Let $ U= \{(x_1,x_2,x_3,x_4)\mid x_2+x_3=0 , x_1-4x_4=0\}$ , similarly $V=\{(x_1,x_2,x_3,x_4)\mid x_1=0\}$.

How can I find a basis for each of the subspaces and how can one determine the dimension of their sum/intersection ?

I just started studying linear algebra and I'm really having a tough time trying to find a basis for a vector space.

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    $\begingroup$ Do you know how to compute the nullspace of a matrix? Can you see how you might reframe this problem in terms of that question? $\endgroup$ – Alex Wertheim Dec 7 '14 at 21:37

You can think about dimension as the minimum number of free parameters with which you can determine every vector of a space. So, for $U$, you know that $x_3=-x_2$ and $x_1 = 4x_4$. Then, $$ U=\{(4x_4,x_2,-x_2,x_4)\,|\,x_2,x_4\in\mathbb{R}\} = \operatorname{Span}\{(4,0,0,1),(0,1,-1,0)\}. $$ Analogously, for $V$: $$ V= \{(0,x_2,x_3,x_4)\,|\, x_2,x_3,x_4\in\mathbb{R}\} = \operatorname{Span}\{(0,1,0,0),(0,0,1,0),(0,0,0,1)\}. $$ If a vector is in the intersection, then its first component must be zero, because it's in $V$. On the other hand, if it's in $U$ its fourth component must also be zero and $x_2=-x_3$. Then: $$ U\cap V = \operatorname{Span}\{(0,1,-1,0)\}. $$


Space $U$ is comprised of all vectors of the form


thus every vectorin $U$ is generated by two basis vectors:

$$(x_1,x_2,-x_2,\frac14x_1) = x_1(1,0,0,\frac14)+x_2(0,1,-1,0)$$

This means that

$e_1 = (1,0,0,\frac14)$ and $e_2 = (0,1,-1,0)$ and dimension of $U$, $dimU=2$.

Similarly the $dimV=3$.


$I = U \cap V = (0,x_2,-x_2,0)$, $dimI=1$, we need only one vector to generate $I$.


$S = U + V$= we need three vectors to generate all vectors of the form $(x_1,x_2,x_3,0)$, but also we need sometimes $x_4$ thus 4th vector. $dimS=4$, we need 4 independent vectors to generate it.


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