The sub-vector spaces $U_1=span\{(0,1,2),\;(1,1,1),\;(3,5,7)\}$ and $U_2=span\{(1,1,0),\;(-1,2,2),\;(2,-13,-10),\;(2,-1,-2)\}$ of the $\mathbb{R}^3$ are given. Determine the dimension $\dim$ and a base of $U_1$,$U_2$, $U_1+U_2$ and $U_1\cap U_2$ in each case.
In order to determine the Basis of $U_1$, we just need to insert all vectors of $U_1$ into a matrix and do the Gauss-algorithm: $$ \begin{pmatrix} 0 & 1 & 3\\ 1 & 1 & 5\\ 2 & 1 & 7 \end{pmatrix} \iff \begin{pmatrix} 1 & 1 & 5\\ 0 & -1 & -3\\ 0 & 0 & 0 \end{pmatrix} $$ Therefore, $\dim(U_1)=2$ and $B_{U_1}=\{(0,1,2),\;(1,1,1)\}$ is a basis of $U_1$
Same procedure for $U_2\dots\implies\dim(U_2)=2$ and $B_{U_2}=\{(1,1,0),\;(-1,2,2)\}$
Now a question: For $U_1+U_2$: Could I use $B_{U_1}$ and $B_{U_2}$ in order to determine the basis and dimension of the sum of both subspaces? (Like that):
$$ \begin{pmatrix} 1 & -1 & 2 & 2\\ 1 & 2 & -13 & -1\\ 0 & 2 & -10 & -2 \end{pmatrix} \iff \begin{pmatrix} 1 & -1 & 2 & 2\\ 0 & 3 & -15 & -3\\ 0 & 0 & 0 & 0 \end{pmatrix} $$ Therefore $\dim(U_1+U_2)=3$ and one basis is $B_{sum}=\{(0,1,2),(1,1,1),(1,1,0)\}$
Now $U_1 \cap U_2 $:
We need to create a linear equation system: \begin{align} \lambda_1(0,1,2)+\lambda_2(1,1,1)+\lambda_3(3,5,7)&=\lambda_4(1,1,0)+\dots+\lambda_7(2,-1,-2)\iff\\ \lambda_1(0,1,2)+\lambda_2(1,1,1)+\lambda_3(3,5,7)-&\lambda_4(1,1,0)-\dots-\lambda_7(2,-1,-2)&=0 \end{align} Because we can multiply the matrix by $(-1)$, we don't need to be aware of the negative signs.
Inserting this into a matrix: $$ \begin{pmatrix} 0 & 1 & 3 & &\dots& & 2\\ 0 & 1 & 5 & &\dots& & -1\\ 2 & 1 & 7 & 0 & 2 & -10 & -2 \end{pmatrix} \iff \begin{pmatrix} 1 & 1 & 5 & &\dots& & 2\\ 0 & -1 & -3 & &\dots& & -1\\ 0 & 0 & 0 & 0 & 2 & -10 & -2 \end{pmatrix} $$ Now we solve the linear equation system and get $\lambda_7=z,\;\lambda_6 = y, \;\lambda_5 = x,\;\lambda_4=-3x+18y+2z,\;\lambda_3=w,\;\lambda_2=-3w+4x-20y-4z,\;\lambda_1=-2w-3x+15y+3z$ We insert those Lambdas on one side of the equation and simplify the term a bit. After that, we get that the basis should be:
$B=\{x(-4,1,2)+y(20,5,-10)+z(4,1,-2)\mid x,y,z\in \mathbb{R}\}$ Therefore, $\dim(intersection)=3$
Are those concepts right? Can you tell me if the dimensions and the basis are right?