The question goes as follows:
Let $x$ be a vector in $\mathbb{R}^n$ written as a column, and define
$$U = (Ax : A \in M_{(mn)}) $$
Show that $U$ is a subspace of $R^m$
I know the three things we test for are if $0 \in U$ and then closure under addition and scalar multiplication but I don't really know how to go about this in the right way.
I'm not sure but I believe that $x$ is a fixed column matrix in this case.
I know the zero vector is in $\mathbb{R}^m$ but how do I go about showing it's also in $U$?
My understanding so far is that U is composed of $Ax$ and that if $A \in M_{mn}$ then there exists an $A$ that is a zero matrix such that $Ax$ will result in the zero vector being in $U$. I don't know if this is correct but this is all I could come up with.
for closure under addition, $A_1x$ and $A_2x$ are in $U$ then $A_1x +A_2x $ must be in $U$ as well since $A$ is an $m*n$ matrix and $A_1 + A_2$ should result in an $m\times n$ matrix as well which should be in $U$ from $x(A_1 + A_2)$
For scalar multiplication closure, I have zero idea on what to do.
To show $0_{m \times 1} \in U$, note that $0_{m \times 1} = 0_{m \times n}x$.
Note taht matrix multiplication is not commutative, that is in general, we do not have $AB=BA$.
To prove closure under addition, $A_1x+A_2x = (A_1+A_2)x$, note that it is not equal to $x(A_1+A_2)$.
To prove closure under scalar multiplication, suppose $v= Ax$, note that $(kv)=(kA)x$, we have $kA \in M_{m \times n}$.