How would you write something like x-y+z=1 as a vector?












1












$begingroup$


Specifically, the question is asking me to prove whether or not the collection of vectors [x,y,z] in R3 forms a subspace of R3 or not. The numbers for one question of this were z=2x and y = 3x. This is easy enough to understand because I can write a vector all in terms of x and see if the properties of a subspace hold for that generalized vector.



But how might I write x-y+z=1 as a vector? The best I can do is write one variable as two of the others, but that doesn't get me anywhere. Is there something I'm missing?



Sorry if this is a stupid question, any help is appreciated.






linear-algebra vector-spaces







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  • $begingroup$
    What are the axoims that a set must satisfy in order to be a subspace?
    $endgroup$
    – Yanko
    Dec 7 '18 at 16:24










  • $begingroup$
    I like to first translate to matrix $[1~-1~1]^t[x~y~z] = 1$, then a can set $a=[1~-1~1]$ and $v = [x~y~z]$, so in a pure vector sense $acdot v = 1$
    $endgroup$
    – Eduardo Elael
    Dec 7 '18 at 16:25










  • $begingroup$
    As @Yanko says, the question asked is "Do the vectors $(x, y, z)$ satisfying $x - y + z = 1$ form a vector space $V$"? To do this, you should look at things like, suppose, $(x_1, y_1, z_1) in V$ and $(x_2, y_2, z_2) in V$, is $(x_1 + x_2, y_1 + y_2, z_1 + z_2) in V$, using the definition of $V$. Similarly the other axioms of vector spaces should also be checked.
    $endgroup$
    – Mriganka Basu Roy Chowdhury
    Dec 7 '18 at 16:28










  • $begingroup$
    It can be easily seen that the plane $x-y+z=1$ is not a subspace of $Bbb R^3$ because it doesn't contain the zero vector, $vec 0$
    $endgroup$
    – Shubham Johri
    Dec 7 '18 at 21:16


















1












$begingroup$


Specifically, the question is asking me to prove whether or not the collection of vectors [x,y,z] in R3 forms a subspace of R3 or not. The numbers for one question of this were z=2x and y = 3x. This is easy enough to understand because I can write a vector all in terms of x and see if the properties of a subspace hold for that generalized vector.



But how might I write x-y+z=1 as a vector? The best I can do is write one variable as two of the others, but that doesn't get me anywhere. Is there something I'm missing?



Sorry if this is a stupid question, any help is appreciated.






linear-algebra vector-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    What are the axoims that a set must satisfy in order to be a subspace?
    $endgroup$
    – Yanko
    Dec 7 '18 at 16:24










  • $begingroup$
    I like to first translate to matrix $[1~-1~1]^t[x~y~z] = 1$, then a can set $a=[1~-1~1]$ and $v = [x~y~z]$, so in a pure vector sense $acdot v = 1$
    $endgroup$
    – Eduardo Elael
    Dec 7 '18 at 16:25










  • $begingroup$
    As @Yanko says, the question asked is "Do the vectors $(x, y, z)$ satisfying $x - y + z = 1$ form a vector space $V$"? To do this, you should look at things like, suppose, $(x_1, y_1, z_1) in V$ and $(x_2, y_2, z_2) in V$, is $(x_1 + x_2, y_1 + y_2, z_1 + z_2) in V$, using the definition of $V$. Similarly the other axioms of vector spaces should also be checked.
    $endgroup$
    – Mriganka Basu Roy Chowdhury
    Dec 7 '18 at 16:28










  • $begingroup$
    It can be easily seen that the plane $x-y+z=1$ is not a subspace of $Bbb R^3$ because it doesn't contain the zero vector, $vec 0$
    $endgroup$
    – Shubham Johri
    Dec 7 '18 at 21:16
















1












1








1





$begingroup$


Specifically, the question is asking me to prove whether or not the collection of vectors [x,y,z] in R3 forms a subspace of R3 or not. The numbers for one question of this were z=2x and y = 3x. This is easy enough to understand because I can write a vector all in terms of x and see if the properties of a subspace hold for that generalized vector.



But how might I write x-y+z=1 as a vector? The best I can do is write one variable as two of the others, but that doesn't get me anywhere. Is there something I'm missing?



Sorry if this is a stupid question, any help is appreciated.






linear-algebra vector-spaces







share|cite|improve this question









$endgroup$




Specifically, the question is asking me to prove whether or not the collection of vectors [x,y,z] in R3 forms a subspace of R3 or not. The numbers for one question of this were z=2x and y = 3x. This is easy enough to understand because I can write a vector all in terms of x and see if the properties of a subspace hold for that generalized vector.



But how might I write x-y+z=1 as a vector? The best I can do is write one variable as two of the others, but that doesn't get me anywhere. Is there something I'm missing?



