Mixing
Smallest dimension of a hyperplane containg 2 hyperplanes
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The hyperplanes $H_1$ and $H_2$ have dimension $p$ and $q$, respectively. What is the smallest dimension which the hyperplane $H_3$ must have in order to be sure to contain both $H_1$ and $H_2$?
Here, a hyperplane is defined to be a set of the form $x+L$ where $L$ is a subspace and $x$ is a vector. The answer is $p+q+1$ unless the dimension of $H_3$ exceeds the dimension of the whole space. However, I am not sure why this is the case.
How about the case for the hyperplane containing $k$ number of hyperplanes?
I tried visualizing in 3D, but even that got complicated (in drawing).
linear-algebra
asked Jan 27 at 13:06
Cute BrownieCute Brownie
1,048417
$endgroup$
add a comment |
$begingroup$
The hyperplanes $H_1$ and $H_2$ have dimension $p$ and $q$, respectively. What is the smallest dimension which the hyperplane $H_3$ must have in order to be sure to contain both $H_1$ and $H_2$?
Here, a hyperplane is defined to be a set of the form $x+L$ where $L$ is a subspace and $x$ is a vector. The answer is $p+q+1$ unless the dimension of $H_3$ exceeds the dimension of the whole space. However, I am not sure why this is the case.
How about the case for the hyperplane containing $k$ number of hyperplanes?
I tried visualizing in 3D, but even that got complicated (in drawing).
linear-algebra
asked Jan 27 at 13:06
Cute BrownieCute Brownie
1,048417
$endgroup$
add a comment |
$begingroup$
The hyperplanes $H_1$ and $H_2$ have dimension $p$ and $q$, respectively. What is the smallest dimension which the hyperplane $H_3$ must have in order to be sure to contain both $H_1$ and $H_2$?
Here, a hyperplane is defined to be a set of the form $x+L$ where $L$ is a subspace and $x$ is a vector. The answer is $p+q+1$ unless the dimension of $H_3$ exceeds the dimension of the whole space. However, I am not sure why this is the case.
How about the case for the hyperplane containing $k$ number of hyperplanes?
I tried visualizing in 3D, but even that got complicated (in drawing).
linear-algebra
asked Jan 27 at 13:06
Cute BrownieCute Brownie
1,048417
$endgroup$
The hyperplanes $H_1$ and $H_2$ have dimension $p$ and $q$, respectively. What is the smallest dimension which the hyperplane $H_3$ must have in order to be sure to contain both $H_1$ and $H_2$?
Here, a hyperplane is defined to be a set of the form $x+L$ where $L$ is a subspace and $x$ is a vector. The answer is $p+q+1$ unless the dimension of $H_3$ exceeds the dimension of the whole space. However, I am not sure why this is the case.
How about the case for the hyperplane containing $k$ number of hyperplanes?
I tried visualizing in 3D, but even that got complicated (in drawing).
linear-algebra
linear-algebra
asked Jan 27 at 13:06
Cute BrownieCute Brownie
1,048417
asked Jan 27 at 13:06
Cute BrownieCute Brownie
1,048417
asked Jan 27 at 13:06
Cute BrownieCute Brownie
1,048417
asked Jan 27 at 13:06
Cute BrownieCute Brownie
1,048417
asked Jan 27 at 13:06
Cute BrownieCute Brownie
1,048417
1,048417
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This question must have gotten slightly damaged in the transcription. Consider, in 3-space, the line through $(-1, 0, 0)$ in the $(0,1,0)$ direction as $H_1$, and the point $(3,3,3)$ as $H_2$. The claim is that if $H_3$ has dimension $1 + 0 + 1 = 2$, then it is sure to contain $H_1$ and $H_2$. That's simply false, as the $yz$-plane has dimension $2$ but contains neither $H_1$ nor $H_2$.
Probably the correct form of the question is rather messier, and looks something like this: "For each pair of hyperplanes $H_1$ and $H_2$ of dimensions $p$ and $q$ respectively, there's a smallest hyperplane $Q(H_1, H_2)$ containing both. What is $max(dim Q(H_1, H_2) )$ over all hyperplanes $H_1$ and $H_2$ of the specified dimensions?"
(An interesting related exercise, as a warmup, is "What is $min dim Q(H_1, H_2)$?", but I'll leave that for beginners to think about.)
