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Examples of dim(V+W)+dim(V∩W)=dim(V)+dim(W).

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Examples of dim(V+W)+dim(V∩W)=dim(V)+dim(W).

Dimension of the sum of two vector subspacesIn SVD, why $u_i=Av_i/sigma_i$?Prove that the Cholesky Decomposition for a positive definite matrix is uniqueThe set of all points $in mathbb{R}^3$ twice as far from $mathbf{a}$ as they are from $mathbf{b}$?Prove that dim(T) = dim(V) dim(W) − dim(X)(dim(W) − dim(Y)).Linear transformation rank and kernel proofProve or disprove the claim: $Ker(T^{n}) = Ker(T^{n+1})$ for infinite dimensiondim(Col(A)) ≤ dim(Nul(A))How to prove the dimension identity of subspace? i.e. $dim(V_1) + dim(V_2) = dim(V_1 + V_2) + dim(V_1 cap V_2)$Direct sum of two subspaceExamples of when the Riesz representation theorem doesn't hold

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1

$begingroup$

I saw a few posts that ask to prove this identity, but I'm still trying to understand what it means intuitively. Can anyone show a few examples of it?

linear-algebra vector-spaces

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asked 8 hours ago

David RefaeliDavid Refaeli

12877 bronze badges

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Take $W=0$ and $V=Bbb R^n$, for example. Find an example yourself with $V$ and $W$ subspaces of $Bbb R^2$. More intuition is given here.
  $endgroup$
  – Dietrich Burde
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

I saw a few posts that ask to prove this identity, but I'm still trying to understand what it means intuitively. Can anyone show a few examples of it?

linear-algebra vector-spaces

share|cite|improve this question

asked 8 hours ago

David RefaeliDavid Refaeli

12877 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Take $W=0$ and $V=Bbb R^n$, for example. Find an example yourself with $V$ and $W$ subspaces of $Bbb R^2$. More intuition is given here.
  $endgroup$
  – Dietrich Burde
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

I saw a few posts that ask to prove this identity, but I'm still trying to understand what it means intuitively. Can anyone show a few examples of it?

linear-algebra vector-spaces

share|cite|improve this question

asked 8 hours ago

David RefaeliDavid Refaeli

12877 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

David RefaeliDavid Refaeli

12877 bronze badges

share|cite|improve this question

asked 8 hours ago

David RefaeliDavid Refaeli

12877 bronze badges

asked 8 hours ago

David RefaeliDavid Refaeli

12877 bronze badges

asked 8 hours ago

David RefaeliDavid Refaeli

12877 bronze badges

2 Answers
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4

$begingroup$

Think about the $x-y$ and $y-z$ coordinate planes in space. Each has dimension $2$. Their intersection is the $1$ dimensional $x$ axis. Their sum is the full $3$ dimensional space.

share|cite|improve this answer

answered 8 hours ago

Ethan BolkerEthan Bolker

52.1k55 gold badges6161 silver badges130130 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks! any more examples ? :-)
  $endgroup$
  – David Refaeli
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, you could probably also use 2 lines intersecting (e.g. x-axis, and y-axis) each has dimension of 1, and their intersection has a dimension of 0, so dim(U+V) = dim(U) + dim(V) = 1 + 1 = 2. Am I close?
  $endgroup$
  – David Refaeli
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DavidRefaeli Yes, that's an example too. I think mine is the simplest one where none of the four dimensions is $0$ and neither $V$ nor $W$ is a subspace of the other, and essentially the only such that fits in three space. You can use coordinate subspaces of different sizes in more dimensions to create as much overlap in the intersection as you like.
  $endgroup$
  – Ethan Bolker
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yeah, and it can be expanded to 4 dimensions, say with U = span of (1,0,0,0), (0,1,0,0), (0,0,1,0), V = span of (0,1,0,0), (0,0,1,0), (0,0,0,1): dim(U) = dim(V) = 3. dim(U+V) = dim(U) + dim(V) - dim(U∩V) = 3 + 3 - 2 = 4.
  $endgroup$
  – David Refaeli
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You've got it. When you learn about bases of vector spaces and subspaces you will see that those are actually all the examples.
  $endgroup$
  – Ethan Bolker
  7 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Let $V$ be a finite dimensional $k$-vector space and let $U_1,U_2 subset V$ be subspaces. Then one can consider the vector space $U = U_1 + U_2$. Now a natural question is which dimension $U$ has and how that dimension depends on the dimensions of $U_1$ and $U_2$. The observation is, that in general $text{dim}(U) neq text{dim}(U_1) + text{dim}(U_2)$. Consider for example:

The dimension of a subspace is always bounded by the dimension of the parent space, such that we have $text{dim}(U) leq 3$ in the case of $V = mathbb{R}^3$. Therefore the dimension of the two planes cannot add up. One has to take the intersection into account and substract the dimension of the intersection.

