How to calculate subspace of a set of solutions of matrix Ax=b
0
$begingroup$
I am looking through some old linear algebra exam papers. However i do not understand how to calculate whether a set of solutions is within a certain subspace R. This is the problem in question:
I think i understand how to check whether vectors are within a subspace R, but how would i calculate this?
Thanks a lot, really hope you can help me out!
linear-algebra matrices vector-spaces invariant-subspace
edited Dec 28 '16 at 18:59
Dan Rust
23k114984
asked Dec 28 '16 at 18:54
user102937user102937
258
$endgroup$
|
show 1 more comment
0
$begingroup$
I am looking through some old linear algebra exam papers. However i do not understand how to calculate whether a set of solutions is within a certain subspace R. This is the problem in question:
I think i understand how to check whether vectors are within a subspace R, but how would i calculate this?
Thanks a lot, really hope you can help me out!
linear-algebra matrices vector-spaces invariant-subspace
edited Dec 28 '16 at 18:59
Dan Rust
23k114984
asked Dec 28 '16 at 18:54
user102937user102937
258
$endgroup$
1
$begingroup$
What is $mathcal{R}$?
$endgroup$
– Dan Rust
Dec 28 '16 at 19:04
$begingroup$
What have you tried? What is your understanding of how to determine if a set is a subspace of the vector space $mathbb{R} $?
$endgroup$
– user23793
Dec 28 '16 at 19:04
$begingroup$
I tried to follow this video:
$endgroup$
– user102937
Dec 28 '16 at 19:09
$begingroup$
I tried to follow this video: youtube.com/watch?v=q97HmMdD8ZM And to me it seems that the subspace is equal to the amount of rows of the vectors within the span. Am i completely off track or?
$endgroup$
– user102937
Dec 28 '16 at 19:10
$begingroup$
I don't understand the question. What do you want to calculate?
$endgroup$
– Jack
Dec 28 '16 at 19:47
|
show 1 more comment
0
0
0
$begingroup$
I am looking through some old linear algebra exam papers. However i do not understand how to calculate whether a set of solutions is within a certain subspace R. This is the problem in question:
I think i understand how to check whether vectors are within a subspace R, but how would i calculate this?
Thanks a lot, really hope you can help me out!
linear-algebra matrices vector-spaces invariant-subspace
edited Dec 28 '16 at 18:59
Dan Rust
23k114984
asked Dec 28 '16 at 18:54
user102937user102937
258
$endgroup$
I am looking through some old linear algebra exam papers. However i do not understand how to calculate whether a set of solutions is within a certain subspace R. This is the problem in question:
I think i understand how to check whether vectors are within a subspace R, but how would i calculate this?
Thanks a lot, really hope you can help me out!
linear-algebra matrices vector-spaces invariant-subspace
linear-algebra matrices vector-spaces invariant-subspace
edited Dec 28 '16 at 18:59
Dan Rust
23k114984
asked Dec 28 '16 at 18:54
user102937user102937
258
edited Dec 28 '16 at 18:59
Dan Rust
23k114984
asked Dec 28 '16 at 18:54
user102937user102937
258
edited Dec 28 '16 at 18:59
Dan Rust
23k114984
edited Dec 28 '16 at 18:59
Dan Rust
23k114984
edited Dec 28 '16 at 18:59
Dan Rust
23k114984
23k114984
asked Dec 28 '16 at 18:54
user102937user102937
258
asked Dec 28 '16 at 18:54
user102937user102937
258
asked Dec 28 '16 at 18:54
user102937user102937
258
258
1
$begingroup$
What is $mathcal{R}$?
$endgroup$
– Dan Rust
Dec 28 '16 at 19:04
$begingroup$
What have you tried? What is your understanding of how to determine if a set is a subspace of the vector space $mathbb{R} $?
$endgroup$
– user23793
Dec 28 '16 at 19:04
$begingroup$
I tried to follow this video:
$endgroup$
– user102937
Dec 28 '16 at 19:09
$begingroup$
I tried to follow this video: youtube.com/watch?v=q97HmMdD8ZM And to me it seems that the subspace is equal to the amount of rows of the vectors within the span. Am i completely off track or?
$endgroup$
– user102937
Dec 28 '16 at 19:10
$begingroup$
I don't understand the question. What do you want to calculate?
$endgroup$
– Jack
Dec 28 '16 at 19:47
|
show 1 more comment
1
$begingroup$
What is $mathcal{R}$?
$endgroup$
– Dan Rust
Dec 28 '16 at 19:04
$begingroup$
What have you tried? What is your understanding of how to determine if a set is a subspace of the vector space $mathbb{R} $?