The secret in this situation is to think about embedding your hyperplanes is a slightly higher dimension, which reduces the problem to one about vector subspaces.
Step 1: Let's suppose we're working in $Bbb R^n$. Then add an extra coordinate to get $E = Bbb R^{n+1} = Bbb R^n times Bbb R$, and for a hyperplane $H$ in $R^n$, define
$$
J(H) = { (tx, t) mid x in H, t in Bbb R }
$$
It's not hard to check that $J(H)$ is a vector subspace of $E$, and $dim(J(H) = dim H + 1$. Furthermore, $J$ is a one-to-one mapping between subspaces of $Bbb R^n$ and subspaces of $Bbb R^{n+1}$ that are not entirely contained in $Bbb R^n times {0}$ (or, alternatively, subspaces that meet $Bbb R^n times {1}$). [You should check these claims!]
Example: If $n = 2$ and $L$ is the line $x + y = 1$, then $J(L)$ is the plane through the origin in 3-space that contains the line $x + y = 1, z = 1$. In short, we put $R^2$ into $R^3$ as the $z = 1$ plane, and then connect every point $P$ of our subspace to the origin by including all multiples $tP$ for any $t in Bbb R$.
Now the question becomes: we've got subspaces $A$ and $B$ (instead of $H_1$ and $H_2$, which are a pain to write) and a minimal subspace $C$ containing them. We apply $J$ to get vector subspaces (of $R^{n+1}$!) that we'll call $A' = J(A), B' = J(B), C' = J(C)$, with $dim A' = p + 1$ and $dim B' = q + 1$.
What choice of $A'$ and $B'$ will make $dim C'$ as large as possible?
Let's think about a simple case: what choice of two lines-through-the-origin in 3-space will make the largest subspace containing both? Answer: choosing them non-parallel! In that case, you need a plane to contain both, while if the two lines are parallel, then they are actually identical, and you can contain both in a line.
Back to $A'$ and $B'$: pick a basis for each, say $v_1, ldots, v_{p+1}$ and $w_1, ldots, w_{q+1}$. To contain both $A'$ and $B'$, the space $C'$ must contain the span of all these vectors. If the vectors are all independent, then $dim C' ge (p+1) + (q+1)$. But a minimal choice for $C'$ is simply $$
span(v_1, ldots, v_{p+1}, w_1, ldots, w_{q+1})
$$
which has dimension exactly $p+1 + q+1$.
There's a slight hitch here: we find the largest $C'$ when all the vectors are independent, but what if there are too many of them? What if $p+1 + q+1 > n+1$? Then the dimension of $C'$ can be forced to be as large as $n+1$ (i.e., $C$ is all of $E$), but no larger.
So the summary is that
$$
dim C' le min(n+1, (p+1) + (q+1))
$$
and that by picking $A'$ to be the span of $e_1, ldots e_p$, and $B'$ to be the span of $e_{n+1}, e_n, ldots, e_{n+1-q}$, we get that $dim C'$ is exactly the value given. (You might object to the fact that $A'$ lies entirely in the $Bbb R^n$ subspace of $Bbb R^{n+1}$, and you're right. If we change the last basis vector of $A$ to $e_p + e_{n+1}$, this problem goes away.)
What's that say about $dim C$? It's one less than $dim C'$, so
$$
dim C = min (n, p + q + 1).
$$
and that's the theorem you wanted to prove.
Working through this in every possible case that you can visualize with $n = 1$ or $2$ (i.e., $(p,q) = (0,0), (0, 1), (1, 1), (2, 0), (2, 1)$) will help make it all make sense.
answered Jan 27 at 13:46
John HughesJohn Hughes
64.8k24292
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For any hyperplane $H subseteq mathbb R^n,$ the set of vectors you obtain by taking differences between all pairs of vectors in $H$ is a subspace of $mathbb R^n.$
(It is the same subspace you get by translating $H$ onto the origin and it is parallel to $H$.)
The dimension of the hyperplane $H$ is the number of vectors in the basis of that subspace.
