# Base of subspaces and intersections

Let $V$ be a vectorial space with base $[v_1,v_2,v_3,v_4]$. Let $U \subset V$ be the subspace generated by $u_1 = v_1-v_2+v_4$, $u_2 = v_3+v_4$,$u_3 = v_1-v_2-v_3$ and $W \subset V$ the subspace generted by $w_1 = v_1 - 2v_3$,$w_2 = v_2 +v_4$,$w_3 = v_1 + 2v_3-v_4$,$w_4 = 2v_1 + v_2$.

a) Find a base of the subspaces $U,W$.

b) Find a base of $U \cap W$, $U + W$.

c) Find a base of a subspace $L$ such that $U\oplus L = V$.

d) Let $f:V\rightarrow V$ the linear application defined by $f(v_1)=v_1$,$f(v_2)=v_2$,$f(v_3)=0$,$f(v_4)=0.$ Find a base of $f(U)$,$f(W)$.

a) I have: $u_1 = v_1-v_2+v_4$, $u_2 = v_3+v_4$,$u_3 = v_1-v_2-v_3$ so $U = <(1,-1,0,1),(0,0,1,1),(1,-1,-1,0)>$. The three vectors are not linearly indipendent, the second is combination of the other two so $$U = <(1,-1,0,1),(0,0,1,1)>.$$ For $W$ I have $W = <(1,0,-2,0),(0,1,0,1),(1,0,2,-1),(2,1,0,0)>$ They are not indipendent, one is combination of the others thus $$W = <(1,0,-2,0),(0,1,0,1),(1,0,2,-1)>$$

b) Now a base of $U \cap W$. $v \in U$ then

$v = x(1,-1,0,1)+y(0,0,1,1) = (x,-x,y,x+y)$ and $v \in W$ then $v = a(1,0,-2,0)+b(0,1,0,1)+c(1,0,2,-1) = (a+c,b,-2a+2c,b-c)$ the system yields the solution $x = a, b = -a, y = -2a, c = 0$ Thus $v = (x,-x,y,x+y)=(a+c,b,-2a+2c,b-c) = (a,-a,-2a,-a)$

So a base of $U \cap W$ is:

$U \cap W = <(1,-1,-2,-1)>$

Now I have: $\dim U + \dim W = \dim U\cap W + \dim U+W$ so

$\dim U+W = 4$ so a base of $U+W$ is any base of $R^4$ like $U+W = <(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)>$

c) The vectors that generate $L$ must be indipendent with the vectors of $U$ but together they must be a base of $V$. So $u_1 = v_1-v_2+v_4$, $u_2 = v_3+v_4$ I can pick $u_3 = v_4$, $u_4 = v_1$ in this way the vectors are linearly indipendent and they span $V$ because $v_1 = u_4$, $v_2 = -u_1 + u_4 + u_3$, $v_3 = u_2 -u_3$ and $v_4 = u_3$

So $L = <(0,0,0,1),(1,0,0,0)>$

d)$U = <u_1,u_2>$ So $f(U) = < f(u_1),f(u_2)> = <v1-v2> = <(1,-1,0,0)>$

$W = <w_1,w_2,w_3>$ So $f(W) = < f(w_1),f(w_2),f(w_3)> = <v1,v2> = <(1,0,0,0),(0,1,0,0)>$

I would like to know if everything is right.

• Try using \langle . . . \rangle to get $\langle . . . \rangle$ instead of < . . . > in this context :) – Shaun Jun 24 '14 at 15:03
• I guess that $\cup$ should be $\cap$ throughout. – egreg Jun 24 '14 at 15:22
• yes $\cap$ thanks – user144037 Jun 24 '14 at 15:48

• That $V$ has a $4$-element basis doesn't mean that $V=\mathbb{R}^4$, just that $V\cong \mathbb{R}^4$. I suggest that you right in the beginning something like "Consider the isomorphism $V\to \mathbb{R}^4, v_i\mapsto e_i$." Then you can transfer every question to $\mathbb{R}^4$ and solve it there instead.
• For c) using the dimension of $L$ you only need to show "linearly independent" or "spanning" and don't need to show both.
• As pointed out by Shaun in the comments, it looks nicer if you use $\langle\dots\rangle$ instead of $<...>$.
• "indipendent" should spell "independent".