Sorry if this is a stupid question, any help is appreciated.






linear-algebra vector-spaces




linear-algebra vector-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 7 '18 at 16:20









James RonaldJames Ronald

1057




1057












  • $begingroup$
    What are the axoims that a set must satisfy in order to be a subspace?
    $endgroup$
    – Yanko
    Dec 7 '18 at 16:24










  • $begingroup$
    I like to first translate to matrix $[1~-1~1]^t[x~y~z] = 1$, then a can set $a=[1~-1~1]$ and $v = [x~y~z]$, so in a pure vector sense $acdot v = 1$
    $endgroup$
    – Eduardo Elael
    Dec 7 '18 at 16:25










  • $begingroup$
    As @Yanko says, the question asked is "Do the vectors $(x, y, z)$ satisfying $x - y + z = 1$ form a vector space $V$"? To do this, you should look at things like, suppose, $(x_1, y_1, z_1) in V$ and $(x_2, y_2, z_2) in V$, is $(x_1 + x_2, y_1 + y_2, z_1 + z_2) in V$, using the definition of $V$. Similarly the other axioms of vector spaces should also be checked.
    $endgroup$
    – Mriganka Basu Roy Chowdhury
    Dec 7 '18 at 16:28










  • $begingroup$
    It can be easily seen that the plane $x-y+z=1$ is not a subspace of $Bbb R^3$ because it doesn't contain the zero vector, $vec 0$
    $endgroup$
    – Shubham Johri
    Dec 7 '18 at 21:16




















  • $begingroup$
    What are the axoims that a set must satisfy in order to be a subspace?
    $endgroup$
    – Yanko
    Dec 7 '18 at 16:24










  • $begingroup$
    I like to first translate to matrix $[1~-1~1]^t[x~y~z] = 1$, then a can set $a=[1~-1~1]$ and $v = [x~y~z]$, so in a pure vector sense $acdot v = 1$
    $endgroup$
    – Eduardo Elael
    Dec 7 '18 at 16:25










  • $begingroup$
    As @Yanko says, the question asked is "Do the vectors $(x, y, z)$ satisfying $x - y + z = 1$ form a vector space $V$"? To do this, you should look at things like, suppose, $(x_1, y_1, z_1) in V$ and $(x_2, y_2, z_2) in V$, is $(x_1 + x_2, y_1 + y_2, z_1 + z_2) in V$, using the definition of $V$. Similarly the other axioms of vector spaces should also be checked.
    $endgroup$
    – Mriganka Basu Roy Chowdhury
    Dec 7 '18 at 16:28










  • $begingroup$
    It can be easily seen that the plane $x-y+z=1$ is not a subspace of $Bbb R^3$ because it doesn't contain the zero vector, $vec 0$
    $endgroup$
    – Shubham Johri
    Dec 7 '18 at 21:16


















$begingroup$
What are the axoims that a set must satisfy in order to be a subspace?
$endgroup$
– Yanko
Dec 7 '18 at 16:24




$begingroup$
What are the axoims that a set must satisfy in order to be a subspace?
$endgroup$
– Yanko
Dec 7 '18 at 16:24












$begingroup$
I like to first translate to matrix $[1~-1~1]^t[x~y~z] = 1$, then a can set $a=[1~-1~1]$ and $v = [x~y~z]$, so in a pure vector sense $acdot v = 1$
$endgroup$
– Eduardo Elael
Dec 7 '18 at 16:25




$begingroup$
I like to first translate to matrix $[1~-1~1]^t[x~y~z] = 1$, then a can set $a=[1~-1~1]$ and $v = [x~y~z]$, so in a pure vector sense $acdot v = 1$
$endgroup$
– Eduardo Elael
Dec 7 '18 at 16:25












$begingroup$
As @Yanko says, the question asked is "Do the vectors $(x, y, z)$ satisfying $x - y + z = 1$ form a vector space $V$"? To do this, you should look at things like, suppose, $(x_1, y_1, z_1) in V$ and $(x_2, y_2, z_2) in V$, is $(x_1 + x_2, y_1 + y_2, z_1 + z_2) in V$, using the definition of $V$. Similarly the other axioms of vector spaces should also be checked.
$endgroup$
– Mriganka Basu Roy Chowdhury
Dec 7 '18 at 16:28




$begingroup$
As @Yanko says, the question asked is "Do the vectors $(x, y, z)$ satisfying $x - y + z = 1$ form a vector space $V$"? To do this, you should look at things like, suppose, $(x_1, y_1, z_1) in V$ and $(x_2, y_2, z_2) in V$, is $(x_1 + x_2, y_1 + y_2, z_1 + z_2) in V$, using the definition of $V$. Similarly the other axioms of vector spaces should also be checked.
$endgroup$
– Mriganka Basu Roy Chowdhury
Dec 7 '18 at 16:28












$begingroup$
It can be easily seen that the plane $x-y+z=1$ is not a subspace of $Bbb R^3$ because it doesn't contain the zero vector, $vec 0$
$endgroup$
– Shubham Johri
Dec 7 '18 at 21:16






$begingroup$
It can be easily seen that the plane $x-y+z=1$ is not a subspace of $Bbb R^3$ because it doesn't contain the zero vector, $vec 0$
$endgroup$
– Shubham Johri
Dec 7 '18 at 21:16