Another way to put it is that any hyperplane in $mathbb R^n$ can be written $v+S$ where $vin mathbb R^n$ and $S$ is a subspace of $mathbb R^n.$
If we have two hyperplanes $H_1=v_1+S_1$ and $H_2=v_2+S_2$
such that the subspaces $S_1$ (with $p$ dimensions) and $S_2$ (with $q$ dimensions) are completely independent and the vector $v_2-v_1$ is independent of both subspaces,
then one can show that any vector between the hyperplanes is a linear combination of the bases of the two subspaces and the vector $v_2-v_1,$
and if any vectors are omitted from that set of $p+q+1$ vectors,
the span of the remaining vectors covers only a proper subset of the vectors between the two hyperplanes.
The minimal hyperplane containing $H_1$ and $H_2$ is therefore $v_1+t(v_2-v_1)+S_1+S_2,$ which has $p+q+1$ dimensions.
An example of this in $mathbb R^3$ is two skew lines, that is, non-parallel non-intersecting lines.
answered Jan 28 at 4:32
David KDavid K
55.4k344120
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2 Answers
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2 Answers
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$begingroup$
This question must have gotten slightly damaged in the transcription. Consider, in 3-space, the line through $(-1, 0, 0)$ in the $(0,1,0)$ direction as $H_1$, and the point $(3,3,3)$ as $H_2$. The claim is that if $H_3$ has dimension $1 + 0 + 1 = 2$, then it is sure to contain $H_1$ and $H_2$. That's simply false, as the $yz$-plane has dimension $2$ but contains neither $H_1$ nor $H_2$.
Probably the correct form of the question is rather messier, and looks something like this: "For each pair of hyperplanes $H_1$ and $H_2$ of dimensions $p$ and $q$ respectively, there's a smallest hyperplane $Q(H_1, H_2)$ containing both. What is $max(dim Q(H_1, H_2) )$ over all hyperplanes $H_1$ and $H_2$ of the specified dimensions?"
(An interesting related exercise, as a warmup, is "What is $min dim Q(H_1, H_2)$?", but I'll leave that for beginners to think about.)
The secret in this situation is to think about embedding your hyperplanes is a slightly higher dimension, which reduces the problem to one about vector subspaces.
Step 1: Let's suppose we're working in $Bbb R^n$. Then add an extra coordinate to get $E = Bbb R^{n+1} = Bbb R^n times Bbb R$, and for a hyperplane $H$ in $R^n$, define
$$
J(H) = { (tx, t) mid x in H, t in Bbb R }
$$
It's not hard to check that $J(H)$ is a vector subspace of $E$, and $dim(J(H) = dim H + 1$. Furthermore, $J$ is a one-to-one mapping between subspaces of $Bbb R^n$ and subspaces of $Bbb R^{n+1}$ that are not entirely contained in $Bbb R^n times {0}$ (or, alternatively, subspaces that meet $Bbb R^n times {1}$). [You should check these claims!]
Example: If $n = 2$ and $L$ is the line $x + y = 1$, then $J(L)$ is the plane through the origin in 3-space that contains the line $x + y = 1, z = 1$. In short, we put $R^2$ into $R^3$ as the $z = 1$ plane, and then connect every point $P$ of our subspace to the origin by including all multiples $tP$ for any $t in Bbb R$.
Now the question becomes: we've got subspaces $A$ and $B$ (instead of $H_1$ and $H_2$, which are a pain to write) and a minimal subspace $C$ containing them. We apply $J$ to get vector subspaces (of $R^{n+1}$!) that we'll call $A' = J(A), B' = J(B), C' = J(C)$, with $dim A' = p + 1$ and $dim B' = q + 1$.
What choice of $A'$ and $B'$ will make $dim C'$ as large as possible?
Let's think about a simple case: what choice of two lines-through-the-origin in 3-space will make the largest subspace containing both? Answer: choosing them non-parallel! In that case, you need a plane to contain both, while if the two lines are parallel, then they are actually identical, and you can contain both in a line.
Back to $A'$ and $B'$: pick a basis for each, say $v_1, ldots, v_{p+1}$ and $w_1, ldots, w_{q+1}$. To contain both $A'$ and $B'$, the space $C'$ must contain the span of all these vectors. If the vectors are all independent, then $dim C' ge (p+1) + (q+1)$. But a minimal choice for $C'$ is simply $$
span(v_1, ldots, v_{p+1}, w_1, ldots, w_{q+1})
$$
which has dimension exactly $p+1 + q+1$.
