Range

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Another subspace associated to a matrix is its range.

Let $A$ be as above. Then the codomain or range of $A$ (viewed as a linear transformation, defined below) is $Range(A) := \{{\bf b}\in \mathbb R^m\ |\ A*{\bf x} = {\bf b}\ \text { is consistent}\}\subseteq \mathbb R^m$

In analogy to the nullspace, we have

For any real $m\times n$ matrix $A$ , $Range(A)$ is a subspace of $\mathbb R^m$ .

Proof

Again, we first note that ${\bf 0}\in Range(A)$ , as $A*{\bf 0} = {\bf 0}$ . As for the closure axioms,

Suppose ${\bf v}, {\bf w}\in Range(A)$ . Choose ${\bf x},{\bf y}\in \mathbb R^n$ satsifying $A*{\bf x} = {\bf v}, A*{\bf y} = {\bf w}$ . Then $A*({\bf x} + {\bf y}) = A*{\bf x} + A*{\bf y} = {\bf v} + {\bf w}$ implying $Range(A)$ is closed under addition.

Suppose $\alpha \in \mathbb R$ and $A*{\bf x} = {\bf v}\in Range(A)$ . Then $A*(\alpha {\bf x}) = \alpha (A*{\bf x}) = \alpha {\bf v}$ implying $Range(A)$ is closed under scalar multiplication. $\blacksquare$

In fact, we can identify the range precisely in terms of a subspace already defined.

For any $m\times n$ real matrix $A$ , $Range(A) = C(A)$ , the column space of $A$ .

Proof: The consistency theorem for matrix equations tells us that the matrix equation $A*{\bf x} = {\bf b}$ is consistent if and only if $\bf b$ can be written as a linear combination of the columns of $A$ . But this is equivalent to saying that $\bf b$ is in the span of the columns of $A$ ; that is, ${\bf b}\in C(A)$ . $\blacksquare$

Since we have already established an algorithm for computing a minimal spanning set for $C(A)$ , that same algorithm applies to produce a minimal spanning set for the range of $A$ .

If $A = \begin {bmatrix} 1 & 0 & 3 & 0 & 4\\ 2 & 5 & -3 & 0 & -2\\ -1 & 3 & 7 & 1 & 4\\ 2 & 6 & 9 & 8 & -3 \end {bmatrix}$ , compute a minimal spanning set for $Range(A)\subseteq \mathbb R^4$ . Determine if $Range(A) = \mathbb R^4$ , or if it is a proper subset. You may use MATLAB or Octave.