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If three subspaces in $R^3$ $w_1 = \{ (x,y,z): x+y-z=0\} , w_2 = \{ (x,y,z): 3x+y-2z=0\}, w_3 = \{ (x,y,z): x-7y+3z=0\} $ then find $dim(w_1 \cap w_2 \cap w_3), dim(w_1+w_2)$

My attempt : basis of 3 subspaces $B_{w_1} = (1,0,1)(0,1,1)\\B_{w_2} = (2,0,3)(0,2,1)\\ B_{w_3} = (3,0,-1)(0,3,7)\\$

Now $dim(w_1 + w_2) $ is obtained by row reduction of $w_1 \cup w_2$ and i got this reduced to (1,0,0) (0,1,0)(0,0,1). So $dim (w_1 + w_2) = 3$

This implies $dim(w_1 \cap w_2) = 2+2-3 = 1$ ---> (A)

But if i take in following way , i am getting different answer for $dim(w_1 \cap w_2)$

Let any arbitrary vector in $w_1$ is (a,b,a+b). If this is present in $w_2$ then

$(a,b,a+b ) = x(2,0,3)+y(0,2,1) \\ = (2x, 2y,3x+y) \\ \implies inconsistency$ for some scalars x,y

So $w_1 \cap w_2 = \phi \implies dim (w_1 \cap w_2) = 0$ But by set theory, i am getting it as 1. (from (A))

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I'd just use some shortcuts.

For example, $\dim(W_1 + W_2) = 3$ because clearly, $\dim(W_1) = 2$ and there is an element in $W_2$ that is not in $W_1$, which one may adjoint to a basis of $W_1$ to obtain a basis of $\mathbb R^3$.

Your first computation of $\dim(W_1 \cap W_2)$ is based on the morphism $\mathbb R^3 \to (\mathbb R^3 / W_1) \times (\mathbb R^3 / W_2)$, whose kernel is indeed $W_1 \cap W_2$. The image is two-dimensional, and thus the kernel must be one-dimensional. Your solution is hence correct.

I don't see though how you derive your "inconsistency". For example, take $x = y$ and $a = b = x/2$.

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    $\begingroup$ got it.... Tq ... i did not solve the equations completely, i just saw x=a/2, y = b/2, then 3x+y = a+b does not satisfy, with x=a/2, y=b/2. In fact if i proceed futher, i get 3x+y = a+b; 2x+2y = a+b. implies x=y=a/2. $\endgroup$ – Magneto Aug 2 '18 at 14:26
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    $\begingroup$ You got four unkowns. Plenty of wiggle room. $\endgroup$ – AlgebraicsAnonymous Aug 2 '18 at 14:27
  • $\begingroup$ is there any shorter way to do this? $\endgroup$ – Magneto Aug 2 '18 at 14:29
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    $\begingroup$ I'm not sure what you mean, but as it stands in my post, I don't see any even faster methods. Though as far as I can see, the first part of the exercise isn't even required for the second. It is a general theorem that the intersection of two two-dimensional subspaces of 3-space is 1-dimensional. (Unless of course they are equal.) $\endgroup$ – AlgebraicsAnonymous Aug 2 '18 at 14:32
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    $\begingroup$ GOt it ... so can we use this $dim(w_1 \cap w_2 \ cap w_3) = dim(w_1)+dim(w_2)+dim(w_3)+dim(w_1 \cup w_2 \ cup w_3)-dim (w_1 \cap w_2) - dim(w_2 \cap w_3) - dim(w_1 \cap w_3) $ ? it seems we can not apply. Actually (1,1,2) is present in all subspaces. It means dim of intersection of 3 subspaces = 1 $\endgroup$ – Magneto Aug 2 '18 at 14:36
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As noted by Grassmann we have

$$\dim(w_1 \cap w_2)=1$$

then find the basis for that subspace $w_4$ (line) and then apply the same method to find $$\dim(w_4 \cap w_3)=\dim(w_1 \cap w_2 \cap w_3)$$

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  • $\begingroup$ so can we use this $dim(w_1 \cap w_2 \ cap w_3) = dim(w_1)+dim(w_2)+dim(w_3)+dim(w_1 \cup w_2 \ cup w_3)-dim (w_1 \cap w_2) - dim(w_2 \cap w_3) - dim(w_1 \cap w_3) $ ? it seems we can not apply. Actually (1,1,2) is present in all subspaces. It means dim of intersection of 3 subspaces = 1 $\endgroup$ – Magneto Aug 2 '18 at 14:34
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    $\begingroup$ @Magneto Your second way to obtain $\dim(w_1 \cap w_2)$. The result obtained bu Grassman is the correct one and $\dim(w_1 \cap w_2)=1$. To obtain a basis we need to solve the system $$av_1+bu_1=cv_2+du_2$$ with $(v_1,u_1)$ basis of $W_1$ and $(v_2,u_2)$ basis of $W_2$. $\endgroup$ – user Aug 2 '18 at 14:39
  • $\begingroup$ In case of 2 subspaces $dim(w1+ w2) = dim(w_1) + dim(w_2)-dim(w_1 \cap w_2)$. But why this representation in 3 subspaces (like above statement) is not working? can u pls explain? $\endgroup$ – Magneto Aug 2 '18 at 14:44
  • $\begingroup$ sir pls explain $\endgroup$ – Magneto Aug 2 '18 at 14:48

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