How to show that the column spaces of two matrices is equal? The linear transformations $R^3 \rightarrow R^3$ have the matrices $A= \begin{bmatrix} 1&2&0 \\ 2&3&3\\ 1&a&3 \end {bmatrix}$ and $B = \begin{bmatrix} 4&1&3 \\ 7&1&b\\ 3&0&2 \end {bmatrix}$ where a and b are constants.
Find the values of a and b so that the columnspace of A and B is equal.
2 things can happen, if det A and det B are non zero then they will span the entire $R^n$ so just need to find the determinants, but when it comes det A = det B  = 0 that is where I am stuck. I know that I need to show that every column of B need to be a linear comb of A or vice versa but that is a long process to do on an exam? In the answers they just said do the same row operation I did on A on B to see if B's column vectors are linearly dependent? Why? 
 A: If $Ax = v$, then row operations on matrix $A$ cause $v$ to transform with same row operations. So, column space of a matrix gets transformed by same row operations as the row operations performed on the matrix. So, if two matrices have same column space, the row operations would lead to same column space. Also, since you can invert a row operation by a row operation, the converse is also true.
But, for $3\times3$ matrices, I think there is even more direct computation. Take cross product of two column vectors of A $([4,7,3]\times[1,1,0] = [-3,3,-3])$. Similarly, compute cross product of column vectors of B $([1,2,1] \times [0,3,3] = [3,-3,3])$. Clearly, they are parallel. So, when determinant of A and B are 0, both have same plane(with same normal) as column space. So, compute $a$ and $b$ such $det A$ and $det B$ are 0 and ensure values of $a$ and $b$ are such that both determinants are zero or both of them are non-zero.
A: Noticing that $$a\in {\bf R}\, ,\quad\det\begin{pmatrix}1&2&0\\2&3&3\\1&a&2\end{pmatrix}=3(1-a)\not=0 \iff \boxed{a\not=1},$$
and $$b\in {\bf R}\,,\quad\det\begin{pmatrix}4&1&3\\7&1&b\\3&0&2\end{pmatrix}=3(b-15)\not=0\iff \boxed{b\not=5}. $$
Thus casework shows:

*

*If $(a,b)\not=(1,5)$, then the column spaces ${\rm col}(A)={\rm col}(B)={\bf R}^{3}$, since we have three vector linearly independent in a space of dimension  three.


*If $(a,b)=(1,5)$, then we can find the equations describing explicitly the column spaces using  elimination by row. Finally,  we get in this case ${\rm col}(A)={\rm col}(B)\not={\bf R}^{3}$. Indeed,  $${\rm col}(A)=\{(x,y,z)\in {\bf R}^3\colon \quad x-y+z=0\},$$ $${\rm col}(B)=\{(x,y,z)\in {\bf R}^{3}\colon \quad x-y+z=0\}$$
Therefore

*

*If $a=1$ and $b=5$, then ${\rm col}(A)={\rm col}(B)$.

*If $a\not=1$ and $b\not=5$, then ${\rm col}(A)={\rm col}(B)$.

Now, consider (thank you @user1551 for the remark):

*

*If $a=1$ and $b\not=5$, $\dim {\rm col}(A)=2$ and $\dim {\rm col}(B)=3$. Thus,  ${\rm col}(A)\not={\rm col}(B)$.


*If $a\not=1$ and $b=5$, then $\dim {\rm col}(A)=3$ and $\dim {\rm col}(B)=2$. Thus, ${\rm col}(A)\not={\rm col}(B)$.
Therefore, for all real number $a$ and $b$ we have
$${\rm col}(A)={\rm col}(B)\iff (a,b)=(1,5) \quad \vee \quad  (a,b)\not=(1,5)$$
A: Let $V=\operatorname{span}\{(1,1,0)^T,(0,1,1)^T\}$. Note that the subspace spanned by two constant columns (the 1st and the 3rd) of $A$ is precisely $V$. So is the subspace spanned by the two constant columns (the 1st and the 2nd) of $B$. Therefore $A$ and $B$ have the same column space if and only if the two matrices are both singular (in which case their column spaces are both equal to $V$) or both nonsingular (in which case their column spaces are both equal to $\mathbb R^3$), i.e., if and only if $(a\ne1\text{ and }b\ne5)$ or $(a=1\text{ and }b=5)$.
As for the official answer you received, note that by applying the same finite sequence of elementary row operations to both $A$ and $B$, one is essentially left-multiplying both $A$ and $B$ by some invertible matrix $P$. Thus the author was using the fact that $A$ and $B$ have the same column space if and only if $PA$ and $PB$ have the same column space. However, as this or other answers show, there is no need to appeal to elementary row operations to solve this problem.
