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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))

Magneto
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2 Answers2

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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 $$\dim(w_4 \cap w_3)=\dim(w_1 \cap w_2 \cap w_3)$$

user
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  • 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 – Magneto Aug 02 '18 at 14:34
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    @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 $W_2$. – user Aug 02 '18 at 14:39
  • 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? – Magneto Aug 02 '18 at 14:44
  • sir pls explain – Magneto Aug 02 '18 at 14:48
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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 $a = b = x/2$.

Cloudscape
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    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. – Magneto Aug 02 '18 at 14:26
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    You got four unkowns. Plenty of wiggle room. – Cloudscape Aug 02 '18 at 14:27
  • is there any shorter way to do this? – Magneto Aug 02 '18 at 14:29
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    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.) – Cloudscape Aug 02 '18 at 14:32
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    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 – Magneto Aug 02 '18 at 14:36
  • If i use above statement (like in sets) i get intersction of 3 subspaces = 2*3 - 1-1-1+3 = 6-3+3 = 6 which is wrong – Magneto Aug 02 '18 at 14:38
  • The image was only two-dimensional because the spaces weren't coincident. – Cloudscape Aug 02 '18 at 14:39
  • But why this statement is not working(in abve 1 -$ dim(w_1 \cap w_2 \ cap w_3) = ....$). We have similar statement which is valid for 2 subspaces. – Magneto Aug 02 '18 at 14:47
  • sir pls explain – Magneto Aug 02 '18 at 14:48
  • got it from this post --https://math.stackexchange.com/questions/875433/dimension-of-the-sum-of-three-subspaces ......... Thank you – Magneto Aug 02 '18 at 14:53