# Proving that the range of linear transformation is a linear subspace

I need some help figuring out how to prove this question.

True or false, the set $S = \left \{ A\mathbf{y}: \mathbf{y} \in \mathbb{R}^4\right \}$ is a subspace of $\mathbb{R}^3$ where A is a fixed $3\times4$ real matrix.

Well I will need to show that the zero vector is in the set S. Then show the closure axioms hold. Or I can show for $A\mathbf{u}, A\mathbf{v}$ and some scalar $c$ in our field, $A\mathbf{u}+cA\mathbf{v} \in S$

What I have so far:

For $\mathbf{y} = \begin{pmatrix} 0\\ 0\\ 0\\ 0 \end{pmatrix} \in \mathbb{R}^4$. It is clear that, $A\mathbf{y} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} \in S$. (Hopefully this is right so far)

Now, we need to prove the closure axioms.

Suppose $A\mathbf{u}, A\mathbf{v} \in S$ where $\mathbf{u} = \begin{pmatrix} u_{1}\\ u_{2}\\ u_{3}\\ u_{4} \end{pmatrix}, \mathbf{v} = \begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\\ v_{4} \end{pmatrix} \in \mathbb{R}^4$

Here is where I'm stuck, I know I am suppose to show $A\mathbf{u} + cA\mathbf{v} \in S$

Any clues or hints would be appreciated. Thanks. Please try to only post partial solutions or hints to get me going.

• use linearity of matrix multiplication. – Max Mar 17 '14 at 10:19

I think I figured it on, now I just need some confirmation so continuing from what I have above, we have $A\mathbf{u}+cA\mathbf{v} = A\begin{pmatrix} u_{1}\\ u_{2}\\ u_{3}\\ u_{4} \end{pmatrix} + cA\begin{pmatrix} v_{1}\\ v_{2}\\ v_{3}\\ v_{4} \end{pmatrix}$

$= A\begin{pmatrix}u_{1}\\ u_{2}\\ u_{3}\\ u_{4} \end{pmatrix} + A\begin{pmatrix} cv_{1}\\ cv_{2}\\ cv_{3}\\ cv_{4} \end{pmatrix}$

$= A\begin{pmatrix} u_{1}+cv_{1}\\ u_{2}+cv_{2}\\ u_{3}+cv_{3}\\ u_{4}+cv_{4} \end{pmatrix} \in S$ as $\begin{pmatrix} u_{1}+cv_{1}\\ u_{2}+cv_{2}\\ u_{3}+cv_{3}\\ u_{4}+cv_{4} \end{pmatrix} \in \mathbb{R}^4$

Hint, for matrix $A$, field coefficient $\alpha$ and vectors $u,u'$, we have $(\alpha A)(u+u')=\alpha (Au)+\alpha (Au')$. This is a corollary of the result that matrix multiplication is linear.

Take $u,v\in S$, then $u=Ax$ and $v=Ay$ for $x,y\in\mathbb R^4$. This means that $$u+cv = Ax + cAy = Ax+A(cy) = A(x+cy)\in S$$

This is similar to what you wrote in your own answer, but written in a way that is more general. Here, you see that all you need for the set $$\{f(y)|y\in V\}$$ to be a linear space is that

1. $V$ is a linear space
2. $f$ is a linear operator.