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Let

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

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?

Guerlando OCs
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1 Answers1

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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 $U\cap V$.

Ben Grossmann
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  • How is this method called? I need to study why it works. By the way, thank you by this :) – Guerlando OCs Dec 09 '14 at 22:06
  • I don't know of any name for it. The reason it works is that row-reduction preserves the null-space. – Ben Grossmann Dec 09 '14 at 22:08
  • Nice. When I solved for the $\beta$s, I got they all $0$. This means there's no intersection? Then, what is the basis? – Guerlando OCs Dec 09 '14 at 22:14
  • This means that the intersection is ${0}$, which has the empty set as its basis. – Ben Grossmann Dec 09 '14 at 22:15
  • Thank you so much! You said the other form a basis for the intersection. What would be, then, the others, in this rref? – Guerlando OCs Dec 09 '14 at 22:16
  • If the $\beta$s had to all be zero, then your rref must be the identity. So, every column is a pivot column. So, there are no other columns. So, the basis of $U \cap V$ consists of no vectors. – Ben Grossmann Dec 09 '14 at 22:19
  • Hi @BenGrossmann would you mind explaining why your method works? I recently asked a similar (more general) question here: https://math.stackexchange.com/questions/3960942/basis-of-u-cap-v-via-row-echelon-form – jlammy Dec 29 '20 at 20:21
  • @jlammy There is a list of a lot of similar questions given in the comment here, perhaps you will find one of this posts helfpul – Ben Grossmann Dec 29 '20 at 20:24