There's a slight hitch here: we find the largest $C'$ when all the vectors are independent, but what if there are too many of them? What if $p+1 + q+1 > n+1$? Then the dimension of $C'$ can be forced to be as large as $n+1$ (i.e., $C$ is all of $E$), but no larger.
So the summary is that
$$
dim C' le min(n+1, (p+1) + (q+1))
$$
and that by picking $A'$ to be the span of $e_1, ldots e_p$, and $B'$ to be the span of $e_{n+1}, e_n, ldots, e_{n+1-q}$, we get that $dim C'$ is exactly the value given. (You might object to the fact that $A'$ lies entirely in the $Bbb R^n$ subspace of $Bbb R^{n+1}$, and you're right. If we change the last basis vector of $A$ to $e_p + e_{n+1}$, this problem goes away.)
What's that say about $dim C$? It's one less than $dim C'$, so
$$
dim C = min (n, p + q + 1).
$$
and that's the theorem you wanted to prove.
Working through this in every possible case that you can visualize with $n = 1$ or $2$ (i.e., $(p,q) = (0,0), (0, 1), (1, 1), (2, 0), (2, 1)$) will help make it all make sense.
answered Jan 27 at 13:46
John HughesJohn Hughes
64.8k24292
$endgroup$
add a comment |
$begingroup$
This question must have gotten slightly damaged in the transcription. Consider, in 3-space, the line through $(-1, 0, 0)$ in the $(0,1,0)$ direction as $H_1$, and the point $(3,3,3)$ as $H_2$. The claim is that if $H_3$ has dimension $1 + 0 + 1 = 2$, then it is sure to contain $H_1$ and $H_2$. That's simply false, as the $yz$-plane has dimension $2$ but contains neither $H_1$ nor $H_2$.
Probably the correct form of the question is rather messier, and looks something like this: "For each pair of hyperplanes $H_1$ and $H_2$ of dimensions $p$ and $q$ respectively, there's a smallest hyperplane $Q(H_1, H_2)$ containing both. What is $max(dim Q(H_1, H_2) )$ over all hyperplanes $H_1$ and $H_2$ of the specified dimensions?"
(An interesting related exercise, as a warmup, is "What is $min dim Q(H_1, H_2)$?", but I'll leave that for beginners to think about.)
The secret in this situation is to think about embedding your hyperplanes is a slightly higher dimension, which reduces the problem to one about vector subspaces.
Step 1: Let's suppose we're working in $Bbb R^n$. Then add an extra coordinate to get $E = Bbb R^{n+1} = Bbb R^n times Bbb R$, and for a hyperplane $H$ in $R^n$, define
$$
J(H) = { (tx, t) mid x in H, t in Bbb R }
$$
It's not hard to check that $J(H)$ is a vector subspace of $E$, and $dim(J(H) = dim H + 1$. Furthermore, $J$ is a one-to-one mapping between subspaces of $Bbb R^n$ and subspaces of $Bbb R^{n+1}$ that are not entirely contained in $Bbb R^n times {0}$ (or, alternatively, subspaces that meet $Bbb R^n times {1}$). [You should check these claims!]
Example: If $n = 2$ and $L$ is the line $x + y = 1$, then $J(L)$ is the plane through the origin in 3-space that contains the line $x + y = 1, z = 1$. In short, we put $R^2$ into $R^3$ as the $z = 1$ plane, and then connect every point $P$ of our subspace to the origin by including all multiples $tP$ for any $t in Bbb R$.
Now the question becomes: we've got subspaces $A$ and $B$ (instead of $H_1$ and $H_2$, which are a pain to write) and a minimal subspace $C$ containing them. We apply $J$ to get vector subspaces (of $R^{n+1}$!) that we'll call $A' = J(A), B' = J(B), C' = J(C)$, with $dim A' = p + 1$ and $dim B' = q + 1$.
What choice of $A'$ and $B'$ will make $dim C'$ as large as possible?
Let's think about a simple case: what choice of two lines-through-the-origin in 3-space will make the largest subspace containing both? Answer: choosing them non-parallel! In that case, you need a plane to contain both, while if the two lines are parallel, then they are actually identical, and you can contain both in a line.
