$$W = \operatorname{span}([2,1,0,1], [0,0,1,0]) \\V = \operatorname{span}([1,2,1,3], [3,1,-1,4])$$

I need to find a basis and the dimension for $U+V$ and $U\cap V$. For $U+V$ I tried:

$$U+V = \{u+v|u\in U, v\in V\} = \alpha_1[2,1,0,1]+\alpha_2[0,0,1,0] + \alpha_3[1,2,1,3] + \alpha_4[3,1,-1,4]$$

Therefore I have to find if this set is linearly independent or not. If it is, then it's a basis for $U+V$.

By transforming this to a system, we have:

$$\begin{cases}2\alpha_1 + 0\alpha_2 + 1\alpha_3 + 3\alpha_4 = 0\\1\alpha_1 + 0\alpha_2 + 2\alpha_3 + 1\alpha_4 = 0\\0\alpha_1 + 1\alpha_2 + 1\alpha_3 -1\alpha_4 = 0\\1\alpha_1 + 0\alpha_2 + 3\alpha_3 + 4\alpha_4 = 0\end{cases}$$

By solving this system, we should get the answer. Is there a easy way to solve it?

For the case $U\cap W$ I can't see how to act. Maybe if i find $\beta_1, \beta_2, \beta_3, \beta_4$ such that $$\beta_1[2,1,0,1]+\beta_2[0,0,1,0] = \beta_3[1,2,1,3] + \beta_4[3,1,-1,4]$$ I should get what the intersection is?


1 Answer 1


Here's an efficient approach: row reduce the matrix $$ \pmatrix{ 2&0&1&3\\ 1&0&2&1\\ 0&1&1&-1\\ 1&0&3&4 } $$ The columns corresponding to the pivots form a basis of $U+V$. The others form a basis of $U\cap V$.

  • $\begingroup$ How is this method called? I need to study why it works. By the way, thank you by this :) $\endgroup$ Dec 9, 2014 at 22:06
  • $\begingroup$ I don't know of any name for it. The reason it works is that row-reduction preserves the null-space. $\endgroup$ Dec 9, 2014 at 22:08
  • $\begingroup$ Nice. When I solved for the $\beta$s, I got they all $0$. This means there's no intersection? Then, what is the basis? $\endgroup$ Dec 9, 2014 at 22:14
  • $\begingroup$ This means that the intersection is $\{0\}$, which has the empty set as its basis. $\endgroup$ Dec 9, 2014 at 22:15
  • $\begingroup$ Thank you so much! You said the other form a basis for the intersection. What would be, then, the others, in this rref? $\endgroup$ Dec 9, 2014 at 22:16

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