Span of a Vector Space in $mathbb{R}^3$












1












$begingroup$


Consider the subspaces $W_1$ and $W_2$ of $mathbb{R}^3$ given by
$W_1= {(x,y,z) in mathbb{R}^3:x+y+z=0 }$ and $W_2={(x,y,z) in mathbb{R}^3:x-y+z=0 }$.



If $W$ is a subspace of $mathbb{R}^3$ such that




  • $W cap W_2= mathrm{span}bigl{(0,1,1)bigr}$


  • $W cap W_1$ is orthogonal to $W cap W_2$ with respect to the usual inner product of $mathbb{R}^3$



then which of these are true?




  1. $W = mathrm{span} bigl{ (0,1,-1),(0,1,1) bigr}$


  2. $W = mathrm{span} bigl{ (1,0,-1),(0,1,-1) bigr}$


  3. $W = mathrm{span} bigl{ (1,0,-1),(0,1,1) bigr}$


  4. $W = mathrm{span} bigl{ (1,0,-1),(1,0,1) bigr}$



My Attempt:
$x+y+z=0 implies x+y=-z$ so that free variables are two so $mathrm{dim}(W_1)=2$ and similarly $x-y+z=0 implies x+z=y$ so that $mathrm{dim}(W_2)=2$.



Also $W cap W_2 = mathrm{span}bigl{(0,1,1)
bigr}$ implies $(0,1,1)$ is one element of $W$ so options 2,4 discarded.



How to approach this type of problems in general?






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Please format your question using MathJax. See here for a tutorial: math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Dave
    Dec 24 '18 at 23:35










  • $begingroup$
    I edited it but I use the symbol $ then curly braces removed.
    $endgroup$
    – Mathforjob
    Dec 25 '18 at 0:40










  • $begingroup$
    You have to "escape" the braces, by typing {, since they are usually used for something else.
    $endgroup$
    – Chris Custer
    Dec 25 '18 at 1:09












  • $begingroup$
    Thanks for the hint
    $endgroup$
    – Mathforjob
    Dec 25 '18 at 5:57
















1












$begingroup$


Consider the subspaces $W_1$ and $W_2$ of $mathbb{R}^3$ given by
$W_1= {(x,y,z) in mathbb{R}^3:x+y+z=0 }$ and $W_2={(x,y,z) in mathbb{R}^3:x-y+z=0 }$.



If $W$ is a subspace of $mathbb{R}^3$ such that




  • $W cap W_2= mathrm{span}bigl{(0,1,1)bigr}$


  • $W cap W_1$ is orthogonal to $W cap W_2$ with respect to the usual inner product of $mathbb{R}^3$



then which of these are true?




  1. $W = mathrm{span} bigl{ (0,1,-1),(0,1,1) bigr}$


  2. $W = mathrm{span} bigl{ (1,0,-1),(0,1,-1) bigr}$


  3. $W = mathrm{span} bigl{ (1,0,-1),(0,1,1) bigr}$


  4. $W = mathrm{span} bigl{ (1,0,-1),(1,0,1) bigr}$



My Attempt:
$x+y+z=0 implies x+y=-z$ so that free variables are two so $mathrm{dim}(W_1)=2$ and similarly $x-y+z=0 implies x+z=y$ so that $mathrm{dim}(W_2)=2$.



Also $W cap W_2 = mathrm{span}bigl{(0,1,1)
bigr}$ implies $(0,1,1)$ is one element of $W$ so options 2,4 discarded.



How to approach this type of problems in general?






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Please format your question using MathJax. See here for a tutorial: math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Dave
    Dec 24 '18 at 23:35










  • $begingroup$
    I edited it but I use the symbol $ then curly braces removed.
    $endgroup$
    – Mathforjob
    Dec 25 '18 at 0:40










  • $begingroup$
    You have to "escape" the braces, by typing {, since they are usually used for something else.
    $endgroup$
    – Chris Custer
    Dec 25 '18 at 1:09












  • $begingroup$
    Thanks for the hint
    $endgroup$
    – Mathforjob
    Dec 25 '18 at 5:57














1












1








1





$begingroup$


Consider the subspaces $W_1$ and $W_2$ of $mathbb{R}^3$ given by
$W_1= {(x,y,z) in mathbb{R}^3:x+y+z=0 }$ and $W_2={(x,y,z) in mathbb{R}^3:x-y+z=0 }$.



If $W$ is a subspace of $mathbb{R}^3$ such that




  • $W cap W_2= mathrm{span}bigl{(0,1,1)bigr}$


  • $W cap W_1$ is orthogonal to $W cap W_2$ with respect to the usual inner product of $mathbb{R}^3$



then which of these are true?




  1. $W = mathrm{span} bigl{ (0,1,-1),(0,1,1) bigr}$


  2. $W = mathrm{span} bigl{ (1,0,-1),(0,1,-1) bigr}$


  3. $W = mathrm{span} bigl{ (1,0,-1),(0,1,1) bigr}$


  4. $W = mathrm{span} bigl{ (1,0,-1),(1,0,1) bigr}$



My Attempt:
$x+y+z=0 implies x+y=-z$ so that free variables are two so $mathrm{dim}(W_1)=2$ and similarly $x-y+z=0 implies x+z=y$ so that $mathrm{dim}(W_2)=2$.



Also $W cap W_2 = mathrm{span}bigl{(0,1,1)
bigr}$ implies $(0,1,1)$ is one element of $W$ so options 2,4 discarded.



How to approach this type of problems in general?