Back to $A'$ and $B'$: pick a basis for each, say $v_1, ldots, v_{p+1}$ and $w_1, ldots, w_{q+1}$. To contain both $A'$ and $B'$, the space $C'$ must contain the span of all these vectors. If the vectors are all independent, then $dim C' ge (p+1) + (q+1)$. But a minimal choice for $C'$ is simply $$
span(v_1, ldots, v_{p+1}, w_1, ldots, w_{q+1})
$$
which has dimension exactly $p+1 + q+1$.
There's a slight hitch here: we find the largest $C'$ when all the vectors are independent, but what if there are too many of them? What if $p+1 + q+1 > n+1$? Then the dimension of $C'$ can be forced to be as large as $n+1$ (i.e., $C$ is all of $E$), but no larger.
So the summary is that
$$
dim C' le min(n+1, (p+1) + (q+1))
$$
and that by picking $A'$ to be the span of $e_1, ldots e_p$, and $B'$ to be the span of $e_{n+1}, e_n, ldots, e_{n+1-q}$, we get that $dim C'$ is exactly the value given. (You might object to the fact that $A'$ lies entirely in the $Bbb R^n$ subspace of $Bbb R^{n+1}$, and you're right. If we change the last basis vector of $A$ to $e_p + e_{n+1}$, this problem goes away.)
What's that say about $dim C$? It's one less than $dim C'$, so
$$
dim C = min (n, p + q + 1).
$$
and that's the theorem you wanted to prove.
Working through this in every possible case that you can visualize with $n = 1$ or $2$ (i.e., $(p,q) = (0,0), (0, 1), (1, 1), (2, 0), (2, 1)$) will help make it all make sense.
answered Jan 27 at 13:46
John HughesJohn Hughes
64.8k24292
$endgroup$
add a comment |
$begingroup$
This question must have gotten slightly damaged in the transcription. Consider, in 3-space, the line through $(-1, 0, 0)$ in the $(0,1,0)$ direction as $H_1$, and the point $(3,3,3)$ as $H_2$. The claim is that if $H_3$ has dimension $1 + 0 + 1 = 2$, then it is sure to contain $H_1$ and $H_2$. That's simply false, as the $yz$-plane has dimension $2$ but contains neither $H_1$ nor $H_2$.
Probably the correct form of the question is rather messier, and looks something like this: "For each pair of hyperplanes $H_1$ and $H_2$ of dimensions $p$ and $q$ respectively, there's a smallest hyperplane $Q(H_1, H_2)$ containing both. What is $max(dim Q(H_1, H_2) )$ over all hyperplanes $H_1$ and $H_2$ of the specified dimensions?"
(An interesting related exercise, as a warmup, is "What is $min dim Q(H_1, H_2)$?", but I'll leave that for beginners to think about.)
The secret in this situation is to think about embedding your hyperplanes is a slightly higher dimension, which reduces the problem to one about vector subspaces.
Step 1: Let's suppose we're working in $Bbb R^n$. Then add an extra coordinate to get $E = Bbb R^{n+1} = Bbb R^n times Bbb R$, and for a hyperplane $H$ in $R^n$, define
$$
J(H) = { (tx, t) mid x in H, t in Bbb R }
$$
It's not hard to check that $J(H)$ is a vector subspace of $E$, and $dim(J(H) = dim H + 1$. Furthermore, $J$ is a one-to-one mapping between subspaces of $Bbb R^n$ and subspaces of $Bbb R^{n+1}$ that are not entirely contained in $Bbb R^n times {0}$ (or, alternatively, subspaces that meet $Bbb R^n times {1}$). [You should check these claims!]
Example: If $n = 2$ and $L$ is the line $x + y = 1$, then $J(L)$ is the plane through the origin in 3-space that contains the line $x + y = 1, z = 1$. In short, we put $R^2$ into $R^3$ as the $z = 1$ plane, and then connect every point $P$ of our subspace to the origin by including all multiples $tP$ for any $t in Bbb R$.
Now the question becomes: we've got subspaces $A$ and $B$ (instead of $H_1$ and $H_2$, which are a pain to write) and a minimal subspace $C$ containing them. We apply $J$ to get vector subspaces (of $R^{n+1}$!) that we'll call $A' = J(A), B' = J(B), C' = J(C)$, with $dim A' = p + 1$ and $dim B' = q + 1$.
What choice of $A'$ and $B'$ will make $dim C'$ as large as possible?
