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 $A\in V$.

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.

The Subset Consisting of the Zero Vector is a Subspace and its Dimension is Zero
Let $V$ be a subset of the vector space $\R^n$ consisting only of the zero vector of $\R^n$. Namely $V=\{\mathbf{0}\}$.
Then prove that $V$ is a subspace of $\R^n$.
Proof.
To prove that $V=\{\mathbf{0}\}$ is a subspace of $\R^n$, we check the following subspace […]

Every Basis of a Subspace Has the Same Number of Vectors
Let $V$ be a subspace of $\R^n$.
Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k\}$ is a basis of the subspace $V$.
Prove that every basis of $V$ consists of $k$ vectors in $V$.
Hint.
You may use the following fact:
Fact.
If […]

Dimension of the Sum of Two Subspaces
Let $U$ and $V$ be finite dimensional subspaces in a vector space over a scalar field $K$.
Then prove that
\[\dim(U+V) \leq \dim(U)+\dim(V).\]
Definition (The sum of subspaces).
Recall that the sum of subspaces $U$ and $V$ is
\[U+V=\{\mathbf{x}+\mathbf{y} \mid […]

Linear Transformation and a Basis of the Vector Space $\R^3$
Let $T$ be a linear transformation from the vector space $\R^3$ to $\R^3$.
Suppose that $k=3$ is the smallest positive integer such that $T^k=\mathbf{0}$ (the zero linear transformation) and suppose that we have $\mathbf{x}\in \R^3$ such that $T^2\mathbf{x}\neq \mathbf{0}$.
Show […]

Prove a Given Subset is a Subspace and Find a Basis and Dimension
Let
\[A=\begin{bmatrix}
4 & 1\\
3& 2
\end{bmatrix}\]
and consider the following subset $V$ of the 2-dimensional vector space $\R^2$.
\[V=\{\mathbf{x}\in \R^2 \mid A\mathbf{x}=5\mathbf{x}\}.\]
(a) Prove that the subset $V$ is a subspace of $\R^2$.
(b) Find a basis for […]

Any Vector is a Linear Combination of Basis Vectors Uniquely
Let $B=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ be a basis for a vector space $V$ over a scalar field $K$. Then show that any vector $\mathbf{v}\in V$ can be written uniquely as
\[\mathbf{v}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3,\]
where $c_1, c_2, c_3$ are […]