# 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)$$U = $$ So $$f(U) = < f(u_1),f(u_2)> = = <(1,-1,0,0)>$$

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

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".