Let's think about a simple case: what choice of two lines-through-the-origin in 3-space will make the largest subspace containing both? Answer: choosing them non-parallel! In that case, you need a plane to contain both, while if the two lines are parallel, then they are actually identical, and you can contain both in a line.
Back to $A'$ and $B'$: pick a basis for each, say $v_1, ldots, v_{p+1}$ and $w_1, ldots, w_{q+1}$. To contain both $A'$ and $B'$, the space $C'$ must contain the span of all these vectors. If the vectors are all independent, then $dim C' ge (p+1) + (q+1)$. But a minimal choice for $C'$ is simply $$
span(v_1, ldots, v_{p+1}, w_1, ldots, w_{q+1})
$$
which has dimension exactly $p+1 + q+1$.
There's a slight hitch here: we find the largest $C'$ when all the vectors are independent, but what if there are too many of them? What if $p+1 + q+1 > n+1$? Then the dimension of $C'$ can be forced to be as large as $n+1$ (i.e., $C$ is all of $E$), but no larger.
So the summary is that
$$
dim C' le min(n+1, (p+1) + (q+1))
$$
and that by picking $A'$ to be the span of $e_1, ldots e_p$, and $B'$ to be the span of $e_{n+1}, e_n, ldots, e_{n+1-q}$, we get that $dim C'$ is exactly the value given. (You might object to the fact that $A'$ lies entirely in the $Bbb R^n$ subspace of $Bbb R^{n+1}$, and you're right. If we change the last basis vector of $A$ to $e_p + e_{n+1}$, this problem goes away.)
What's that say about $dim C$? It's one less than $dim C'$, so
$$
dim C = min (n, p + q + 1).
$$
and that's the theorem you wanted to prove.
Working through this in every possible case that you can visualize with $n = 1$ or $2$ (i.e., $(p,q) = (0,0), (0, 1), (1, 1), (2, 0), (2, 1)$) will help make it all make sense.
answered Jan 27 at 13:46
John HughesJohn Hughes
64.8k24292
$endgroup$
This question must have gotten slightly damaged in the transcription. Consider, in 3-space, the line through $(-1, 0, 0)$ in the $(0,1,0)$ direction as $H_1$, and the point $(3,3,3)$ as $H_2$. The claim is that if $H_3$ has dimension $1 + 0 + 1 = 2$, then it is sure to contain $H_1$ and $H_2$. That's simply false, as the $yz$-plane has dimension $2$ but contains neither $H_1$ nor $H_2$.
Probably the correct form of the question is rather messier, and looks something like this: "For each pair of hyperplanes $H_1$ and $H_2$ of dimensions $p$ and $q$ respectively, there's a smallest hyperplane $Q(H_1, H_2)$ containing both. What is $max(dim Q(H_1, H_2) )$ over all hyperplanes $H_1$ and $H_2$ of the specified dimensions?"
(An interesting related exercise, as a warmup, is "What is $min dim Q(H_1, H_2)$?", but I'll leave that for beginners to think about.)
The secret in this situation is to think about embedding your hyperplanes is a slightly higher dimension, which reduces the problem to one about vector subspaces.
Step 1: Let's suppose we're working in $Bbb R^n$. Then add an extra coordinate to get $E = Bbb R^{n+1} = Bbb R^n times Bbb R$, and for a hyperplane $H$ in $R^n$, define
$$
J(H) = { (tx, t) mid x in H, t in Bbb R }
$$
It's not hard to check that $J(H)$ is a vector subspace of $E$, and $dim(J(H) = dim H + 1$. Furthermore, $J$ is a one-to-one mapping between subspaces of $Bbb R^n$ and subspaces of $Bbb R^{n+1}$ that are not entirely contained in $Bbb R^n times {0}$ (or, alternatively, subspaces that meet $Bbb R^n times {1}$). [You should check these claims!]
Example: If $n = 2$ and $L$ is the line $x + y = 1$, then $J(L)$ is the plane through the origin in 3-space that contains the line $x + y = 1, z = 1$. In short, we put $R^2$ into $R^3$ as the $z = 1$ plane, and then connect every point $P$ of our subspace to the origin by including all multiples $tP$ for any $t in Bbb R$.
Now the question becomes: we've got subspaces $A$ and $B$ (instead of $H_1$ and $H_2$, which are a pain to write) and a minimal subspace $C$ containing them. We apply $J$ to get vector subspaces (of $R^{n+1}$!) that we'll call $A' = J(A), B' = J(B), C' = J(C)$, with $dim A' = p + 1$ and $dim B' = q + 1$.
