Let $A$ be an $m \times n$ matrix.
Let $\calN(A)$ be the null space of $A$. Suppose that $\mathbf{u} \in \calN(A)$ and $\mathbf{v} \in \calN(A)$.
Let $\mathbf{w}=3\mathbf{u}-5\mathbf{v}$.

Recall that the null space of an $m\times n$ matrix $A$ is a subspace of $\R^n$ defined by
\[ \calN(A)=\{ \mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{0}_m\}.\]
Here $\mathbf{0}_m$ is the $m$-dimensional zero vector in $\R^m$.

Solution.

Since $\mathbf{u}, \mathbf{v} \in \calN(A)$, we have
\[A\mathbf{u}=\mathbf{0}_m \text{ and } A\mathbf{v}=\mathbf{0}_m,\]
where $\mathbf{0}_m$ is the $m$-dimensional zero vector in $\R^m$.

Now using the properties of the matrix multiplication, we have
\begin{align*}
A\mathbf{w}&=A(3\mathbf{u}-5\mathbf{v})\\
&=A(3\mathbf{u})+A(-5\mathbf{v})\\
&=3A\mathbf{u}-5A\mathbf{v}\\
&=3\mathbf{0}_m-5\mathbf{0}_m=\mathbf{0}_m.
\end{align*}
Therefore we obtained
\[A\mathbf{w}=\mathbf{0}_m\in \R^m.\]

Remark.

Note that a map $T:\R^n \to \R^m$ defined by $T(\mathbf{x})=A\mathbf{x}$, where $A$ is an $m\times n$ matrix is a linear transformation.
That is the map $T$ satisfies:

$T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})$ for any $\mathbf{u}, \mathbf{v} \in \R^n$, and

$T(c\mathbf{v})=cT(\mathbf{v})$ for any $\mathbf{v}\in \R^n$ and $c \in \R$.

From this point of view, the above problem can be classified into a problem of linear transformation.

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