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$




Consider the subspaces $W_1$ and $W_2$ of $mathbb{R}^3$ given by
$W_1= {(x,y,z) in mathbb{R}^3:x+y+z=0 }$ and $W_2={(x,y,z) in mathbb{R}^3:x-y+z=0 }$.



If $W$ is a subspace of $mathbb{R}^3$ such that




  • $W cap W_2= mathrm{span}bigl{(0,1,1)bigr}$


  • $W cap W_1$ is orthogonal to $W cap W_2$ with respect to the usual inner product of $mathbb{R}^3$



then which of these are true?




  1. $W = mathrm{span} bigl{ (0,1,-1),(0,1,1) bigr}$


  2. $W = mathrm{span} bigl{ (1,0,-1),(0,1,-1) bigr}$


  3. $W = mathrm{span} bigl{ (1,0,-1),(0,1,1) bigr}$


  4. $W = mathrm{span} bigl{ (1,0,-1),(1,0,1) bigr}$



My Attempt:
$x+y+z=0 implies x+y=-z$ so that free variables are two so $mathrm{dim}(W_1)=2$ and similarly $x-y+z=0 implies x+z=y$ so that $mathrm{dim}(W_2)=2$.



Also $W cap W_2 = mathrm{span}bigl{(0,1,1)
bigr}$ implies $(0,1,1)$ is one element of $W$ so options 2,4 discarded.



How to approach this type of problems in general?






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








edited Dec 25 '18 at 11:04









Lee David Chung Lin

4,47841242




4,47841242










asked Dec 24 '18 at 23:23









MathforjobMathforjob

164




164








  • 2




    $begingroup$
    Please format your question using MathJax. See here for a tutorial: math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Dave
    Dec 24 '18 at 23:35










  • $begingroup$
    I edited it but I use the symbol $ then curly braces removed.
    $endgroup$
    – Mathforjob
    Dec 25 '18 at 0:40










  • $begingroup$
    You have to "escape" the braces, by typing {, since they are usually used for something else.
    $endgroup$
    – Chris Custer
    Dec 25 '18 at 1:09












  • $begingroup$
    Thanks for the hint
    $endgroup$
    – Mathforjob
    Dec 25 '18 at 5:57














  • 2




    $begingroup$
    Please format your question using MathJax. See here for a tutorial: math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – Dave
    Dec 24 '18 at 23:35










  • $begingroup$
    I edited it but I use the symbol $ then curly braces removed.
    $endgroup$
    – Mathforjob
    Dec 25 '18 at 0:40










  • $begingroup$
    You have to "escape" the braces, by typing {, since they are usually used for something else.
    $endgroup$
    – Chris Custer
    Dec 25 '18 at 1:09












  • $begingroup$
    Thanks for the hint
    $endgroup$
    – Mathforjob
    Dec 25 '18 at 5:57








2




2




$begingroup$
Please format your question using MathJax. See here for a tutorial: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Dave
Dec 24 '18 at 23:35




$begingroup$
Please format your question using MathJax. See here for a tutorial: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Dave
Dec 24 '18 at 23:35












$begingroup$
I edited it but I use the symbol $ then curly braces removed.
$endgroup$
– Mathforjob
Dec 25 '18 at 0:40




$begingroup$
I edited it but I use the symbol $ then curly braces removed.
$endgroup$
– Mathforjob
Dec 25 '18 at 0:40












$begingroup$
You have to "escape" the braces, by typing {, since they are usually used for something else.
$endgroup$
– Chris Custer
Dec 25 '18 at 1:09






$begingroup$
You have to "escape" the braces, by typing {, since they are usually used for something else.
$endgroup$
– Chris Custer
Dec 25 '18 at 1:09














$begingroup$
Thanks for the hint
$endgroup$
– Mathforjob
Dec 25 '18 at 5:57




$begingroup$
Thanks for the hint
$endgroup$
– Mathforjob
Dec 25 '18 at 5:57










1 Answer
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$begingroup$

You reasoned correctly and discarded $2$ and $4$. It must be $1$, since $(1,0,-1)$ isn't orthogonal to $(0,1,1)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the help
    $endgroup$
    – Mathforjob
    Dec 25 '18 at 6:52












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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









0












$begingroup$

You reasoned correctly and discarded $2$ and $4$. It must be $1$, since $(1,0,-1)$ isn't orthogonal to $(0,1,1)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the help
    $endgroup$
    – Mathforjob
    Dec 25 '18 at 6:52
















0












$begingroup$

You reasoned correctly and discarded $2$ and $4$. It must be $1$, since $(1,0,-1)$ isn't orthogonal to $(0,1,1)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for the help
    $endgroup$
    – Mathforjob
    Dec 25 '18 at 6:52














0












0








0





$begingroup$

You reasoned correctly and discarded $2$ and $4$. It must be $1$, since $(1,0,-1)$ isn't orthogonal to $(0,1,1)$.






share|cite|improve this answer









$endgroup$



You reasoned correctly and discarded $2$ and $4$. It must be $1$, since $(1,0,-1)$ isn't orthogonal to $(0,1,1)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 25 '18 at 1:27









Chris CusterChris Custer

14.3k3827




14.3k3827












  • $begingroup$
    Thank you for the help
    $endgroup$
    – Mathforjob
    Dec 25 '18 at 6:52


















  • $begingroup$
    Thank you for the help
    $endgroup$
    – Mathforjob
    Dec 25 '18 at 6:52
















$begingroup$
Thank you for the help
$endgroup$
– Mathforjob
Dec 25 '18 at 6:52




$begingroup$
Thank you for the help
$endgroup$
– Mathforjob
Dec 25 '18 at 6:52


















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