MHB Intersection of all subspace of V is the empty set

[mathmari]

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)

 

[GJA]

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?

 

[mathmari]

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)

 

[GJA]

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 :)

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.

 

[mathmari]

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)

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)

 

[johng1]

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.

 

[HallsofIvy]

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

 

[Greg]

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

 

[HallsofIvy]

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}.