Linear Algebra

Powers of a Matrix Cannot be a Basis of the Vector Space of Matrices

Problem 375

Let $n>1$ be a positive integer. Let $V=M_{n\times n}(\C)$ be the vector space over the complex numbers $\C$ consisting of all complex $n\times n$ matrices. The dimension of $V$ is $n^2$.
Let $A \in V$ and consider the set
\[S_A=\{I=A^0, A, A^2, \dots, A^{n^2-1}\}\] of $n^2$ elements.
Prove that the set $S_A$ cannot be a basis of the vector space $V$ for any $A\in V$.

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Proof.

We prove that the set $S_A$ is linearly dependent, hence it cannot be a basis of $V$.
Since $A$ is an $n\times n$ matrix, its characteristic polynomial $p(t)=\det(tI-A)$ is a degree $n$ polynomial.

(Your preferred definition of the characteristic polynomial might be $\det(A-tI)$. It is straight forward to modify the following proof with this definition.)

Let us write it as
\[p(t)=t^n+a_{n-1}t^{n-1}+\cdots+a_1x+a_0.\] Then the Cayley-Hamilton theorem states that
\[p(A)=A^n+a_{n-1}A^{n-1}+\cdots+a_1A+a_0I=O\] is the zero matrix.

Since the coefficient of $A^n$ is $1$, this gives a non-trivial linear combination of $I, A, \dots, A^n$. Therefore the set
\[T:=\{I, A, \dots, A^n\}\] is linearly dependent.

As $T$ is a subset of $S_A$, the set $S_A$ is also linearly dependent.
Therefore, $S_A$ is not a basis of $V$. This completes the proof.

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