Given 3 subspaces, Find dimension of intersection of subspaces

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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))

linear-algebra vector-spaces

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asked Aug 2 at 13:53

Magneto

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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))

linear-algebra vector-spaces

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asked Aug 2 at 13:53

Magneto

722213

add a comment | 

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up vote
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1

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))

linear-algebra vector-spaces

share|cite|improve this question

asked Aug 2 at 13:53

Magneto

722213

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))

linear-algebra vector-spaces

share|cite|improve this question

asked Aug 2 at 13:53

Magneto

722213

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share|cite|improve this question

asked Aug 2 at 13:53

Magneto

722213

asked Aug 2 at 13:53

Magneto

722213

asked Aug 2 at 13:53

Magneto

722213

722213

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2 Answers
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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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answered Aug 2 at 14:09

AlgebraicsAnonymous

66111

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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 2 at 14:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You got four unkowns. Plenty of wiggle room.
  – AlgebraicsAnonymous
  Aug 2 at 14:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  is there any shorter way to do this?
  – Magneto
  Aug 2 at 14:29
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  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.)
  – AlgebraicsAnonymous
  Aug 2 at 14:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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 2 at 14:36
  

  
  
  
  

 | 
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1
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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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answered Aug 2 at 14:06

gimusi

63.8k73480

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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 2 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 $W_2$.
  – gimusi
  Aug 2 at 14:39
  
  

  
  
  
  


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  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 2 at 14:44
  

  
  
  
  


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  sir pls explain
  – Magneto
  Aug 2 at 14:48
  

  
  
  
  

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

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 $a = b = x/2$.

share|cite|improve this answer

answered Aug 2 at 14:09

AlgebraicsAnonymous

66111

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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 2 at 14:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You got four unkowns. Plenty of wiggle room.
  – AlgebraicsAnonymous
  Aug 2 at 14:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  is there any shorter way to do this?
  – Magneto
  Aug 2 at 14:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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.)
  – AlgebraicsAnonymous
  Aug 2 at 14:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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 2 at 14:36
  

  
  
  
  

 | 
show 5 more comments

up vote
1
down vote



accepted

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 $a = b = x/2$.

share|cite|improve this answer

answered Aug 2 at 14:09

AlgebraicsAnonymous

66111

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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 2 at 14:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You got four unkowns. Plenty of wiggle room.
  – AlgebraicsAnonymous
  Aug 2 at 14:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  is there any shorter way to do this?
  – Magneto
  Aug 2 at 14:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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.)
  – AlgebraicsAnonymous
  Aug 2 at 14:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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 2 at 14:36
  

  
  
  
  

 | 
show 5 more comments

up vote
1
down vote



accepted

up vote
1
down vote



accepted

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 $a = b = x/2$.

share|cite|improve this answer

answered Aug 2 at 14:09

AlgebraicsAnonymous

66111

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 $a = b = x/2$.

share|cite|improve this answer

answered Aug 2 at 14:09

AlgebraicsAnonymous

66111

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 14:09

AlgebraicsAnonymous

66111

answered Aug 2 at 14:09

AlgebraicsAnonymous

66111

answered Aug 2 at 14:09

AlgebraicsAnonymous

66111

66111

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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 2 at 14:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You got four unkowns. Plenty of wiggle room.
  – AlgebraicsAnonymous
  Aug 2 at 14:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  is there any shorter way to do this?
  – Magneto
  Aug 2 at 14:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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.)
  – AlgebraicsAnonymous
  Aug 2 at 14:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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 2 at 14:36
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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 2 at 14:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You got four unkowns. Plenty of wiggle room.
  – AlgebraicsAnonymous
  Aug 2 at 14:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  is there any shorter way to do this?
  – Magneto
  Aug 2 at 14:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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.)
  – AlgebraicsAnonymous
  Aug 2 at 14:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  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 2 at 14:36
  

  
  
  
  

1

1

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 2 at 14:26

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 2 at 14:26

1

1

You got four unkowns. Plenty of wiggle room.
– AlgebraicsAnonymous
Aug 2 at 14:27

You got four unkowns. Plenty of wiggle room.
– AlgebraicsAnonymous
Aug 2 at 14:27

is there any shorter way to do this?
– Magneto
Aug 2 at 14:29

is there any shorter way to do this?
– Magneto
Aug 2 at 14:29

1

1

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.)
– AlgebraicsAnonymous
Aug 2 at 14:32

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.)
– AlgebraicsAnonymous
Aug 2 at 14:32

1

1

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 2 at 14:36

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 2 at 14:36

 | 
show 5 more comments

up vote
1
down vote

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

share|cite|improve this answer

answered Aug 2 at 14:06

gimusi

63.8k73480

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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 2 at 14:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @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 $W_2$.
  – gimusi
  Aug 2 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 2 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  sir pls explain
  – Magneto
  Aug 2 at 14:48
  

  
  
  
  

add a comment | 

up vote
1
down vote

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

share|cite|improve this answer

answered Aug 2 at 14:06

gimusi

63.8k73480

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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 2 at 14:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @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 $W_2$.
  – gimusi
  Aug 2 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 2 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  sir pls explain
  – Magneto
  Aug 2 at 14:48
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

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

share|cite|improve this answer

answered Aug 2 at 14:06

gimusi

63.8k73480

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

share|cite|improve this answer

answered Aug 2 at 14:06

gimusi

63.8k73480

share|cite|improve this answer

share|cite|improve this answer

answered Aug 2 at 14:06

gimusi

63.8k73480

answered Aug 2 at 14:06

gimusi

63.8k73480

answered Aug 2 at 14:06

gimusi

63.8k73480

63.8k73480

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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 2 at 14:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @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 $W_2$.
  – gimusi
  Aug 2 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 2 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  sir pls explain
  – Magneto
  Aug 2 at 14:48
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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 2 at 14:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @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 $W_2$.
  – gimusi
  Aug 2 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 2 at 14:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  sir pls explain
  – Magneto
  Aug 2 at 14:48
  

  
  
  
  

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 2 at 14:34

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 2 at 14:34

1

1

@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 $W_2$.
– gimusi
Aug 2 at 14:39

@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 $W_2$.
– gimusi
Aug 2 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 2 at 14:44

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 2 at 14:44

sir pls explain
– Magneto
Aug 2 at 14:48

sir pls explain
– Magneto
Aug 2 at 14:48

add a comment | 

 

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