# Intersection of all subspace of V is the empty set

• MHB
• mathmari
In summary: Yes, the new vectors must be linearly independent in order for the new set to be a basis. Thank you for the explanation! In summary, we discussed the concept of exchanging a vector in a basis with another element in order to obtain a new basis in a subspace. Using this fact, we were able to prove that the intersection of all subspaces of a given dimension in a vector space is the empty set. We also looked at an example in $\mathbb{R}^{3}$ and examined a particular subcollection of 2-dimensional subspaces to illustrate this concept.
mathmari
Gold Member
MHB
Hey!

Let $V$ be a $\mathbb{R}$-subspace with basis $B=\{v_1 ,v_2, \ldots , v_n\}$ and $\overline{v}\in V$, $\overline{v}\neq 0$.

I have shown that if we exchange $\overline{v}$ with an element $v_i\in B$ we get again a basis.

How can we show, using this fact, that the intersection of all subspace of $V$ of dimension $n-1$ is $\{0\}$ ?

Could you give me a hint? (Wondering)

Hi mathmari,

Trying to argue by contradiction can be useful here. Suppose there is a non-zero vector, say $u$, common to all $n-1$-dimensional subspaces. Can you construct an $n-1$-dimensional subspace to which $u$ cannot belong?

We have that $B=\{v_1, \ldots , v_{j-1}, v_j,v_{j+1},\ldots ,v_n\}$ is a basis of $V$.

We can exchange $\overline{v}$ with an element $v_j \in B$ and we get again a basis. So, we have that $B'=\{\bar{v},v_1,\dots,v_{j-1},v_{j+1},v_n\}$ is also a basis of $V$.

We have that $\bar{v}$ doesn't belong to the span of $\{v_1,\dots,v_{j-1},v_{j+1},v_n\}$ that has dimension $n-1$, otherwise the vectors $\bar{v}, v_1,\dots,v_{j-1},v_{j+1},v_n$ would be linear dependent, a contradiction to the fact that $B'=\{\bar{v},v_1,\dots,v_{j-1},v_{j+1},v_n\}$ is a basis.

So, we have that $\bar{v}$ doesn't belong to a $(n-1)$-dimensional subspace, and so it cannot belong to the intersection of all subspaces of $V$ of dimension $n-1$.

It follows that the intersection of all subspaces of $V$ of dimension $n-1$ is the empty set, $\{0\}$. Is everything correct? (Wondering)
Can we make it specific when we have a specific vector space? If we take for example $V=\mathbb{R}^3$, we have the basis $B=\{(1,0,0), (0,1,0), (0,0,1)\}$. We want the intersection of alle subspaces of dimension 2. Can we find them? (Wondering)

mathmari said:
We have that $B=\{v_1, \ldots , v_{j-1}, v_j,v_{j+1},\ldots ,v_n\}$ is a basis of $V$.

We can exchange $\overline{v}$ with an element $v_j \in B$ and we get again a basis. So, we have that $B'=\{\bar{v},v_1,\dots,v_{j-1},v_{j+1},v_n\}$ is also a basis of $V$.

We have that $\bar{v}$ doesn't belong to the span of $\{v_1,\dots,v_{j-1},v_{j+1},v_n\}$ that has dimension $n-1$, otherwise the vectors $\bar{v}, v_1,\dots,v_{j-1},v_{j+1},v_n$ would be linear dependent, a contradiction to the fact that $B'=\{\bar{v},v_1,\dots,v_{j-1},v_{j+1},v_n\}$ is a basis.

So, we have that $\bar{v}$ doesn't belong to a $(n-1)$-dimensional subspace, and so it cannot belong to the intersection of all subspaces of $V$ of dimension $n-1$.

It follows that the intersection of all subspaces of $V$ of dimension $n-1$ is the empty set, $\{0\}$. Is everything correct? (Wondering)

Argument looks good. I would add one small comment: you cannot exchange an arbitrary element of your basis with $\bar{v}$ to obtain a new basis; i.e., there is a specific $v_{j}$ which can be replaced by $\bar{v}$. For example, if $\bar{v}=[2, 0, 0]^{T}$, you cannot replace $[0, 1, 0]^{T}$ or $[0, 0, 1]^{T}$ by $\bar{v}$ and obtain a new basis for $\mathbb{R}^{3}$. Only by replacing $[1, 0 , 0]^{T}$ with $\bar{v}$ do we obtain a "new" basis for $\mathbb{R}^{3}$. Not saying you weren't aware of this, just wanted to make sure it was said explicitly just in case :)

mathmari said:
Can we make it specific when we have a specific vector space? If we take for example $V=\mathbb{R}^3$, we have the basis $B=\{(1,0,0), (0,1,0), (0,0,1)\}$. We want the intersection of all subspaces of dimension 2. Can we find them? (Wondering)

As we now know, the intersection of all 2-dimensional subspaces in $\mathbb{R}^{3}$ will be $0=[0,0,0]^{T}.$ There are infinitely many subspaces of dimension 2 in $\mathbb{R}^{3}$ since there are infinitely many planes passing through the origin. Nevertheless, we can examine a particular subcollection of these to the same end.

