Special Subspaces

From elementary linear algebra, recall that there are special subspaces that arise as special cases of the images and kernels of module homomorphisms.

Definition. The range of an $m\times n$ matrix $A$ is the set

\[\mathrm{range}(A) := \{Ax \mid x \in \mathbb{F}^n\}.\]

Theorem. The range of $A$ is the subspace of $\mathbb{F}^m$ spanned by the columns of $A$.

Proof. Recall that we noted

\[Av = v_1a_1 + \cdots + v_na_n\]

where $v\in \mathbb{F}^n$. So for any $b \in \mathrm{range}(A)$, we there is some $x\in\mathbb{F}^n$ such that $b = Ax$. So the claim follows by the formula above and also noting that $0\in\mathrm{range}(A)$.

Definition. The nullspace of $A$ is the set

\[\mathrm{null}(A) = \{x \in \mathbb{F}^n \mid Ax = 0\}.\]

We will skip the proof that the nullspace of a matrix is a subspace.

Another important topic related to the dimensions of these subspaces is the rank of a matrix.

Definition. The column rank of a matrix is the dimension of its column space and the row rank of a matrix is the dimension of its row space.

Theorem. Column rank = row rank. Accordingly, we refer to the column rank simply as the rank of the matrix.

Proof. It is immediate that column rank and row rank is preserved by elementary row/column operations. Accordingly, $A$ can be written in the form of $A = PBQ$ where $P$ is the matrix obtained via elementary row operations to obtain the RREF of $A$ and $Q$ is the matrix obtained via elementary column operations to obtain the RCEF of $PA$. Clearly, $B$ contains an identity matrix as a block whose size is the column rank and row rank.

Definition. An $m \times n$ matrix is said to be full rank provided that the rank of the matrix is $\min(m, n)$.

Notice that if $m \ge n$, then the columns of the matrix are necessarily linearly independent. Since the matrix has $n$ columns and the range of the matrix is dimension $n$, it seems only natural that $A$ induces an embedding into $\mathbb{F}^m$. This is, in fact, true.

Theorem. Let $A \in \mathbb{F}^{m\times n}$ with $m\ge n$. Then $A$ is full rank if and only if its induced linear transformation is injective.

Proof. ($\implies$) Assume that $A$ is full rank and that $Av = 0$. Since matrix-vector multiplication can be realized as a linear combination of the columns of $A$ with weights $v$, the independence of the columns imply that $v = 0$. This implies the nullspace of $A$ is trivial so the induced map must be injective.

($\impliedby$) We proceed by contrapositive and assume that $A$ is not full rank. Accordingly, the columns of $A$ are linearly dependent. Thus, there is some nontrivial linear combination of the columns that is equal to $0$. Using the weights of the linear combination as the appropriate components of a vector $v$, we have that $Av = 0$ for a nontrivial $v$. Thus, $A$ has a nontrivial nullspace and is not injective.