$endgroup$
– user23793
Dec 28 '16 at 19:04
$begingroup$
I tried to follow this video:
$endgroup$
– user102937
Dec 28 '16 at 19:09
$begingroup$
I tried to follow this video: youtube.com/watch?v=q97HmMdD8ZM And to me it seems that the subspace is equal to the amount of rows of the vectors within the span. Am i completely off track or?
$endgroup$
– user102937
Dec 28 '16 at 19:10
$begingroup$
I don't understand the question. What do you want to calculate?
$endgroup$
– Jack
Dec 28 '16 at 19:47
1
1
$begingroup$
What is $mathcal{R}$?
$endgroup$
– Dan Rust
Dec 28 '16 at 19:04
$begingroup$
What is $mathcal{R}$?
$endgroup$
– Dan Rust
Dec 28 '16 at 19:04
$begingroup$
What have you tried? What is your understanding of how to determine if a set is a subspace of the vector space $mathbb{R} $?
$endgroup$
– user23793
Dec 28 '16 at 19:04
$begingroup$
What have you tried? What is your understanding of how to determine if a set is a subspace of the vector space $mathbb{R} $?
$endgroup$
– user23793
Dec 28 '16 at 19:04
$begingroup$
I tried to follow this video:
$endgroup$
– user102937
Dec 28 '16 at 19:09
$begingroup$
I tried to follow this video:
$endgroup$
– user102937
Dec 28 '16 at 19:09
$begingroup$
I tried to follow this video: youtube.com/watch?v=q97HmMdD8ZM And to me it seems that the subspace is equal to the amount of rows of the vectors within the span. Am i completely off track or?
$endgroup$
– user102937
Dec 28 '16 at 19:10
$begingroup$
I tried to follow this video: youtube.com/watch?v=q97HmMdD8ZM And to me it seems that the subspace is equal to the amount of rows of the vectors within the span. Am i completely off track or?
$endgroup$
– user102937
Dec 28 '16 at 19:10
$begingroup$
I don't understand the question. What do you want to calculate?
$endgroup$
– Jack
Dec 28 '16 at 19:47
$begingroup$
I don't understand the question. What do you want to calculate?
$endgroup$
– Jack
Dec 28 '16 at 19:47
|
show 1 more comment
1 Answer
1
active
oldest
votes
1
$begingroup$
The set of solutions of $Amathbf{x}=mathbf{b}$ is not a subespace of $mathbb{R}^4$ becasuse the null vector $mathbf{0}=begin{bmatrix}{0}\{0}\{0}\{0}end{bmatrix}$ does not satisfy $Amathbf{0}=mathbf{b}$.
answered Dec 28 '16 at 19:11
Fernando RevillaFernando Revilla
3,332520
$endgroup$
$begingroup$
Thank you. Is this the same as saying that it is because the last row does not equal the zero vector?
$endgroup$
– user102937
Dec 28 '16 at 19:14
$begingroup$
- Of the RREF of Ax=B, i mean. Sorry i do not understand how to edit my comments.
$endgroup$
– user102937
Dec 28 '16 at 19:15
$begingroup$
No, nothing to do. For example, $Amathbf{x}=mathbf{0}$ would be a subspace of $mathbb{R}^4$ with $A$ the same matrix.
$endgroup$
– Fernando Revilla
Dec 28 '16 at 19:17
$begingroup$
@user102937 No, it’s the same as saying that $mathbf bnemathbf 0$. The solution set to $Amathbf x=mathbf b$ can only be a vector space when $mathbf b=mathbf 0$. The rref in the problem is sort of a red herring.
$endgroup$
– amd
Dec 28 '16 at 19:22
$begingroup$
Thanks again Alone and amd. So just to be completely sure i get it: If and only if b=0, can said problem be true? And if b=0 is it then always within the subspace R^n?
$endgroup$
– user102937
Dec 28 '16 at 20:16
|
show 1 more comment
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1 Answer
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oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
The set of solutions of $Amathbf{x}=mathbf{b}$ is not a subespace of $mathbb{R}^4$ becasuse the null vector $mathbf{0}=begin{bmatrix}{0}\{0}\{0}\{0}end{bmatrix}$ does not satisfy $Amathbf{0}=mathbf{b}$.
answered Dec 28 '16 at 19:11
Fernando RevillaFernando Revilla
3,332520
$endgroup$
$begingroup$
Thank you. Is this the same as saying that it is because the last row does not equal the zero vector?
$endgroup$
– user102937
Dec 28 '16 at 19:14
$begingroup$
- Of the RREF of Ax=B, i mean. Sorry i do not understand how to edit my comments.
