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I am given two subspaces in $\mathbb R^4$ $v_1 = \{x_1 + x_2 = 0\}$ and $v_2 = \{x_3 + x_4 = 0 \}$

How do I find the intersection $v_1\cap v_2$?

Am I correct in thinking that I need to find a basis? And if so, how do I go about finding that in the first place.

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    Are we to guess that you're working with $\;\Bbb R^4\;$ ? And that the sets you're defining are *actually* sets of vectors such that the sum of the first two coordinates equal zero and ...etc. ? – 2017-01-20
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    Yes it is in $\mathbb R^4$ I have corrected that – 2017-01-20
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    Check this out. Linear Algebra, Vector Space: how to find intersection of two subspaces ? "https://math.stackexchange.com/a/2179047/423856" – 2017-12-07

4 Answers 4

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One way would be to find a basis first, you are most right.

One can also reason about the two subspaces. $V_1$ is the subspace where all vectors have $x_1 = -x_2$; $V_2$ is the subspace where all vectors have $x_3 = -x_4$. Thus, if a vector exists in both subspaces at the same time, it must be that it has both $x_1 = -x_2$ and $x_3 = -x_4$. Therefore the intersection is

$$V_1 \cap V_2 = \{v = (x_1, x_2, x_3, x_4) \in V: x_1 = -x_2 \wedge x_3 = -x_4\}$$

Working with basis one would proceed as follows:

$V_1$ only has restrictions in the first two coordinates, thus the vectors $e_3 = (0,0,1,0), e_4 = (0,0,0,1)$ are part of $V_1$; but we also have that $v_1 = (1, -1, 0,0)$ is in $V_1$. For $V_2$, the vectors $e_1 = (1,0,0,0), e_2 = (0,1,0,0)$ are part of $V_2$ and we also have $v_2 = (0,0,1,-1) \in V_2$. Can you show that these two sets form two basis?

Then we just have to see what is the intersection of the two sets of vectors. We want to find

$$(\text{span}\{v_1, e_3, e_4\}) \cap (\text{span}\{v_2, e_1, e_2\})$$

Can you do it?

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    So is it safe to say that the vectors $(1,-1,0,0), (0,0,1,0), (0,0,0,1) = span(v_1)$ and thus are a basis for $V_1$? – 2017-01-20
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    @Torched90 it is the other way around, the subspace $V_1$ is equal to the span of the three vectors: $$\text{span}\{(1,−1,0,0),(0,0,1,0),(0,0,0,1)\} = V_1$$ – 2017-01-20
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    is the answer then just the six vectors that we have found? Namely the interesection $v_1 \cap v_2 = (1,-1,0,0),(0,0,1,0),(0,0,0,1),(1,0,0,0),(0,1,0,0,),(0,0,1,-1)$ – 2017-01-20
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    @Torched90 no, not really! You just have that $V_1$ is created by 3 vectors and that $V_2$ is created by 3 other vectors. You have to find what vectors build the subspace that is the intersection of $V_1$ with $V_2$. – 2017-01-20
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    How do I find the vectors that build the subspace given the information I have? – 2017-01-20
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    You can either reason about it or check, for example, the accepted answer and its first comment, of this question here http://math.stackexchange.com/questions/767882/linear-algebra-vector-space-how-to-find-intersection-of-two-subspaces – 2017-01-20
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GENERAL HINT:

Rewrite $v= \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix}$ by rearranging the given equations, substituting, then separating the resulting vector into a sum with each individual term having only one of the $x_i$'s and factor out each variable from each vector in the sum.

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In this case, it is not too hard to find the intersection. If you want to compute a basis for the given intersection, it implies that those basis vectors (or basis vector) obey the two given equations. So the basis of the intersection is the basis for the following span: $\text{vct}\{(x_1, x_2, x_3, x_4): x_1 + x_2 =0 \ \text{and } x_3 + x_4 =0\}$.

From this point on, I believe you can solve it. Mind that this basis consists of two vectors.

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The intersection is formed by all the vectors of the form

$$\begin{pmatrix}a\\\!\!-a\\b\\\!\!-b\end{pmatrix}$$

Is it now easier to see what the dimension is and even to get a basis for the intersection?

Another approach: observe that

$$\dim v_1=\dim v_2=3\;,\;\;\text{so}\;\;4\ge\dim(v_1+v_2)=\dim v_1+\dim v_2-\dim (v_1\cap v_2)$$

and we get the intersection's dimension is at least $\;2\;$ ...