Having a problem proving $operatorname{rank}T = m$.
$begingroup$
So, I have managed to do parts a,b and the first half of c, However I am struggling with showing that $operatorname{rank} T = m$.
Any tips are really appreciated.
Thanks
linear-algebra vector-spaces linear-transformations
edited Dec 25 '18 at 19:18
Bernard
123k741116
asked Dec 25 '18 at 19:14
SulSul
332114
$endgroup$
|
show 3 more comments
$begingroup$
So, I have managed to do parts a,b and the first half of c, However I am struggling with showing that $operatorname{rank} T = m$.
Any tips are really appreciated.
Thanks
linear-algebra vector-spaces linear-transformations
edited Dec 25 '18 at 19:18
Bernard
123k741116
asked Dec 25 '18 at 19:14
SulSul
332114
$endgroup$
$begingroup$
In re. (3e): what is the inner product defined in Q3(b)?
$endgroup$
– Robert Lewis
Dec 25 '18 at 19:20
1
$begingroup$
I take it that you do not have access to the Rank-Nullity Theorem?
$endgroup$
– ImNotTheGuy
Dec 25 '18 at 19:20
$begingroup$
@ImNotTheGuy I imagine he does but that's not what is required for 3(c).
$endgroup$
– AlephNull
Dec 25 '18 at 19:22
$begingroup$
@AlephNull Except that is exactly answers the part of (c) he is stuck on.
$endgroup$
– ImNotTheGuy
Dec 25 '18 at 19:22
$begingroup$
Except that the 'deduction' in the next part obviously requires Rank-Nullity Theorem.
$endgroup$
– AlephNull
Dec 25 '18 at 19:24
|
show 3 more comments
$begingroup$
So, I have managed to do parts a,b and the first half of c, However I am struggling with showing that $operatorname{rank} T = m$.
Any tips are really appreciated.
Thanks
linear-algebra vector-spaces linear-transformations
edited Dec 25 '18 at 19:18
Bernard
123k741116
asked Dec 25 '18 at 19:14
SulSul
332114
$endgroup$
So, I have managed to do parts a,b and the first half of c, However I am struggling with showing that $operatorname{rank} T = m$.
Any tips are really appreciated.
Thanks
linear-algebra vector-spaces linear-transformations
linear-algebra vector-spaces linear-transformations
edited Dec 25 '18 at 19:18
Bernard
123k741116
asked Dec 25 '18 at 19:14
SulSul
332114
edited Dec 25 '18 at 19:18
Bernard
123k741116
asked Dec 25 '18 at 19:14
SulSul
332114
edited Dec 25 '18 at 19:18
Bernard
123k741116
edited Dec 25 '18 at 19:18
Bernard
123k741116
edited Dec 25 '18 at 19:18
Bernard
123k741116
123k741116
asked Dec 25 '18 at 19:14
SulSul
332114
asked Dec 25 '18 at 19:14
SulSul
332114
asked Dec 25 '18 at 19:14
SulSul
332114
332114
$begingroup$
In re. (3e): what is the inner product defined in Q3(b)?
$endgroup$
– Robert Lewis
Dec 25 '18 at 19:20
1
$begingroup$
I take it that you do not have access to the Rank-Nullity Theorem?
$endgroup$
– ImNotTheGuy
Dec 25 '18 at 19:20
$begingroup$
@ImNotTheGuy I imagine he does but that's not what is required for 3(c).
$endgroup$
– AlephNull
Dec 25 '18 at 19:22
$begingroup$
@AlephNull Except that is exactly answers the part of (c) he is stuck on.
$endgroup$
– ImNotTheGuy
Dec 25 '18 at 19:22
$begingroup$
Except that the 'deduction' in the next part obviously requires Rank-Nullity Theorem.
$endgroup$
– AlephNull
Dec 25 '18 at 19:24
|
show 3 more comments
$begingroup$
In re. (3e): what is the inner product defined in Q3(b)?
$endgroup$
– Robert Lewis
Dec 25 '18 at 19:20
1
$begingroup$
I take it that you do not have access to the Rank-Nullity Theorem?