Consider the set of planes/2-dimensional subspaces obtained by rotating the $xz$-plane in $\mathbb{R}^{3}$ around the $z$-axis by $\theta$, where $0\leq \theta <\pi$. For fixed $\theta$, $\left\{[\cos\theta, \sin\theta, 0]^{T}, [0, 0, 1]^{T} \right\}$ is a basis for the 2-dimensional subspace, say $V_{\theta}$, obtained by rotating the $xz$-plane by $\theta.$ We then have $\cap_{0\leq \theta<\pi}V_{\theta}=0$. Since $V_{\theta}$ is just a particular subcollection of all the 2-dimensional subspaces of $\mathbb{R}^{3}$ (and since $0=[0,0,0]^{T}$ must be in all subspaces of $\mathbb{R}^{3}$, regardless of dimension), the intersection of all 2-dimensional subspaces must be $0$.

Try forming the analogous argument (with pictures!) in $\mathbb{R}^{2}$ using lines passing through the origin for another example. Hope this helps.

GJA said:
Argument looks good. I would add one small comment: you cannot exchange an arbitrary element of your basis with $\bar{v}$ to obtain a new basis; i.e., there is a specific $v_{j}$ which can be replaced by $\bar{v}$. For example, if $\bar{v}=[2, 0, 0]^{T}$, you cannot replace $[0, 1, 0]^{T}$ or $[0, 0, 1]^{T}$ by $\bar{v}$ and obtain a new basis for $\mathbb{R}^{3}$. Only by replacing $[1, 0 , 0]^{T}$ with $\bar{v}$ do we obtain a "new" basis for $\mathbb{R}^{3}$. Not saying you weren't aware of this, just wanted to make sure it was said explicitly just in case :)

It is because the new vectors must again be linearly independet, right? (Wondering)
GJA said:
As we now know, the intersection of all 2-dimensional subspaces in $\mathbb{R}^{3}$ will be $0=[0,0,0]^{T}.$ There are infinitely many subspaces of dimension 2 in $\mathbb{R}^{3}$ since there are infinitely many planes passing through the origin. Nevertheless, we can examine a particular subcollection of these to the same end.

Consider the set of planes/2-dimensional subspaces obtained by rotating the $xz$-plane in $\mathbb{R}^{3}$ around the $z$-axis by $\theta$, where $0\leq \theta <\pi$. For fixed $\theta$, $\left\{[\cos\theta, \sin\theta, 0]^{T}, [0, 0, 1]^{T} \right\}$ is a basis for the 2-dimensional subspace, say $V_{\theta}$, obtained by rotating the $xz$-plane by $\theta.$ We then have $\cap_{0\leq \theta<\pi}V_{\theta}=0$. Since $V_{\theta}$ is just a particular subcollection of all the 2-dimensional subspaces of $\mathbb{R}^{3}$ (and since $0=[0,0,0]^{T}$ must be in all subspaces of $\mathbb{R}^{3}$, regardless of dimension), the intersection of all 2-dimensional subspaces must be $0$.

Try forming the analogous argument (with pictures!) in $\mathbb{R}^{2}$ using lines passing through the origin for another example. Hope this helps.

Ah ok! (Thinking)

I think you are working too hard.

Let $V$ be a vector space of finite dimension $n$ over a field $K$. The intersection of all subspaces of $V$ of dimension $n-1$ is $<0>$.

Let $v\neq 0\in V$. Then there exists a basis $B$ of $V$ with $v\in B$ (for any set $S$ of independent vectors of $V$, there is a basis $B$ of $V$ with $S\subseteq B$). Clearly the span $W$ of $B\setminus \{v\}$ is a subspace of dimension $n-1$ with $v\notin W$. QED.

By the way, the title to this thread said "empty set". Obviously, {0} is not the "empty set".

Use \{\varnothing\} for $\{\varnothing\}$.

greg1313 said:
Use \{\varnothing\} for $\{\varnothing\}$.
But that isn't what we want. That is "the set whose only member is the empty set". The "intersection of all subspaces of V" is the set whose only member is the 0 vector, {0}.

## 1. What is the intersection of all subspaces of V?

The intersection of all subspaces of V is the set of all elements that are common to every subspace in V. In other words, it is the set of all elements that belong to every subspace in V.

## 2. Why is the intersection of all subspaces of V often referred to as the empty set?

The intersection of all subspaces of V is often referred to as the empty set because it is a set that contains no elements. This is because there may not be any common elements among all subspaces in V, resulting in an empty set.

## 3. How is the intersection of all subspaces of V related to linear independence?

The intersection of all subspaces of V is closely related to linear independence. If the intersection of all subspaces of V is non-empty, then the set of vectors that span these subspaces is linearly dependent. If the intersection is empty, then the set of vectors is linearly independent.

## 4. Can the intersection of all subspaces of V be non-empty?

Yes, the intersection of all subspaces of V can be non-empty if there is at least one element that is common to all subspaces in V. This can happen if V is a finite-dimensional vector space and some of its subspaces have a non-empty intersection.

## 5. How can the intersection of all subspaces of V be used in linear algebra?

The intersection of all subspaces of V can be used in linear algebra to determine the linear dependence or independence of a set of vectors that span these subspaces. It can also help in understanding the structure and properties of a vector space, as well as in solving systems of linear equations.

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