Intersection of two column spaces Let the matrices A and B be:
$$A = \begin{bmatrix}
2 & 3 & 0 \\
2 & 3 & 1 \\
2 & 3 & 1
\end{bmatrix} and \space
B = \begin{bmatrix}
1 & 0 \\
1 & 0 \\
0 & 1
\end{bmatrix}$$
If $C(A)$ and $C(B)$ are the vector spaces generated by the columns of A and B, show that $C(A) \cap C(B)$ is a vector space and describe it.
I started to tackle this problem by transforming the two matrices in:
$$A' = \begin{bmatrix}
1 & 3/2 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{bmatrix} and \space
B' = \begin{bmatrix}
1 & 0 \\
0 & 1 \\
0 & 0
\end{bmatrix}$$
and so the base of the column spaces are:
$$Basis(A) = \{\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix} \}$$ and $$Basis(B) = \{\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix} \}$$
And so the intersection between the column spaces is:
$$C(A) \cap C(B) = Basis(A) = Basis(B)$$
Am I right? I'm not sure this makes sense.
Thanks!
 A: This doesn't make sense: You've reduced $A$ to $A'$ via row reduction, but this operation preserves the row space, not the column space. Indeed, we can see that the proposed $Basis(A)$ cannot span the column space of $A$, as the last entry of any element in the span of $Basis(A)$ is zero, but this is not even true of the first column of $A$.
One can modify your approach, though, by carrying out column reduction instead (equivalently, row-reducing $A^T$ and then taking the transpose). In this case, though, the matrices are simple enough that one can carry this out more or less by inspection. (In fact, $B$ itself is already column-reduced.)
A: A basis of $A$ is $\left\{\left[\begin{array}{c}1 \\ 1\\ 1\end{array} \right] ,\left[\begin{array}{c}0 \\ 1\\ 1\end{array} \right]\right\}$. 
Given a vector, a quick way to know that it is in the column space of $A$ is to check that the second and third component are equal.
A basis of $B$ is $\left\{\left[\begin{array}{c}1 \\ 1\\ 1\end{array} \right] ,\left[\begin{array}{c}1 \\ 1\\ 0\end{array} \right]\right\}$.
Given a vector, a quick way to know that it is in the column space of $A$ is to check that the first and second component are equal.
What about $C(A) \cap C(B)$? 
It is the space where there all the components are equal. 
$$C(A) \cap C(B)=span((1,1,1)^T)$$