$endgroup$
– ImNotTheGuy
Dec 25 '18 at 19:20
$begingroup$
@ImNotTheGuy I imagine he does but that's not what is required for 3(c).
$endgroup$
– AlephNull
Dec 25 '18 at 19:22
$begingroup$
@AlephNull Except that is exactly answers the part of (c) he is stuck on.
$endgroup$
– ImNotTheGuy
Dec 25 '18 at 19:22
$begingroup$
Except that the 'deduction' in the next part obviously requires Rank-Nullity Theorem.
$endgroup$
– AlephNull
Dec 25 '18 at 19:24
$begingroup$
In re. (3e): what is the inner product defined in Q3(b)?
$endgroup$
– Robert Lewis
Dec 25 '18 at 19:20
$begingroup$
In re. (3e): what is the inner product defined in Q3(b)?
$endgroup$
– Robert Lewis
Dec 25 '18 at 19:20
1
1
$begingroup$
I take it that you do not have access to the Rank-Nullity Theorem?
$endgroup$
– ImNotTheGuy
Dec 25 '18 at 19:20
$begingroup$
I take it that you do not have access to the Rank-Nullity Theorem?
$endgroup$
– ImNotTheGuy
Dec 25 '18 at 19:20
$begingroup$
@ImNotTheGuy I imagine he does but that's not what is required for 3(c).
$endgroup$
– AlephNull
Dec 25 '18 at 19:22
$begingroup$
@ImNotTheGuy I imagine he does but that's not what is required for 3(c).
$endgroup$
– AlephNull
Dec 25 '18 at 19:22
$begingroup$
@AlephNull Except that is exactly answers the part of (c) he is stuck on.
$endgroup$
– ImNotTheGuy
Dec 25 '18 at 19:22
$begingroup$
@AlephNull Except that is exactly answers the part of (c) he is stuck on.
$endgroup$
– ImNotTheGuy
Dec 25 '18 at 19:22
$begingroup$
Except that the 'deduction' in the next part obviously requires Rank-Nullity Theorem.
$endgroup$
– AlephNull
Dec 25 '18 at 19:24
$begingroup$
Except that the 'deduction' in the next part obviously requires Rank-Nullity Theorem.
$endgroup$
– AlephNull
Dec 25 '18 at 19:24
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
So I will attempt to prove what you need without invoking the Rank-Nullity Theorem, as I feel that the whole point of this exercise is to prove it. If you already have access to it, then I think you can be more direct in your argumentation.
You are given an $n$-dimensional real vector space $V$, an $m$-dimensional subspace $U$, and you are told there is an inner product at your disposal. You are also given a basis ${u_1,cdots,u_m}$ of $U$.
Now, choose a vector in $Vsetminus U$, say $w_1$. Define $v_1=w_1-sum u_ifrac{langle u_i,w_1rangle}{langle u_i,u_irangle}$. Then it is clear that $v_1neq vec{0}$ (since that would imply that $w_1in U$), and by construction $v_1in U^perp$. Now, choose a vector in $Vsetminus text{span}(U,{v_1})$, say $w_2ldots$
Basically, repeat the arguemntation that gives you the Gram-Schmidt process, but use it to create new basis vectors that are orthogonal to everything you have so far. This process ends because $V$ is finite dimensional, and you will be left with a basis of $V$ of the form ${u_1,ldots,u_m,v_1,ldots,v_{n-m}}$ with precisely the properties you need.
If you have the Rank-Nullity Theorem you are then done, as you have pinned down the dimensions of the ambient space and the kernel. If you don't have that theorem, then you need to argue about the dimensions of the image of your transformation $T$, but that is probably even easier than what I have done here; the dimension of the image is the size of a basis for the image, and you need only show that the transformation is a surjection.
EDIT: I don't think I got enough sleep last night, I appear to have wandered a lot in my answer, and accidentally answered part (d) as well.
answered Dec 25 '18 at 19:47
ImNotTheGuyImNotTheGuy
38516
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add a comment |
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$begingroup$
So I will attempt to prove what you need without invoking the Rank-Nullity Theorem, as I feel that the whole point of this exercise is to prove it. If you already have access to it, then I think you can be more direct in your argumentation.
