\}$ is linearly dependent.

For proof, use the Cayley-Hamilton theorem.

Answer 3

Consider $V$ a vector space over a field $\mathbb{K}$. Then, if you take any matrix $M \in \mathcal{M}_n(\mathbb{K})$ you can compute its characteristic polynomial which you know has degree $n$. Then, notice this:

Let $m_A(t)$ be the minimal polynomial of $A$. Then, if you call $k=deg(m_A(t))$, $k$ is the dimension of the vector space generated by the powers of the matrix (i.e. Span$(I,A,A^2,\dots,A^{k-1})$.

Conclude by saying that $deg(m_A(t)) \leq deg(p_A(t)) = n$ where $p_A(t)$ is the characteristic polynomial.