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:



enter image description here



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







share|cite|improve this question











$endgroup$








  • 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
















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:



enter image description here



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







share|cite|improve this question











$endgroup$








  • 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














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:



enter image description here



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







share|cite|improve this question











$endgroup$




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:



enter image description here



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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '16 at 18:59









Dan Rust

23k114984




23k114984










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














  • 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










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 $Amathbf{0}=mathbf{b}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












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1 Answer
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active

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 $Amathbf{0}=mathbf{b}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















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 $Amathbf{0}=mathbf{b}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














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 $Amathbf{0}=mathbf{b}$.






share|cite|improve this answer









$endgroup$



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 $Amathbf{0}=mathbf{b}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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