You are given an $n$-dimensional real vector space $V$, an $m$-dimensional subspace $U$, and you are told there is an inner product at your disposal. You are also given a basis ${u_1,cdots,u_m}$ of $U$.
Now, choose a vector in $Vsetminus U$, say $w_1$. Define $v_1=w_1-sum u_ifrac{langle u_i,w_1rangle}{langle u_i,u_irangle}$. Then it is clear that $v_1neq vec{0}$ (since that would imply that $w_1in U$), and by construction $v_1in U^perp$. Now, choose a vector in $Vsetminus text{span}(U,{v_1})$, say $w_2ldots$
Basically, repeat the arguemntation that gives you the Gram-Schmidt process, but use it to create new basis vectors that are orthogonal to everything you have so far. This process ends because $V$ is finite dimensional, and you will be left with a basis of $V$ of the form ${u_1,ldots,u_m,v_1,ldots,v_{n-m}}$ with precisely the properties you need.
If you have the Rank-Nullity Theorem you are then done, as you have pinned down the dimensions of the ambient space and the kernel. If you don't have that theorem, then you need to argue about the dimensions of the image of your transformation $T$, but that is probably even easier than what I have done here; the dimension of the image is the size of a basis for the image, and you need only show that the transformation is a surjection.
EDIT: I don't think I got enough sleep last night, I appear to have wandered a lot in my answer, and accidentally answered part (d) as well.
answered Dec 25 '18 at 19:47
ImNotTheGuyImNotTheGuy
38516
$endgroup$
add a comment |
$begingroup$
So I will attempt to prove what you need without invoking the Rank-Nullity Theorem, as I feel that the whole point of this exercise is to prove it. If you already have access to it, then I think you can be more direct in your argumentation.
You are given an $n$-dimensional real vector space $V$, an $m$-dimensional subspace $U$, and you are told there is an inner product at your disposal. You are also given a basis ${u_1,cdots,u_m}$ of $U$.
Now, choose a vector in $Vsetminus U$, say $w_1$. Define $v_1=w_1-sum u_ifrac{langle u_i,w_1rangle}{langle u_i,u_irangle}$. Then it is clear that $v_1neq vec{0}$ (since that would imply that $w_1in U$), and by construction $v_1in U^perp$. Now, choose a vector in $Vsetminus text{span}(U,{v_1})$, say $w_2ldots$
Basically, repeat the arguemntation that gives you the Gram-Schmidt process, but use it to create new basis vectors that are orthogonal to everything you have so far. This process ends because $V$ is finite dimensional, and you will be left with a basis of $V$ of the form ${u_1,ldots,u_m,v_1,ldots,v_{n-m}}$ with precisely the properties you need.
If you have the Rank-Nullity Theorem you are then done, as you have pinned down the dimensions of the ambient space and the kernel. If you don't have that theorem, then you need to argue about the dimensions of the image of your transformation $T$, but that is probably even easier than what I have done here; the dimension of the image is the size of a basis for the image, and you need only show that the transformation is a surjection.
EDIT: I don't think I got enough sleep last night, I appear to have wandered a lot in my answer, and accidentally answered part (d) as well.
answered Dec 25 '18 at 19:47
ImNotTheGuyImNotTheGuy
38516
$endgroup$
add a comment |
$begingroup$
So I will attempt to prove what you need without invoking the Rank-Nullity Theorem, as I feel that the whole point of this exercise is to prove it. If you already have access to it, then I think you can be more direct in your argumentation.
You are given an $n$-dimensional real vector space $V$, an $m$-dimensional subspace $U$, and you are told there is an inner product at your disposal. You are also given a basis ${u_1,cdots,u_m}$ of $U$.
