Is the matrix V in the subspace U? I'm given that $U$ is the subspace of $M(3,2)$ generated by
$A=\begin{bmatrix} 0 & 0 \\ 1 & 1 \\ 0 & 0 \end{bmatrix}$,
$B=\begin{bmatrix} 0 & 1 \\ 0 & -1 \\ 1 & 0 \end{bmatrix}$ and
$C=\begin{bmatrix} 0 & 1 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}$
I'm asked if $V=\begin{bmatrix} 0 & 2 \\ 3 & 4 \\ 5 & 0 \end{bmatrix}$ is in the subspace $U$.
Since I'm used to vectors I treated $A,B, C$ as vectors of $R^6$ and thought that if there are $\alpha, \beta, \gamma$ such as $\alpha A + \beta B + \gamma C = V$ then $V$ is a linear combination of the vectors that generate $U$, thus, it is in $U$. Can I do this? That is, treat $A,B,C$ as vectors of $R^6$? And can I test it using that equation? If not, how should I solve it? Doing like that I got the following system of linear equations:

 $\begin{bmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 5 \\ 0 & 1 & 1 & 2 \\ 1 & -1 & 0 & 4 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$ which simplifies to $\begin{bmatrix} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 5 \\ 0 & 1 & 1 & 2 \\ 1 & -1 & 0 & 4 \\ \end{bmatrix}$ and gives $L_1; \alpha=3 \\ L_2; \beta=5 \\ L_4; 3-5 = 4$ And since $3-5$ is obviously not $4$ then there is no solution for the equation, thus $V$ is not a linear combination of the vectors that generate $U$, i.e. it's not in the subspace $U$.

 A: This problem can be solved explicitly as $V$ is not in $U$. But it might be helpful to outline a strategy. 
$V$ is a vector in $M_{3,2}\cong \mathbb{F}^{6}$. So if you have 3 linearly independent vectors $a,b,c$ and you want to know if $V$ fits in $a,b,c$, then you can project $V$ to $a,b,c$ respectively. For example we can project $V$ to $c$, and we find it must have a factor of $2c$. Similarly it must have a factor of $3a$. But this give us a contradiction, and the problem is solved. 
In general you are asked to solve a system of six equations. Assume $V=xa+yb+zc$, then write out all coordinates $v_{i}=ax_{i}+yb_{i}+zc_{i}$ give you a system of 6 equations and 3 unknowns. This now boils down to linear algebra: If the 6 equations has rank 3 then there is a unique solution by taking the matrix inverse. If it has rank $\ge 3$ (like this case) then there is no solution at all. If it has rank $2$ or less then there should be infinitely many solutions. 
A: if $V$ be in subspace U then  must exist $c_1,c_2,c_3$ such that $V=c_1A+c_2B+c_3C$
so we have to solve this system :
$c_1(0,0)+c_2(0,1)+c_3(0,1)=(0,1)$$\to$$c_2+c_3=2$
$c_1(1,1)+c_2(0,-1)+c_3(0,0)=(3,4)$$\to$$c_1=3$and $c_1-c_2=4$
$c_1(0,0)+c_2(1,0)+c_3(0,0)=(5,0)$$\to$$c_2=5$
but with attention $c_1-c_2=4$
and $3-5\ne4$ this system do not have solution and so V is not in U subspace