What choice of $A'$ and $B'$ will make $dim C'$ as large as possible?
Let's think about a simple case: what choice of two lines-through-the-origin in 3-space will make the largest subspace containing both? Answer: choosing them non-parallel! In that case, you need a plane to contain both, while if the two lines are parallel, then they are actually identical, and you can contain both in a line.
Back to $A'$ and $B'$: pick a basis for each, say $v_1, ldots, v_{p+1}$ and $w_1, ldots, w_{q+1}$. To contain both $A'$ and $B'$, the space $C'$ must contain the span of all these vectors. If the vectors are all independent, then $dim C' ge (p+1) + (q+1)$. But a minimal choice for $C'$ is simply $$
span(v_1, ldots, v_{p+1}, w_1, ldots, w_{q+1})
$$
which has dimension exactly $p+1 + q+1$.
There's a slight hitch here: we find the largest $C'$ when all the vectors are independent, but what if there are too many of them? What if $p+1 + q+1 > n+1$? Then the dimension of $C'$ can be forced to be as large as $n+1$ (i.e., $C$ is all of $E$), but no larger.
So the summary is that
$$
dim C' le min(n+1, (p+1) + (q+1))
$$
and that by picking $A'$ to be the span of $e_1, ldots e_p$, and $B'$ to be the span of $e_{n+1}, e_n, ldots, e_{n+1-q}$, we get that $dim C'$ is exactly the value given. (You might object to the fact that $A'$ lies entirely in the $Bbb R^n$ subspace of $Bbb R^{n+1}$, and you're right. If we change the last basis vector of $A$ to $e_p + e_{n+1}$, this problem goes away.)
What's that say about $dim C$? It's one less than $dim C'$, so
$$
dim C = min (n, p + q + 1).
$$
and that's the theorem you wanted to prove.
Working through this in every possible case that you can visualize with $n = 1$ or $2$ (i.e., $(p,q) = (0,0), (0, 1), (1, 1), (2, 0), (2, 1)$) will help make it all make sense.
answered Jan 27 at 13:46
John HughesJohn Hughes
64.8k24292
answered Jan 27 at 13:46
John HughesJohn Hughes
64.8k24292
answered Jan 27 at 13:46
John HughesJohn Hughes
64.8k24292
answered Jan 27 at 13:46
John HughesJohn Hughes
64.8k24292
64.8k24292
add a comment |
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$begingroup$
For any hyperplane $H subseteq mathbb R^n,$ the set of vectors you obtain by taking differences between all pairs of vectors in $H$ is a subspace of $mathbb R^n.$
(It is the same subspace you get by translating $H$ onto the origin and it is parallel to $H$.)
The dimension of the hyperplane $H$ is the number of vectors in the basis of that subspace.
Another way to put it is that any hyperplane in $mathbb R^n$ can be written $v+S$ where $vin mathbb R^n$ and $S$ is a subspace of $mathbb R^n.$
If we have two hyperplanes $H_1=v_1+S_1$ and $H_2=v_2+S_2$
such that the subspaces $S_1$ (with $p$ dimensions) and $S_2$ (with $q$ dimensions) are completely independent and the vector $v_2-v_1$ is independent of both subspaces,
then one can show that any vector between the hyperplanes is a linear combination of the bases of the two subspaces and the vector $v_2-v_1,$
and if any vectors are omitted from that set of $p+q+1$ vectors,
the span of the remaining vectors covers only a proper subset of the vectors between the two hyperplanes.
The minimal hyperplane containing $H_1$ and $H_2$ is therefore $v_1+t(v_2-v_1)+S_1+S_2,$ which has $p+q+1$ dimensions.
An example of this in $mathbb R^3$ is two skew lines, that is, non-parallel non-intersecting lines.
answered Jan 28 at 4:32
David KDavid K
55.4k344120
$endgroup$
add a comment |
$begingroup$
For any hyperplane $H subseteq mathbb R^n,$ the set of vectors you obtain by taking differences between all pairs of vectors in $H$ is a subspace of $mathbb R^n.$
(It is the same subspace you get by translating $H$ onto the origin and it is parallel to $H$.)
