We have two vector subspaces in $\mathbb R^4$:
$U=sp\{(1,-1,1,-1),(1,3,-1,4)\}$
$V=\{x_1,x_2,x_3,x_4|x_1=x_3, \ and \ x_1+2x_2+2x_3+x_4=0\}$
Find $\dim(U+V)$ and $\dim(U\cap V)$
What I did:
It's easy to see that $U$ is linearly independent, thus it's a basis. So $\dim U=2$.
For V I get:
$S=\{(s,t,s,-3s-2t)|s,t\in \mathbb R\}$ so $sp V = \{(1,0,1,-3),(0,1,0,-2)\}$ again, linearly independent so it's a basis so $\dim V = 2$.
Now if I'll find one of the two dimensions I'm being asked I'll be able to find the other with dimension theorem but I'm not sure how to find $\dim(U+V)$ and $\dim(U\cap V)$ because both subspaces are different.
Best Answer
$\DeclareMathOperator{\span}{span}$ For $U+V$:
By the definition of $U+V$, we know that $$ U+V=\span\left(\{(1,-1,1,-1),(1,3,-1,4)\} \cup \{(1,0,1,-3),(0,1,0,-2)\}\right) $$ We do not know, however, whether $\{(1,-1,1,-1),(1,3,-1,4),(1,0,1,-3),(0,1,0,-2)\}$ forms a basis of this space, since we don't know whether these vectors are linearly independent (hint: they're not). Given a set of vectors, how do we extract a maximal linearly-independent subset?
Hint: given a set of vectors $\{v_1,\dots,v_k\} \subset \mathbb{R}^n$, you can extract a maximal linearly independent subset by row reducing the matrix $$ \pmatrix{|&|&&|\\v_1&v_2&\cdots&v_k\\|&|&&|} $$ and selecting the vectors corresponding to the pivots of the row-reduced matrix.
For $U \cap V$:
Which vectors of the form $a_1\cdot (1,-1,1,-1) + a_2\cdot(1,3,-1,4)$ satisfy the equations defining $V$? Could you solve for $a_1$ and $a_2$ using some sort of matrix equation?
That is, define $u_1 = (1,-1,1,-1)^T$ and $u_2 = (1,3,-1,4)^T$. Any $u \in U$ can be expressed in the form $$ u = a_1u_1 + a_2u_2 = \pmatrix{u_1 & u_2}\pmatrix{a_1\\a_2} = \pmatrix{a_1+a_2 \\ -a_1 + 3a_2 \\ a_1 - a_2 \\ -a_1 + 4a_2} $$ We want to find the values of $u$ (and hence of $a_1$ and $a_2$) that satisfy the equalities defining $V$.
As you found, $V$ is the solution set to the matrix equality $$ \pmatrix{1&0&-1&0\\1&2&2&1} \pmatrix{x_1\\x_2\\x_3\\x_4} = 0 $$ So, in order for $a_1u_1 + a_2 u_2$ to be in $V$, it has to satisfy $$ \pmatrix{1&0&-1&0\\1&2&2&1} (a_1 u_1 + a_2 u_2) = 0 $$ That is, we need to find the solution to the matrix equation $$ \pmatrix{1&0&-1&0\\1&2&2&1} \pmatrix{u_1 & u_2}\pmatrix{a_1\\a_2} = 0 $$