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I'll try to write this as best as I can...

Let the following $U_1, U_2$ be subspaces of $\mathbb{R}^4$

$$ U_1 = \begin{Bmatrix} (x, y, z, w) : z-y+2w = 0 \end{Bmatrix} $$

$$ U_2 = \begin{Bmatrix} (x, y, z, w) : z-y+2w = 0, x=2z \end{Bmatrix} $$

Find a basis for the subspace $(U_1 \cap U_2)$

I have found the bases

$$ B_1 = \begin{Bmatrix} (1, 1, 0, 0), (0, 2, 0, 1), (0, 0, 1, 0) \end{Bmatrix} $$ $$ B_2 = \begin{Bmatrix} (2, 2, 1, 0), (0, 2, 0, 1) \end{Bmatrix} $$

for $U_1, U_2$ respectively, but do not know where to go from here, any help would be greatly appreciated.

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Hint: $U_2\subset U_1$. – Did Jan 10 '12 at 14:45
@DidierPiau I see that $U_2$ is really $U_1$ with the extra condition that $x=2z$. So the elements of $(U_1 \cap U_2)$ are of the form $(2z, 2z+2w, z, w)$. So a basis for $(U_1 \cap U_2)$ is {(2, 2, 1, 0), (0, 2, 0, 1)} is this correct? Also, for subspaces with much more complicated rules, is there a way to derive a basis for the intersection, with the bases for each subspace? thanks – Sam Jan 10 '12 at 14:51
A tip: In general (unlike in this specific artificial problem), the bases for $U_1$ and $U_2$ do not directly help in finding a basis for $U_1 \cap U_2$. (In contrast, it is easier to write down a basis for $U_1 + U_2$ given bases for $U_1$ and $U_2$.) – Srivatsan Jan 10 '12 at 14:53
@Srivatsan I see, I just thought I'd check to see if the bases for each subspace came in handy when working this out, cheers! – Sam Jan 10 '12 at 14:54
The sets $B_1$ and $B_2$ are not bases for $U_1$ and $U_2$. Check your calculations. – Michael Joyce Jan 10 '12 at 15:13
up vote 1 down vote accepted

Since $U_{2}\subset U_{1}$, you have $U_{1}\cap U_{2}=U_{2}$ and for $U_{2}$ for you already found a base (incase that's calculated correctly). I got a different base for $U_{2}$ but there are plenty of different bases for it.

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Not sure why I didn't see it wasn't contained before, thanks! – Sam Jan 10 '12 at 14:55

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