3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ use linearity of matrix multiplication. $\endgroup$ – Max Mar 17 '14 at 10:19
2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.