Finding the basis for the intersection of two subspaces Find a basis for the intersection of the subspaces $X_1$ and $X_2$ of $\mathbb{R^4}$ where $$X_1=\text{span}\left\{ (1,1,0,0), (0,1,1,0), (0,0,1,1)\right\}$$
and $$X_2=\text{span}\left\{ (1,0,t,0), (0,1,0,t)\right\},$$ where $t\in \mathbb{R}$ is given.
Well, I know that if $v\in X_1\cap X_2$, then 
$v=a(1,1,0,0)+ b(0,1,1,0)+ c(0,0,1,1)=d(1,0,t,0)+ m(0,1,0,t)$,
where $a, b, c, d, m \in \mathbb{R}$. Then
this implies that
$(0,0,0,0)=v-v=(a-d,a+b-m,b+c-dt,c-mt)$,
that is $$a-d=0$$ $$a+b-m=0$$ $$b+c-dt=0$$ $$c-mt=0.$$
Since $t\in \mathbb{R}$ is given, I have 5 unknowns and 4 equations. I can't solve this! Is there any other way to solve this problem??
Thank you!
 A: You are on the right track! Well, actually we want to solve the system:
$$\begin{bmatrix} 
1 & 0 & 0 & -1 & 0\\
1 & 1 & 0 &  0 & -1\\ 
0 & 1 & 1 & -t & 0\\ 
0 & 0 & 1 &  0 & -t
\end{bmatrix}\cdot \begin{bmatrix} 
a\\ b\\ c \\ d\\m \end{bmatrix} = \begin{bmatrix} 0 \\0 \\0 \\0  \end{bmatrix}.$$
After some work we can find the reduced row echelon form  of the coefficient matrix (for $t\neq -1$ as @snulty has pointed out), which is:
$$R = \begin{bmatrix}
1 & 0 & 0 & 0 & -1\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & -t\\
0 & 0 & 0 & 1 & -1
\end{bmatrix}.
$$
It is sufficient to solve the system $$R \cdot \begin{bmatrix} 
a\\ b\\ c \\ d\\m \end{bmatrix} = \begin{bmatrix} 0 \\0 \\0 \\0 \end{bmatrix}.$$
Clearly, we have that $ b = 0.$ Also, $ a = d = m$ and $ c= dt$. Combining all the information we have, the solution is:
$$\begin{bmatrix}
a\\ b \\ c\\ d\\m \end{bmatrix} = \begin{bmatrix}a \\ 0 \\ at \\ a\\a \end{bmatrix} = a\cdot \begin{bmatrix} 1\\0\\t\\1\\1 \end{bmatrix}, $$ where $a \in \mathbb R.$
Thus, the vectors included in $X_1\cap X_2$ are of the form:
$$a(1,1,0,0) + a t(0,0,1,1) = a (1,1,t,t).$$
Thus $X_1 \cap X_2 = \operatorname{span}\{ (1,1,t,t)\}.$

For $t = -1$ we can do something similar, or even better as @snulty has pointed out, in that case $X_2 \subset X_1$, so $X_1 \cap X_2 = X_2.$
A: I like the answer given by @thanasissdr and I'll add a simple approach to complement his, which works in this case because the equations aren't too complicated.
You can very easily see that $a=d$ for the first equation, $c=mt$ from the fourth. Using these in the middle two gives you a simultaneous equation:
\begin{align}
&b+a-m=0\\
&b-at+mt=0
\end{align}
Subtracting the two gives that:
$$a(1+t)=m(1+t)$$
Now for $t\neq -1$ we can conclude that $a=m=d$ and $b=0$ and $c=at$
So $(a,b,c,d,m)=a(1,0,t,1,1)$ or since $v=d(1,0,t,0)+m(0,1,0,t)=a(1,1,t,t)$. 
We can conclude that the intersection is one dimensional and taking $a=1$ say:
$$X_1\cap X_2=\operatorname{span}\{(1,1,t,t)\}$$
You can probably justify that five variables and four constraints suggests that there will be one free variable and hence a dimension $1$ space at the end.
Now for the case we excluded $t=-1$ we can see that if $X_1=\operatorname{span}\{e_1,e_2,e_3\}$ then $X_2=\operatorname{span}\{e_1-e_2,e_2-e_3\}$. So $X_2\subset X_1$ and hence $$X_1\cap X_2=X_2$$
Again you could justify this by noticing that when $t=-1$ using equations one and four we see that equations two and three are the same. So five unknowns and three constraints suggests the space will be $2$ dimensional.
Finally if you want to check dimensions you could use that
$$\dim(X_1+X_2)=\dim(X_1)+\dim(X_2)-\dim(X_1\cap X_2)$$
$\dim(X_1+X_2)=3,4$ only so $\dim(X_1\cap X_2)=1,2$ only.