In the case that the intersection is trivial, we call $U$ the direct sum of $U_1$ and $U_2$, which basically means that $U$ has a decomposition into $U_1$ and $U_2$. An equivalent way of stating that is that each vector $u in U$ has a unique representation as $u = u_1 + u_2$, where $u_i in U_i$. In that way the notion of a direct sum can be generalized to arbitrary finite sums of vector spaces $U_1 + dots + U_n$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

ThorWittichThorWittich

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4

$begingroup$

Think about the $x-y$ and $y-z$ coordinate planes in space. Each has dimension $2$. Their intersection is the $1$ dimensional $x$ axis. Their sum is the full $3$ dimensional space.

share|cite|improve this answer

answered 8 hours ago

Ethan BolkerEthan Bolker

52.1k55 gold badges6161 silver badges130130 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks! any more examples ? :-)
  $endgroup$
  – David Refaeli
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, you could probably also use 2 lines intersecting (e.g. x-axis, and y-axis) each has dimension of 1, and their intersection has a dimension of 0, so dim(U+V) = dim(U) + dim(V) = 1 + 1 = 2. Am I close?
  $endgroup$
  – David Refaeli
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DavidRefaeli Yes, that's an example too. I think mine is the simplest one where none of the four dimensions is $0$ and neither $V$ nor $W$ is a subspace of the other, and essentially the only such that fits in three space. You can use coordinate subspaces of different sizes in more dimensions to create as much overlap in the intersection as you like.
  $endgroup$
  – Ethan Bolker
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yeah, and it can be expanded to 4 dimensions, say with U = span of (1,0,0,0), (0,1,0,0), (0,0,1,0), V = span of (0,1,0,0), (0,0,1,0), (0,0,0,1): dim(U) = dim(V) = 3. dim(U+V) = dim(U) + dim(V) - dim(U∩V) = 3 + 3 - 2 = 4.
  $endgroup$
  – David Refaeli
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You've got it. When you learn about bases of vector spaces and subspaces you will see that those are actually all the examples.
  $endgroup$
  – Ethan Bolker
  7 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

Think about the $x-y$ and $y-z$ coordinate planes in space. Each has dimension $2$. Their intersection is the $1$ dimensional $x$ axis. Their sum is the full $3$ dimensional space.

share|cite|improve this answer

answered 8 hours ago

Ethan BolkerEthan Bolker

52.1k55 gold badges6161 silver badges130130 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thanks! any more examples ? :-)
  $endgroup$
  – David Refaeli
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, you could probably also use 2 lines intersecting (e.g. x-axis, and y-axis) each has dimension of 1, and their intersection has a dimension of 0, so dim(U+V) = dim(U) + dim(V) = 1 + 1 = 2. Am I close?
  $endgroup$
  – David Refaeli
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @DavidRefaeli Yes, that's an example too. I think mine is the simplest one where none of the four dimensions is $0$ and neither $V$ nor $W$ is a subspace of the other, and essentially the only such that fits in three space. You can use coordinate subspaces of different sizes in more dimensions to create as much overlap in the intersection as you like.
  $endgroup$
  – Ethan Bolker
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yeah, and it can be expanded to 4 dimensions, say with U = span of (1,0,0,0), (0,1,0,0), (0,0,1,0), V = span of (0,1,0,0), (0,0,1,0), (0,0,0,1): dim(U) = dim(V) = 3. dim(U+V) = dim(U) + dim(V) - dim(U∩V) = 3 + 3 - 2 = 4.
  $endgroup$
  – David Refaeli
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You've got it. When you learn about bases of vector spaces and subspaces you will see that those are actually all the examples.
  $endgroup$
  – Ethan Bolker
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Think about the $x-y$ and $y-z$ coordinate planes in space. Each has dimension $2$. Their intersection is the $1$ dimensional $x$ axis. Their sum is the full $3$ dimensional space.

