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Let $$L_{1}(x_{1},x_{2},x_{3},x_{4})=(3x_{1}+x_{2}+2x_{3}-x_{4}, 2x_{1}+4x_{2}+5x_{3}-x_{4})$$ and $$L_{2}(x_{1},x_{2},x_{3},x_{4})=(5x_{1}+7x_{2}+11x_{3}+3_{4}, 2x_{1}+6x_{2}+9x_{3}+4x_{4})$$ Let $U_{1}$ denote the kernel of $L_{1}$ and $U_{2}$ the kernel of $L_{2}$. Construct bases for $U_{1}$,$U_{2}$, $U_{1}\cap U_{2}$ and $U_{1}\cup U_{2}$.

Now I am a little stuck on how to go about $U_{1}\cup U_{2}$, previous workings show the following,

Our bases for $U_{1} $ is $$\{(-3/10,-11/10,1,0), (3/10,1/10,0,1)\}$$ and for $U_{2}$ is $$\{(-3/16,-23/16,1,0),(5/8,-7/8,0,1)\}$$ therefore for $U_{1}\cap U_{2}$ is $$\{(3/10,1,0,0),(3/11,59/18,26/9,1)\}$$ Any help on where to go on $U_{1}\cup U_{2}$ would be most appreciated.

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$U_1 \cup U_2$ will not be a vector space unless $U_1 \subset U_2$ or $U_2 \subset U_1$ (I haven't checked if either is the case here). If it's not a vector space, it's meaningless to ask for a basis of it. – Chris Eagle May 1 '12 at 15:12
up vote 0 down vote accepted

Sorry but I have no clue about the meaning of the two vectors in the set after the implication sign on the right of $U_1\cap U_2$.

A painless method to solve this is to first look for a basis of $U_0=U_1\cap U_2$. The vector $(x_1,x_2,x_3,x_4)$ is in $U_0$ if and only if $L_1(x_1,x_2,x_3,x_4)=L_2(x_1,x_2,x_3,x_4)=0$. This is a priori a system of 4 linear equations, but in the present case, this system yields only 3 independent equations. Hence $U_0$ is a line, that is, $U_0=\text{Vect}(u_0)$ for some $u_0\ne0$.

At this point, one is ready to look for a basis of $U_1$ and a basis of $U_2$. Note that $u_0$ is already in $U_1$ and $U_2$. You should find some $u_1\notin U_0$ and $u_2\notin U_0$ such that $U_1=\text{Vect}(u_0,u_1)$ and $U_2=\text{Vect}(u_0,u_2)$. A bonus will be that $(u_1,u_2)$ is free hence this yields without further ado that $U_1+U_2=\text{Vect}(u_0,u_1,u_2)$.

Good luck.

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As mentioned by @Chris, $U_1\cup U_2$ is not in general, and not in the present case, a linear subspace. – Did May 1 '12 at 18:41

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