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I need to find a basis for the intersection of the following subspace: $$U = \text{Span}\left\{(1, 1, 1, 1),(1, -2, -2, 1)\right\}$$

$$W = \left\{(x,y,z,t) \in \mathbb{R}^4 | x+y+z+t =0\right\}$$

Find a basis of $U \cap W$.

I wrote: $$x=-y-z-t$$ $$W=\text{Span}\left\{(-1,1,0,0), (-1,0,1,0), (-1,0,0,1)\right\}$$ Then I assumed that a vector $v$ is in both spans and solved for $v-v=0$

The result is $(-1,-2,3,3,-3)$ for the scalars respectively. In this way both subspace spans yield $(-3,3,3,-3)$

What now? What is the basis or how do I proceed from here?

Also is there a better, perhaps more methodical way of doing it than I did?

Note: Please don't use matrices, kernels or slightly more advanced techniques, only basic methods, second month of freshman undergraduate.

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  • $\begingroup$ Upon the request in the last paragraph, what book on linear algebra are you reading? $\endgroup$
    – user9464
    Dec 23 '16 at 19:41
  • $\begingroup$ No book, anything I can say about what I know that would make it easier to answer? Because only the very basics of vectors spans etc. $\endgroup$
    – Lumon
    Dec 23 '16 at 19:42
  • $\begingroup$ Also, there is no "the" basis but "a" basis. See a related question: math.stackexchange.com/q/97880/9464 $\endgroup$
    – user9464
    Dec 23 '16 at 19:44
  • $\begingroup$ Corrected, thanks. $\endgroup$
    – Lumon
    Dec 23 '16 at 19:46
  • $\begingroup$ "... was $(-1,-2,3,3,-3)$"? $\endgroup$
    – user9464
    Dec 23 '16 at 19:59
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Hint: every vector $u$ of $U$ can be written as a linear combination $u = a(1,1,1,1) + b(1, -2, -2, 1)$. What condition(s) should $a$ and $b$ satisfy for the vector $u$ to be in $W$?

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  • $\begingroup$ The sum of coordinates ot the resulting vector must be zero. Although this is not a condition for a and b, so I don't know $\endgroup$
    – Lumon
    Dec 23 '16 at 19:57
  • $\begingroup$ Well, $u=(a+b, a-2b, a-2b, a+b)$, so $(a+b)+(a-2b)+(a-2b)+(a+b) = 0$, which is a condition on $a$ and $b$ ... $\endgroup$ Dec 23 '16 at 19:58
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You are doing too much. The dimension of $U\cap W$ must be at most $2$ since $U\cap W\subset U$. Now note that neither $v=(1,1,1,1)$ nor $w=(1,-2,-2,1)$ is in $W$. Hence the dimension must be at most $1$. On the other hand, one has $$ \frac{1}{2}u+v\in W. $$ which implies that $U\cap W$ is of dimension at least $1$. Now, what is the conclusion?

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  • $\begingroup$ Just 1.5,-1.5,-1.5,1.5? $\endgroup$
    – Lumon
    Dec 23 '16 at 20:03
  • $\begingroup$ That is correct. Though, you might want to be careful with the writing: it is $(1.5,-1.5,-1.5,1.5)$. Also, any non-zero multiple of this vector can serve as a basis for $U\cap W$. $\endgroup$
    – user9464
    Dec 23 '16 at 20:12

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