Now, choose a vector in $Vsetminus U$, say $w_1$. Define $v_1=w_1-sum u_ifrac{langle u_i,w_1rangle}{langle u_i,u_irangle}$. Then it is clear that $v_1neq vec{0}$ (since that would imply that $w_1in U$), and by construction $v_1in U^perp$. Now, choose a vector in $Vsetminus text{span}(U,{v_1})$, say $w_2ldots$
Basically, repeat the arguemntation that gives you the Gram-Schmidt process, but use it to create new basis vectors that are orthogonal to everything you have so far. This process ends because $V$ is finite dimensional, and you will be left with a basis of $V$ of the form ${u_1,ldots,u_m,v_1,ldots,v_{n-m}}$ with precisely the properties you need.
If you have the Rank-Nullity Theorem you are then done, as you have pinned down the dimensions of the ambient space and the kernel. If you don't have that theorem, then you need to argue about the dimensions of the image of your transformation $T$, but that is probably even easier than what I have done here; the dimension of the image is the size of a basis for the image, and you need only show that the transformation is a surjection.
EDIT: I don't think I got enough sleep last night, I appear to have wandered a lot in my answer, and accidentally answered part (d) as well.
answered Dec 25 '18 at 19:47
ImNotTheGuyImNotTheGuy
38516
$endgroup$
So I will attempt to prove what you need without invoking the Rank-Nullity Theorem, as I feel that the whole point of this exercise is to prove it. If you already have access to it, then I think you can be more direct in your argumentation.
You are given an $n$-dimensional real vector space $V$, an $m$-dimensional subspace $U$, and you are told there is an inner product at your disposal. You are also given a basis ${u_1,cdots,u_m}$ of $U$.
Now, choose a vector in $Vsetminus U$, say $w_1$. Define $v_1=w_1-sum u_ifrac{langle u_i,w_1rangle}{langle u_i,u_irangle}$. Then it is clear that $v_1neq vec{0}$ (since that would imply that $w_1in U$), and by construction $v_1in U^perp$. Now, choose a vector in $Vsetminus text{span}(U,{v_1})$, say $w_2ldots$
Basically, repeat the arguemntation that gives you the Gram-Schmidt process, but use it to create new basis vectors that are orthogonal to everything you have so far. This process ends because $V$ is finite dimensional, and you will be left with a basis of $V$ of the form ${u_1,ldots,u_m,v_1,ldots,v_{n-m}}$ with precisely the properties you need.
If you have the Rank-Nullity Theorem you are then done, as you have pinned down the dimensions of the ambient space and the kernel. If you don't have that theorem, then you need to argue about the dimensions of the image of your transformation $T$, but that is probably even easier than what I have done here; the dimension of the image is the size of a basis for the image, and you need only show that the transformation is a surjection.
EDIT: I don't think I got enough sleep last night, I appear to have wandered a lot in my answer, and accidentally answered part (d) as well.
answered Dec 25 '18 at 19:47
ImNotTheGuyImNotTheGuy
38516
edited Dec 25 '18 at 19:55
answered Dec 25 '18 at 19:47
ImNotTheGuyImNotTheGuy
38516
answered Dec 25 '18 at 19:47
ImNotTheGuyImNotTheGuy
38516
answered Dec 25 '18 at 19:47
ImNotTheGuyImNotTheGuy
38516
38516
add a comment |
add a comment |
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$begingroup$
In re. (3e): what is the inner product defined in Q3(b)?
$endgroup$
– Robert Lewis
Dec 25 '18 at 19:20
1
$begingroup$
I take it that you do not have access to the Rank-Nullity Theorem?
$endgroup$
– ImNotTheGuy
Dec 25 '18 at 19:20
$begingroup$
@ImNotTheGuy I imagine he does but that's not what is required for 3(c).
$endgroup$
– AlephNull
Dec 25 '18 at 19:22
$begingroup$
@AlephNull Except that is exactly answers the part of (c) he is stuck on.
$endgroup$
– ImNotTheGuy
Dec 25 '18 at 19:22
$begingroup$
Except that the 'deduction' in the next part obviously requires Rank-Nullity Theorem.
$endgroup$
– AlephNull
Dec 25 '18 at 19:24