The dimension of the hyperplane $H$ is the number of vectors in the basis of that subspace.
Another way to put it is that any hyperplane in $mathbb R^n$ can be written $v+S$ where $vin mathbb R^n$ and $S$ is a subspace of $mathbb R^n.$
If we have two hyperplanes $H_1=v_1+S_1$ and $H_2=v_2+S_2$
such that the subspaces $S_1$ (with $p$ dimensions) and $S_2$ (with $q$ dimensions) are completely independent and the vector $v_2-v_1$ is independent of both subspaces,
then one can show that any vector between the hyperplanes is a linear combination of the bases of the two subspaces and the vector $v_2-v_1,$
and if any vectors are omitted from that set of $p+q+1$ vectors,
the span of the remaining vectors covers only a proper subset of the vectors between the two hyperplanes.
The minimal hyperplane containing $H_1$ and $H_2$ is therefore $v_1+t(v_2-v_1)+S_1+S_2,$ which has $p+q+1$ dimensions.
An example of this in $mathbb R^3$ is two skew lines, that is, non-parallel non-intersecting lines.
answered Jan 28 at 4:32
David KDavid K
55.4k344120
$endgroup$
add a comment |
$begingroup$
For any hyperplane $H subseteq mathbb R^n,$ the set of vectors you obtain by taking differences between all pairs of vectors in $H$ is a subspace of $mathbb R^n.$
(It is the same subspace you get by translating $H$ onto the origin and it is parallel to $H$.)
The dimension of the hyperplane $H$ is the number of vectors in the basis of that subspace.
Another way to put it is that any hyperplane in $mathbb R^n$ can be written $v+S$ where $vin mathbb R^n$ and $S$ is a subspace of $mathbb R^n.$
If we have two hyperplanes $H_1=v_1+S_1$ and $H_2=v_2+S_2$
such that the subspaces $S_1$ (with $p$ dimensions) and $S_2$ (with $q$ dimensions) are completely independent and the vector $v_2-v_1$ is independent of both subspaces,
then one can show that any vector between the hyperplanes is a linear combination of the bases of the two subspaces and the vector $v_2-v_1,$
and if any vectors are omitted from that set of $p+q+1$ vectors,
the span of the remaining vectors covers only a proper subset of the vectors between the two hyperplanes.
The minimal hyperplane containing $H_1$ and $H_2$ is therefore $v_1+t(v_2-v_1)+S_1+S_2,$ which has $p+q+1$ dimensions.
An example of this in $mathbb R^3$ is two skew lines, that is, non-parallel non-intersecting lines.
answered Jan 28 at 4:32
David KDavid K
55.4k344120
$endgroup$
For any hyperplane $H subseteq mathbb R^n,$ the set of vectors you obtain by taking differences between all pairs of vectors in $H$ is a subspace of $mathbb R^n.$
(It is the same subspace you get by translating $H$ onto the origin and it is parallel to $H$.)
The dimension of the hyperplane $H$ is the number of vectors in the basis of that subspace.
Another way to put it is that any hyperplane in $mathbb R^n$ can be written $v+S$ where $vin mathbb R^n$ and $S$ is a subspace of $mathbb R^n.$
If we have two hyperplanes $H_1=v_1+S_1$ and $H_2=v_2+S_2$
such that the subspaces $S_1$ (with $p$ dimensions) and $S_2$ (with $q$ dimensions) are completely independent and the vector $v_2-v_1$ is independent of both subspaces,
then one can show that any vector between the hyperplanes is a linear combination of the bases of the two subspaces and the vector $v_2-v_1,$
and if any vectors are omitted from that set of $p+q+1$ vectors,
the span of the remaining vectors covers only a proper subset of the vectors between the two hyperplanes.
The minimal hyperplane containing $H_1$ and $H_2$ is therefore $v_1+t(v_2-v_1)+S_1+S_2,$ which has $p+q+1$ dimensions.
An example of this in $mathbb R^3$ is two skew lines, that is, non-parallel non-intersecting lines.
answered Jan 28 at 4:32
David KDavid K
55.4k344120
answered Jan 28 at 4:32
David KDavid K
55.4k344120
answered Jan 28 at 4:32
David KDavid K
55.4k344120
answered Jan 28 at 4:32
David KDavid K
55.4k344120
55.4k344120
add a comment |
add a comment |
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