Intuition of the column space of a vector $4\times3$ matrix $\mathbf{A}$ Suppose I have this matrix:
$\mathbf{A} = 
\begin{bmatrix}
1&1&2\\
2&1&3\\
3&1&4\\
4&1&5\\
\end{bmatrix}$
My understanding that $\mathbf{A}$ is the equivalent of this system of equations:


*

*$x  + y + 2z$

*$2x + y + 3z$

*$3x + y + 4z$

*$4x + y + 5z$


Professor Strang in his lecture here at 13:26 claims that $\mathbf{A}$ is a subspace of $\mathbb{R^4}$. 
Since this system of equations uses only 3 dimensions ($x, y, z$), could it be said that $\mathbf{A}$ is a subspace of $\mathbb{R^3}$, or am I missing something?
 A: You consider ${\bf A}$ as a set of column vectors, 3 of them in the case given.  The space consisteing of all linear combinations of those 3 4-component vectors is a subspace of the space of all 4-component vectors ($\Bbb{R}^4$).
A: Putting $A$ into reduced row echelon form: 
$$
\mbox{RowReduce}(A)=\begin{bmatrix}
1&0&1\\0&1&1\\0&0&0\\0&0&0
\end{bmatrix}
$$
we can see that the column space of $A$, denoted $\mbox{C}(A)$, is a two-dimensional subspace of $\mathbb{R}^4$.
It is two-dimensional since the last column shows that the third vector can be formed from the other two with coefficients (otherwise known as coordinates) $1$ and $1$:
$$
\begin{bmatrix}
2\\3\\4\\5
\end{bmatrix}
=
1 \begin{bmatrix}
1\\2\\3\\4
\end{bmatrix}
+
1 \begin{bmatrix}
1\\1\\1\\1
\end{bmatrix}
$$
The coefficients (e.g.,  $x$, $y$, and $z$ in your question) in the linear combination are the coordinates of a particular vector using some set of vectors to define an axis.
(If all of the vectors in the set are linearly independent and span the space they are from, the set is called a basis.)
For example, what coefficients are needed to generate $[5,7,9,11]^T$ using the first two columns of $A$ as an axis?
$2$ and $3$:
$$
\begin{bmatrix}
5\\7\\9\\11
\end{bmatrix}
=
2 \begin{bmatrix}
1\\2\\3\\4
\end{bmatrix}
+
3 \begin{bmatrix}
1\\1\\1\\1
\end{bmatrix}
$$
Or in other words, that vector exists at position $(2,3)$ in the two-dimensional subspace using $[1,2,3,4]^T$ and $[1,1,1,1]^T$ to define the axis.
Does the vector $[2,3,3,5]^T$ exist in this two-dimensional space?
No, it does not.
Because the first two elements require the coefficients to be $1$ and $1$ (see above) but the third element cannot be $3$ with those weights.
For the discussing dimension, I tend to be less mathematical (since I am an engineer) and think of it as the length of the vectors I am working with, but it must be done with care since Linear Algebra defines dimension as the number of linearly independent vectors you have to define an axis.
We could also do all of the above with $A$'s row space, which is also a two-dimensional subspace but of $\mathbb{R}^3$ (not $\mathbb{R}^4$).
