Adding the column space of a matrix with another matrix? Supposed I have these matrices:
\begin{gather}
M = \begin{bmatrix}  0 & 1 & 1 \\  0 & 1 & 0 \\  0 & 0 &1 \end{bmatrix}
\hspace{1.5em} 
N := ℝ^3
\end{gather}
\begin{gather}
A = \begin{bmatrix}  1 & 0 \\  0 & 0 \\  0 & 1 \end{bmatrix}\text{.}
\end{gather}
I want to solve $(MN + \text{im}(A))$. How exactly do I do this?
So far, I know that 
\begin{gather}
\text{im}(A) = \text{span}\{ \begin{bmatrix}  1 \\  0 \\  0 \end{bmatrix}, \begin{bmatrix}  0 \\  0 \\  1 \end{bmatrix}\}=\{a_1, a_2\}
\end{gather}
and I'm taking $N$ to be a $3 \times 3$ identity matrix. How do I sum them up, since the sizes of the $MN$ and either column vectors $a_1, a_2$ are different?
Am I doing it right? Please help me! Thanks!
 A: Note that $MN$ is not a matrix. Rather, it is a set.
To be more precise $MN = M\mathbb{R}^3 = \{y \in \mathbb{R}^3 : y = Mx \text{ for some } x \in \mathbb{R}^3\},$ that is, $MN$ is the image of (the associated linear map of) $M$.
Similarly $\operatorname{im}(A)$ is the set which is the image of $A$. (As $A$ is a $3 \times 2$ matrix, the image is indeed a subset of $\mathbb{R}^3$.)  
Both of these sets are in fact subspaces of $\mathbb{R}^3$. Given two subspaces $U$ and $W$ of $\mathbb{R}^3$, their sum is defined as the following:
$$U + W := \{y \in \mathbb{R}^3 : y = u + w \text{ for some } u \in U, w \in W\}.$$
The above is what the question is asking you to find.
That is, the final answer must be a subset (in fact, a subspace) of $\mathbb{R}^3$ and not a matrix.
A: Adding another answer to actually solve the question.
As noted before, $MN = \operatorname{im} M = \operatorname{span}\left\{\begin{bmatrix} 0 \\0\\0\\\end{bmatrix},\begin{bmatrix} 1 \\1\\0\\\end{bmatrix},\begin{bmatrix} 1 \\0\\1\\\end{bmatrix}\right\} = \operatorname{span}\left\{\begin{bmatrix} 1 \\1\\0\\\end{bmatrix},\begin{bmatrix} 1 \\0\\1\\\end{bmatrix}\right\}.$
Where the last equality comes from noting that $[0\; 0\; 0]^T$ belongs to the span of the other two.
You had already figured out that $\operatorname{im} A = \operatorname{span}\left\{\begin{bmatrix} 1 \\0\\0\\\end{bmatrix},\begin{bmatrix} 0 \\0\\1\\\end{bmatrix}\right\}.$
It is not tough to show that if $\operatorname{span} A = U$ and $\operatorname{span} B = W$, then $\operatorname{span} (A \cup B) = U+W$.
This, we get that $MN + \operatorname{im}A = \operatorname{span}\left\{\begin{bmatrix} 1 \\1\\0\\\end{bmatrix},\begin{bmatrix} 1 \\0\\1\\\end{bmatrix},\begin{bmatrix} 1 \\0\\0\\\end{bmatrix},\begin{bmatrix} 0 \\0\\1\\\end{bmatrix}\right\}.$
However, note that the collection of the four vectors written isn't actually a basis as it is clear that the second vector belongs to the span of the last two. Thus, we can drop that without affecting the linear span. Thus, we also have
$$MN + \operatorname{im}A = \operatorname{span}\left\{\begin{bmatrix} 1 \\1\\0\\\end{bmatrix},\begin{bmatrix} 1 \\0\\0\\\end{bmatrix},\begin{bmatrix} 0 \\0\\1\\\end{bmatrix}\right\}.$$
It is now easy to check that the above set is indeed linearly independent. Thus, the desired subspace is a $3$ dimensional subspace of $\mathbb{R}^3$.
However, note that the only three-dimensional subspace of $\mathbb{R}^3$ is $\mathbb{R}^3$ (since its own dimension is $3$) and thus, your answer is simply $\boxed{\mathbb{R}^3}$.
