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In $\mathbb{R}^4$ find a basis of $L_1\cap L_2$ where $$L_1 = \operatorname{span} \{ (1,2,0,3),(1,1,-1,2),(0,1,1,1) \}$$ $$L_2 = \operatorname{span} \{ (2,0,-2,1), (3,1,0,2),(4,2,2,3) \}$$

I tried this: $$\begin{align}v &= a(1,2,0,3) + b(1,1,-1,2) + c(0,1,1,1)\\ &= d(2,0,-2,1) + e(3,1,0,2) + f(4,2,2,3)\\ \ &\ \\ \Rightarrow & a(1,2,0,3) + b(1,1,-1,2) + c(0,1,1,1)\\ &= d(2,0,-2,1) + e(3,1,0,2) + f(4,2,2,3)\\ \ &\ \\ \Rightarrow & a(1,2,0,3) + b(1,1,-1,2) + c(0,1,1,1)\\ & - d(2,0,-2,1) - e(3,1,0,2) - f(4,2,2,3)\\ &=0\end{align}$$

and then Gauss elimination. But I didn't get anywhere. Please help.

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Not sure what "I didn't get anywhere" means. You've reduced it to $4$ homogeneous equations in $6$ unknowns. Gaussian elimination should give you parametric formulas for the $6$ unknowns, which you should be able to plug back in to get your basis. –  Gerry Myerson Jan 30 '13 at 0:24
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3 Answers 3

For a different approach:

By rewriting we see $L_1$ and $L_2$ are from dimension 2. By using the dimension theorem $L_1+L_2=4$ at most and $L_1+L_2=2$ at least. Find the basis of $L_1+L_2$ by using gaussian elimination of the 2x2 vectors you found in $L_1$ and $L_2$. Finding a basis for $L_1 \cap L_2$ should be no problem then.

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I would just like to mention that this kind of problem is easier if one (and only one) of the subspaces is given by equations rather than by spanning vectors. Your first concern is how many equations are necessary; this would be $4-3=1$ for each of the subspaces if their spanning vectors were linearly independent. But they aren't for either space: the linear combinations with coefficients $(-1,1,1)$ respectively $(1,-2,1)$ vanish. So you might as well drop one vector from each spanning set (you are free to choose since no two spanning vectors are dependent). I'll drop the two vectors without zero entries. Each of the subspaces would now require $4-2=2$ equations.

Now find two independent equations satisfied by the (remaining) spanning vectors of (say) $L_1$, for instance those setting the linear forms $\alpha=(1~~1~~0~~-1)$ and $\beta=(0~~3~~-1~~-2)$ to be zero (so for a vector $(x,y,z,t)$ we get equations $x+y-t=0$ and $3y-z-2t=0$). Now write down the equations that a linear combination of the spanning vectors for $L_2$ must satisfy to by in $L_1$, namely $$ \begin{aligned} \alpha(p(2,0,-2,1)+q(3,1,0,2))&=0 \\ \beta( p(2,0,-2,1)+q(3,1,0,2))&=0 \end{aligned} \qquad\text{that is}\qquad \begin{aligned} p+2q&=0 \\ -q&=0 \end{aligned} $$ (the equations were simplified by evaluating the linear forms on the spanning vectors, to $1$, $2$, $0$, $-1$, repsectively). Now it is clear that the only solution is $p=q=0$, to the intersection is $L_1\cap L_2=\{0\}$.

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Perhaps an easier way is to first find a nice basis for $L_1$ and a nice basis for $L_2$. For $L_1$, make a matrix whose rows are the three given vectors, and row-reduce it; the rows in the reduced form still span $L_1$. If I did my arithmetic correctly, $L_1$ has dimension $2$, with a basis of the form $\{{(r_1,0,r_2,r_3),(0,r_4,r_5,r_6)\}}$. Then do the same thing for $L_2$. Again, up to errors in my arithmetic, it's $2$-dimensional, with a basis of the form $\{{(r_1',0,r_2',r_3'),(0,r_4',r_5',r_6')\}}$. Now you can write $$a(r_1,0,r_2,r_3)+b(0,r_4,r_5,r_6)=c(r_1',0,r_2',r_3')+d(0,r_4',r_5',r_6')$$ and those zeros will make life easier.

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