share|cite|improve this answer

answered 8 hours ago

Ethan BolkerEthan Bolker

52.1k55 gold badges6161 silver badges130130 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Ethan BolkerEthan Bolker

52.1k55 gold badges6161 silver badges130130 bronze badges

answered 8 hours ago

Ethan BolkerEthan Bolker

52.1k55 gold badges6161 silver badges130130 bronze badges

answered 8 hours ago

Ethan BolkerEthan Bolker

52.1k55 gold badges6161 silver badges130130 bronze badges

3

$begingroup$

Let $V$ be a finite dimensional $k$-vector space and let $U_1,U_2 subset V$ be subspaces. Then one can consider the vector space $U = U_1 + U_2$. Now a natural question is which dimension $U$ has and how that dimension depends on the dimensions of $U_1$ and $U_2$. The observation is, that in general $text{dim}(U) neq text{dim}(U_1) + text{dim}(U_2)$. Consider for example:

The dimension of a subspace is always bounded by the dimension of the parent space, such that we have $text{dim}(U) leq 3$ in the case of $V = mathbb{R}^3$. Therefore the dimension of the two planes cannot add up. One has to take the intersection into account and substract the dimension of the intersection.

In the case that the intersection is trivial, we call $U$ the direct sum of $U_1$ and $U_2$, which basically means that $U$ has a decomposition into $U_1$ and $U_2$. An equivalent way of stating that is that each vector $u in U$ has a unique representation as $u = u_1 + u_2$, where $u_i in U_i$. In that way the notion of a direct sum can be generalized to arbitrary finite sums of vector spaces $U_1 + dots + U_n$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

ThorWittichThorWittich

2,46011 silver badge1515 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

Let $V$ be a finite dimensional $k$-vector space and let $U_1,U_2 subset V$ be subspaces. Then one can consider the vector space $U = U_1 + U_2$. Now a natural question is which dimension $U$ has and how that dimension depends on the dimensions of $U_1$ and $U_2$. The observation is, that in general $text{dim}(U) neq text{dim}(U_1) + text{dim}(U_2)$. Consider for example:

The dimension of a subspace is always bounded by the dimension of the parent space, such that we have $text{dim}(U) leq 3$ in the case of $V = mathbb{R}^3$. Therefore the dimension of the two planes cannot add up. One has to take the intersection into account and substract the dimension of the intersection.

In the case that the intersection is trivial, we call $U$ the direct sum of $U_1$ and $U_2$, which basically means that $U$ has a decomposition into $U_1$ and $U_2$. An equivalent way of stating that is that each vector $u in U$ has a unique representation as $u = u_1 + u_2$, where $u_i in U_i$. In that way the notion of a direct sum can be generalized to arbitrary finite sums of vector spaces $U_1 + dots + U_n$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

ThorWittichThorWittich

2,46011 silver badge1515 bronze badges

$endgroup$

add a comment | 

$begingroup$

Let $V$ be a finite dimensional $k$-vector space and let $U_1,U_2 subset V$ be subspaces. Then one can consider the vector space $U = U_1 + U_2$. Now a natural question is which dimension $U$ has and how that dimension depends on the dimensions of $U_1$ and $U_2$. The observation is, that in general $text{dim}(U) neq text{dim}(U_1) + text{dim}(U_2)$. Consider for example:

The dimension of a subspace is always bounded by the dimension of the parent space, such that we have $text{dim}(U) leq 3$ in the case of $V = mathbb{R}^3$. Therefore the dimension of the two planes cannot add up. One has to take the intersection into account and substract the dimension of the intersection.

In the case that the intersection is trivial, we call $U$ the direct sum of $U_1$ and $U_2$, which basically means that $U$ has a decomposition into $U_1$ and $U_2$. An equivalent way of stating that is that each vector $u in U$ has a unique representation as $u = u_1 + u_2$, where $u_i in U_i$. In that way the notion of a direct sum can be generalized to arbitrary finite sums of vector spaces $U_1 + dots + U_n$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

ThorWittichThorWittich

2,46011 silver badge1515 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

ThorWittichThorWittich

2,46011 silver badge1515 bronze badges

answered 8 hours ago

ThorWittichThorWittich

2,46011 silver badge1515 bronze badges

answered 8 hours ago

ThorWittichThorWittich

2,46011 silver badge1515 bronze badges