$endgroup$
– user102937
Dec 28 '16 at 19:15
$begingroup$
No, nothing to do. For example, $Amathbf{x}=mathbf{0}$ would be a subspace of $mathbb{R}^4$ with $A$ the same matrix.
$endgroup$
– Fernando Revilla
Dec 28 '16 at 19:17
$begingroup$
@user102937 No, it’s the same as saying that $mathbf bnemathbf 0$. The solution set to $Amathbf x=mathbf b$ can only be a vector space when $mathbf b=mathbf 0$. The rref in the problem is sort of a red herring.
$endgroup$
– amd
Dec 28 '16 at 19:22
$begingroup$
Thanks again Alone and amd. So just to be completely sure i get it: If and only if b=0, can said problem be true? And if b=0 is it then always within the subspace R^n?
$endgroup$
– user102937
Dec 28 '16 at 20:16
|
show 1 more comment
1
$begingroup$
The set of solutions of $Amathbf{x}=mathbf{b}$ is not a subespace of $mathbb{R}^4$ becasuse the null vector $mathbf{0}=begin{bmatrix}{0}\{0}\{0}\{0}end{bmatrix}$ does not satisfy $Amathbf{0}=mathbf{b}$.
answered Dec 28 '16 at 19:11
Fernando RevillaFernando Revilla
3,332520
$endgroup$
$begingroup$
Thank you. Is this the same as saying that it is because the last row does not equal the zero vector?
$endgroup$
– user102937
Dec 28 '16 at 19:14
$begingroup$
- Of the RREF of Ax=B, i mean. Sorry i do not understand how to edit my comments.
$endgroup$
– user102937
Dec 28 '16 at 19:15
$begingroup$
No, nothing to do. For example, $Amathbf{x}=mathbf{0}$ would be a subspace of $mathbb{R}^4$ with $A$ the same matrix.
$endgroup$
– Fernando Revilla
Dec 28 '16 at 19:17
$begingroup$
@user102937 No, it’s the same as saying that $mathbf bnemathbf 0$. The solution set to $Amathbf x=mathbf b$ can only be a vector space when $mathbf b=mathbf 0$. The rref in the problem is sort of a red herring.
$endgroup$
– amd
Dec 28 '16 at 19:22
$begingroup$
Thanks again Alone and amd. So just to be completely sure i get it: If and only if b=0, can said problem be true? And if b=0 is it then always within the subspace R^n?
$endgroup$
– user102937
Dec 28 '16 at 20:16
|
show 1 more comment
1
1
1
$begingroup$
The set of solutions of $Amathbf{x}=mathbf{b}$ is not a subespace of $mathbb{R}^4$ becasuse the null vector $mathbf{0}=begin{bmatrix}{0}\{0}\{0}\{0}end{bmatrix}$ does not satisfy $Amathbf{0}=mathbf{b}$.
answered Dec 28 '16 at 19:11
Fernando RevillaFernando Revilla
3,332520
$endgroup$
The set of solutions of $Amathbf{x}=mathbf{b}$ is not a subespace of $mathbb{R}^4$ becasuse the null vector $mathbf{0}=begin{bmatrix}{0}\{0}\{0}\{0}end{bmatrix}$ does not satisfy $Amathbf{0}=mathbf{b}$.
answered Dec 28 '16 at 19:11
Fernando RevillaFernando Revilla
3,332520
answered Dec 28 '16 at 19:11
Fernando RevillaFernando Revilla
3,332520
answered Dec 28 '16 at 19:11
Fernando RevillaFernando Revilla
3,332520
answered Dec 28 '16 at 19:11
Fernando RevillaFernando Revilla
3,332520
3,332520
$begingroup$
Thank you. Is this the same as saying that it is because the last row does not equal the zero vector?
$endgroup$
– user102937
Dec 28 '16 at 19:14
$begingroup$
- Of the RREF of Ax=B, i mean. Sorry i do not understand how to edit my comments.
$endgroup$
– user102937
Dec 28 '16 at 19:15
$begingroup$
No, nothing to do. For example, $Amathbf{x}=mathbf{0}$ would be a subspace of $mathbb{R}^4$ with $A$ the same matrix.
$endgroup$
– Fernando Revilla
Dec 28 '16 at 19:17
$begingroup$
@user102937 No, it’s the same as saying that $mathbf bnemathbf 0$. The solution set to $Amathbf x=mathbf b$ can only be a vector space when $mathbf b=mathbf 0$. The rref in the problem is sort of a red herring.