1 Answer
1






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oldest

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$begingroup$

Writing the set of all $x,y,z$ such that $x+y+z=1$ as a vector makes no sense mathematically. But I believe you mean the following:



$${[x,y,z]inmathbb{R}^3: x+y+z=1} = { [x,y,1-x-y] : x,y,zinmathbb{R}}$$



Now we can write $[x,y,1-x-y]$ as $[x,0,-x]+[0,y,-y]+[0,0,1]$. In other words you have that



$${[x,y,z]inmathbb{R}^3: x+y+z=1} = {[0,0,1]+xcdot [1,0,-1] + ycdot [0,1,-1] : x,yinmathbb{R}}$$



This is a simple form of the given set which I believe is what you meant by "writing $x+y+z=1$ as a vector". From this expression it might be easier to deduce that the set is not a subspace of $mathbb{R}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much, this helps a lot!
    $endgroup$
    – James Ronald
    Dec 7 '18 at 17:24











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









3












$begingroup$

Writing the set of all $x,y,z$ such that $x+y+z=1$ as a vector makes no sense mathematically. But I believe you mean the following:



$${[x,y,z]inmathbb{R}^3: x+y+z=1} = { [x,y,1-x-y] : x,y,zinmathbb{R}}$$



Now we can write $[x,y,1-x-y]$ as $[x,0,-x]+[0,y,-y]+[0,0,1]$. In other words you have that



$${[x,y,z]inmathbb{R}^3: x+y+z=1} = {[0,0,1]+xcdot [1,0,-1] + ycdot [0,1,-1] : x,yinmathbb{R}}$$



This is a simple form of the given set which I believe is what you meant by "writing $x+y+z=1$ as a vector". From this expression it might be easier to deduce that the set is not a subspace of $mathbb{R}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much, this helps a lot!
    $endgroup$
    – James Ronald
    Dec 7 '18 at 17:24
















3












$begingroup$

Writing the set of all $x,y,z$ such that $x+y+z=1$ as a vector makes no sense mathematically. But I believe you mean the following:



$${[x,y,z]inmathbb{R}^3: x+y+z=1} = { [x,y,1-x-y] : x,y,zinmathbb{R}}$$



Now we can write $[x,y,1-x-y]$ as $[x,0,-x]+[0,y,-y]+[0,0,1]$. In other words you have that



$${[x,y,z]inmathbb{R}^3: x+y+z=1} = {[0,0,1]+xcdot [1,0,-1] + ycdot [0,1,-1] : x,yinmathbb{R}}$$



This is a simple form of the given set which I believe is what you meant by "writing $x+y+z=1$ as a vector". From this expression it might be easier to deduce that the set is not a subspace of $mathbb{R}$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much, this helps a lot!
    $endgroup$
    – James Ronald
    Dec 7 '18 at 17:24














3












3








3





$begingroup$

Writing the set of all $x,y,z$ such that $x+y+z=1$ as a vector makes no sense mathematically. But I believe you mean the following:



$${[x,y,z]inmathbb{R}^3: x+y+z=1} = { [x,y,1-x-y] : x,y,zinmathbb{R}}$$



Now we can write $[x,y,1-x-y]$ as $[x,0,-x]+[0,y,-y]+[0,0,1]$. In other words you have that



$${[x,y,z]inmathbb{R}^3: x+y+z=1} = {[0,0,1]+xcdot [1,0,-1] + ycdot [0,1,-1] : x,yinmathbb{R}}$$



This is a simple form of the given set which I believe is what you meant by "writing $x+y+z=1$ as a vector". From this expression it might be easier to deduce that the set is not a subspace of $mathbb{R}$.






share|cite|improve this answer









$endgroup$



Writing the set of all $x,y,z$ such that $x+y+z=1$ as a vector makes no sense mathematically. But I believe you mean the following:



$${[x,y,z]inmathbb{R}^3: x+y+z=1} = { [x,y,1-x-y] : x,y,zinmathbb{R}}$$



Now we can write $[x,y,1-x-y]$ as $[x,0,-x]+[0,y,-y]+[0,0,1]$. In other words you have that



$${[x,y,z]inmathbb{R}^3: x+y+z=1} = {[0,0,1]+xcdot [1,0,-1] + ycdot [0,1,-1] : x,yinmathbb{R}}$$



This is a simple form of the given set which I believe is what you meant by "writing $x+y+z=1$ as a vector". From this expression it might be easier to deduce that the set is not a subspace of $mathbb{R}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 16:27









YankoYanko

6,5571529




6,5571529












  • $begingroup$
    Thank you very much, this helps a lot!
    $endgroup$
    – James Ronald
    Dec 7 '18 at 17:24


















  • $begingroup$
    Thank you very much, this helps a lot!
    $endgroup$
    – James Ronald
    Dec 7 '18 at 17:24
















$begingroup$
Thank you very much, this helps a lot!
$endgroup$
– James Ronald
Dec 7 '18 at 17:24




$begingroup$
Thank you very much, this helps a lot!
$endgroup$
– James Ronald
Dec 7 '18 at 17:24


















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