I have a $n \times m$ matrix A, can someone explain why the Image of $A$ is a subspace of $\mathbb R^m$ I have a $n \times m$ matrix $A$, can someone explain why the Image of $A$ is a subspace of $\mathbb R^m$?
I know it must meet the three requirements: contains zero vector, closure under addition, closure under multiplication.
 A: For the zero vector it is clear, since $\textit{0}=A\textit{0}\in\operatorname{im}(A)$. Now, suppose we have vectors $x$ and $y$ in the image of $A$, that is, $x=Au$ and $y=Aw$ for some $u,w\in \Bbb{R}^m$. Then, for every $a\in \Bbb R$, we see that
$$ax+y=a(Au)+Aw=A(au+w) \in \operatorname{im}(A).$$
A: Image space $img(A)=\{ Ax:x\in \mathbb{R}^n\} $ is a subset of $\mathbb{R}^m$ is clear.
Now, you can see that, for any two vectors $ u,v \in Img(A)$ and a constant $k \in \mathbb{R}$, $ku+v \in Img(A)$.
A: First, notice that, for any $u,v\in \text{Im}(A)\subseteq\mathbb{R}^m$, we have that
$$
u+v=Ax+Ay=A(x+y)
$$
for some $x,y\in\mathbb{R}^n$, which is a vector space, thus $x+y\in\mathbb{R}^n$. Hence, $A(x+y)\in\text{Im}(A)$.
Similarly, for any $\alpha\in\mathbb{R}$, we have that
$$
\alpha u=\alpha Ax=A(\alpha x)
$$
for some $x\in\mathbb{R}^n$. Again, $\alpha x\in\mathbb{R}^n$ and so $A(\alpha x)\in\text{Im}(A)$. Note that this includes the zero vector case.
Hence, $\text{Im}(A)$ is a subspace of $\mathbb{R}^m$.
Edit: To prove that $\alpha Ax=A(\alpha x)$, notice that
\begin{align}
\alpha Ax&=\alpha
\begin{pmatrix}
a_{11}&a_{12}&\cdots & a_{1m} \\
a_{21}&a_{22}&\cdots & a_{2m} \\
\vdots & \vdots &\ddots &\vdots\\
a_{n1}&a_{n2}&\cdots &a_{nm}
\end{pmatrix}
\begin{pmatrix}
x_1\\x_2\\ \vdots \\x_m
\end{pmatrix}=
\begin{pmatrix}
\alpha a_{11}&\alpha a_{12}&\cdots & \alpha a_{1m} \\
\alpha a_{21}&\alpha a_{22}&\cdots &\alpha  a_{2m} \\
\vdots & \vdots &\ddots &\vdots\\
\alpha a_{n1}&\alpha a_{n2}&\cdots &\alpha a_{nm}
\end{pmatrix}
\begin{pmatrix}
x_1\\x_2\\ \vdots \\x_m.
\end{pmatrix}\\
&=\begin{pmatrix}
\sum_{j=1}^m\alpha a_{1j}x_j\\\sum_{j=1}^m\alpha a_{2j}x_j\\ \vdots \\\sum_{j=1}^m\alpha a_{nj}x_j
\end{pmatrix}=
\begin{pmatrix}
\sum_{j=1}^ma_{1j}(\alpha x_j)\\\sum_{j=1}^ma_{2j}(\alpha x_j)\\ \vdots \\\sum_{j=1}^m a_{nj}(\alpha x_j)
\end{pmatrix}=
\begin{pmatrix}
a_{11}&a_{12}&\cdots & a_{1m} \\
a_{21}&a_{22}&\cdots & a_{2m} \\
\vdots & \vdots &\ddots &\vdots\\
a_{n1}&a_{n2}&\cdots &a_{nm}
\end{pmatrix}
\begin{pmatrix}
\alpha x_1\\\alpha x_2\\ \vdots \\\alpha x_m
\end{pmatrix}\\&=A(\alpha x).
\end{align}