$endgroup$
– amd
Dec 28 '16 at 19:22
$begingroup$
Thanks again Alone and amd. So just to be completely sure i get it: If and only if b=0, can said problem be true? And if b=0 is it then always within the subspace R^n?
$endgroup$
– user102937
Dec 28 '16 at 20:16
|
show 1 more comment
$begingroup$
Thank you. Is this the same as saying that it is because the last row does not equal the zero vector?
$endgroup$
– user102937
Dec 28 '16 at 19:14
$begingroup$
- Of the RREF of Ax=B, i mean. Sorry i do not understand how to edit my comments.
$endgroup$
– user102937
Dec 28 '16 at 19:15
$begingroup$
No, nothing to do. For example, $Amathbf{x}=mathbf{0}$ would be a subspace of $mathbb{R}^4$ with $A$ the same matrix.
$endgroup$
– Fernando Revilla
Dec 28 '16 at 19:17
$begingroup$
@user102937 No, it’s the same as saying that $mathbf bnemathbf 0$. The solution set to $Amathbf x=mathbf b$ can only be a vector space when $mathbf b=mathbf 0$. The rref in the problem is sort of a red herring.
$endgroup$
– amd
Dec 28 '16 at 19:22
$begingroup$
Thanks again Alone and amd. So just to be completely sure i get it: If and only if b=0, can said problem be true? And if b=0 is it then always within the subspace R^n?
$endgroup$
– user102937
Dec 28 '16 at 20:16
$begingroup$
Thank you. Is this the same as saying that it is because the last row does not equal the zero vector?
$endgroup$
– user102937
Dec 28 '16 at 19:14
$begingroup$
Thank you. Is this the same as saying that it is because the last row does not equal the zero vector?
$endgroup$
– user102937
Dec 28 '16 at 19:14
$begingroup$
- Of the RREF of Ax=B, i mean. Sorry i do not understand how to edit my comments.
$endgroup$
– user102937
Dec 28 '16 at 19:15
$begingroup$
- Of the RREF of Ax=B, i mean. Sorry i do not understand how to edit my comments.
$endgroup$
– user102937
Dec 28 '16 at 19:15
$begingroup$
No, nothing to do. For example, $Amathbf{x}=mathbf{0}$ would be a subspace of $mathbb{R}^4$ with $A$ the same matrix.
$endgroup$
– Fernando Revilla
Dec 28 '16 at 19:17
$begingroup$
No, nothing to do. For example, $Amathbf{x}=mathbf{0}$ would be a subspace of $mathbb{R}^4$ with $A$ the same matrix.
$endgroup$
– Fernando Revilla
Dec 28 '16 at 19:17
$begingroup$
@user102937 No, it’s the same as saying that $mathbf bnemathbf 0$. The solution set to $Amathbf x=mathbf b$ can only be a vector space when $mathbf b=mathbf 0$. The rref in the problem is sort of a red herring.
$endgroup$
– amd
Dec 28 '16 at 19:22
$begingroup$
@user102937 No, it’s the same as saying that $mathbf bnemathbf 0$. The solution set to $Amathbf x=mathbf b$ can only be a vector space when $mathbf b=mathbf 0$. The rref in the problem is sort of a red herring.
$endgroup$
– amd
Dec 28 '16 at 19:22
$begingroup$
Thanks again Alone and amd. So just to be completely sure i get it: If and only if b=0, can said problem be true? And if b=0 is it then always within the subspace R^n?
$endgroup$
– user102937
Dec 28 '16 at 20:16
$begingroup$
Thanks again Alone and amd. So just to be completely sure i get it: If and only if b=0, can said problem be true? And if b=0 is it then always within the subspace R^n?
$endgroup$
– user102937
Dec 28 '16 at 20:16
|
show 1 more comment
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1
$begingroup$
What is $mathcal{R}$?
$endgroup$
– Dan Rust
Dec 28 '16 at 19:04
$begingroup$
What have you tried? What is your understanding of how to determine if a set is a subspace of the vector space $mathbb{R} $?
$endgroup$
– user23793
Dec 28 '16 at 19:04
$begingroup$
I tried to follow this video:
$endgroup$
– user102937
Dec 28 '16 at 19:09
$begingroup$
I tried to follow this video: youtube.com/watch?v=q97HmMdD8ZM And to me it seems that the subspace is equal to the amount of rows of the vectors within the span. Am i completely off track or?
$endgroup$
– user102937
Dec 28 '16 at 19:10
$begingroup$
I don't understand the question. What do you want to calculate?
$endgroup$
– Jack
Dec 28 '16 at 19:47