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Let two subspaces of $V=\mathbb{R}^4$:
$$w1 = \left\{ {\left( {\matrix{ 1 \cr 1 \cr 1 \cr 1 \cr } } \right),\left( {\matrix{ 1 \cr 0 \cr 2 \cr 0 \cr } } \right),\left( {\matrix{ 0 \cr 2 \cr 1 \cr 1 \cr } } \right)} \right\},w2 = \left\{ {\left( {\matrix{ 1 \cr 1 \cr 1 \cr 1 \cr } } \right),\left( {\matrix{ 3 \cr 2 \cr 3 \cr 2 \cr } } \right),\left( {\matrix{ 2 \cr { - 1} \cr 2 \cr 0 \cr } } \right)} \right\}$$

I row-reduced them and compared the generalized form (because those vectors are in $w_1 \cap w_2$:

$$\left( {\matrix{ {{\alpha _1}} \cr {{\alpha _1} + {\alpha _2}} \cr { - 2{\alpha _2} + 5{\alpha _3}} \cr {{\alpha _1} + {\alpha _2} + {\alpha _3}} \cr } } \right) = \left( {\matrix{ {{\beta _1}} \cr {{\beta _1} + {\beta _2}} \cr {{\beta _1}} \cr {{\beta _1} + {\beta _2} + {\beta _3}} \cr } } \right)$$

What action should I take from here to find the basis of $w_1 \cap w_2$?
Thanks.

Gerry Myerson
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AnnieOK
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    A small notational point that lots of people get wrong - the sets you have written are not subspaces. You probably mean that $w_1$ and $w_2$ are the spans of those sets. Also, you should say "a basis", rather than "the basis". – mdp May 20 '14 at 13:33
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    A different way to do it is to note that $$\begin{pmatrix}1 \1 \1 \1 \end{pmatrix}\in \langle w_1\rangle\cap \langle w_2\rangle,$$ $$\begin{pmatrix} 2\ -1\ 2\ 0\end{pmatrix}=\begin{pmatrix} 1\ 1\ 1\ 1\end{pmatrix}+\begin{pmatrix} 1\ 0\ 2\ 0\end{pmatrix}-\begin{pmatrix} 0\ 2\ 1\ 1\end{pmatrix}\in \langle w_1\rangle\cap \langle w_2\rangle$$ and to prove that $\begin{pmatrix} 3\ 2\ 3\ 2\end{pmatrix}\not \in \langle w_1\rangle$. – Git Gud May 20 '14 at 13:40

1 Answers1

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One way to do this is to find the vectors in your basis of $W_2$ that are in (the span of your basis for) $W_1$. That is, take the matrix $$ [v_1 \ v_2\ v_3\ w_1 \ w_2 \ w_3] = \pmatrix{ 1&1&0&1&3&2\\ 1&0&2&1&2&-1\\ 1&2&1&1&3&2\\ 1&0&1&1&2&0 } $$ and row-reduce it. The free columns (columns without a pivot) will correspond to non-trivial relations between these basis elements, and we can use these to form a basis for the intersection $W_1 \cap W_2$.

In this particular example, row-reducing yields $$ \pmatrix{1& 0& 0& 1& 0& 1 \\ 0 & 1 & 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 & 1 & 0} $$ Reading the free columns indicates that we have the relations $$ v_1 = w_1\\ v_1 + v_2 - v_3 = w_3. $$ Since these equations have the $v$'s on the left and the $w$'s on the right, we may say that $w_1$ will be the first basis element of our intersection, and $w_3$ will be the second.

Ben Grossmann
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  • Wait, shouldn't you transpose the matrix before row-reduce it? – AnnieOK May 20 '14 at 13:56
  • No. Why would you do that? – Ben Grossmann May 20 '14 at 14:12
  • You right. No need to. By the way, I actually thought that would be the way to find a basis for $w_1 + w_2$. What would you do for that case? – AnnieOK May 20 '14 at 14:17
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    If you want a basis for $W_1 + W_2$, then instead of looking for the free columns, take the pivot columns – Ben Grossmann May 20 '14 at 14:18
  • Right. Now it's 100% clear. Thank you! – AnnieOK May 20 '14 at 14:33
  • (Sorry for reviving old post) But does your method imply that the vector corresponding to 1st column will never form the basis? Since in RRE form, 1st column is unlikely to correspond to a free column. – AJ_ Sep 27 '19 at 11:23
  • @AJ_ Indeed: if you follow this method, then the first column never will be one of the elements for the selected basis of $W_1 \cap W_2$. – Ben Grossmann Sep 27 '19 at 12:34
  • @Omnomnomnom Which means your method will fail for this question? https://math.stackexchange.com/questions/25371/how-to-find-basis-for-intersection-of-two-vector-spaces-in-mathbbrn – AJ_ Sep 27 '19 at 12:47
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    @AJ_ looking at the rref we see that the basis vectors satisfy the relation $$ (1,2,2,-2) = 1 \cdot (1,1,0,-1) + (-1)\cdot (0,-1,-2,1) $$ and indeed, this is not enough to deduce that $(1,2,2,-2)$ spans the intersection. So it seems that my method (while it does indicate the correct dimension) does not give a basis for the intersection. Thanks for pointing out the example. – Ben Grossmann Sep 28 '19 at 11:37
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    @AJ_ in order to correct the method, I would need to add a step wherein we take the relation indicated and rewrite it with vectors from the first basis on the left and vectors from the other on the right. In this case, rewriting the relation yields $$ 1 \cdot (1,1,0,-1) = 1 \cdot (1,2,2,-2) + 1 \cdot (0,-1,-2,1) $$ The vector attained (on each side) after this step (I believe) is the vector we should use to form our basis of $W_1 \cap W_2$. – Ben Grossmann Sep 28 '19 at 11:53
  • @AJ_ I'll write a more thorough explanation of what I mean in an edit to my original answer some time in the next few days. – Ben Grossmann Sep 28 '19 at 12:00
  • @AJ_ see my latest edit – Ben Grossmann Sep 28 '19 at 13:29
  • Makes sense now. Thanks for clarifying! – AJ_ Sep 28 '19 at 15:54