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












1














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







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  • 2




    Please format your question using MathJax. See here for a tutorial: math.meta.stackexchange.com/questions/5020/…
    – Dave
    2 days ago










  • I edited it but I use the symbol $ then curly braces removed.
    – Mathforjob
    2 days ago










  • You have to "escape" the braces, by typing {, since they are usually used for something else.
    – Chris Custer
    2 days ago












  • Thanks for the hint
    – Mathforjob
    yesterday
















1














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









New contributor




Mathforjob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2




    Please format your question using MathJax. See here for a tutorial: math.meta.stackexchange.com/questions/5020/…
    – Dave
    2 days ago










  • I edited it but I use the symbol $ then curly braces removed.
    – Mathforjob
    2 days ago










  • You have to "escape" the braces, by typing {, since they are usually used for something else.
    – Chris Custer
    2 days ago












  • Thanks for the hint
    – Mathforjob
    yesterday














1












1








1







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









New contributor




Mathforjob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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









New contributor




Mathforjob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Mathforjob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




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edited yesterday









Lee David Chung Lin

3,54531039




3,54531039






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asked 2 days ago









Mathforjob

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63




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New contributor





Mathforjob is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 2




    Please format your question using MathJax. See here for a tutorial: math.meta.stackexchange.com/questions/5020/…
    – Dave
    2 days ago










  • I edited it but I use the symbol $ then curly braces removed.
    – Mathforjob
    2 days ago










  • You have to "escape" the braces, by typing {, since they are usually used for something else.
    – Chris Custer
    2 days ago












  • Thanks for the hint
    – Mathforjob
    yesterday














  • 2




    Please format your question using MathJax. See here for a tutorial: math.meta.stackexchange.com/questions/5020/…
    – Dave
    2 days ago










  • I edited it but I use the symbol $ then curly braces removed.
    – Mathforjob
    2 days ago










  • You have to "escape" the braces, by typing {, since they are usually used for something else.
    – Chris Custer
    2 days ago












  • Thanks for the hint
    – Mathforjob
    yesterday








2




2




Please format your question using MathJax. See here for a tutorial: math.meta.stackexchange.com/questions/5020/…
– Dave
2 days ago




Please format your question using MathJax. See here for a tutorial: math.meta.stackexchange.com/questions/5020/…
– Dave
2 days ago












I edited it but I use the symbol $ then curly braces removed.
– Mathforjob
2 days ago




I edited it but I use the symbol $ then curly braces removed.
– Mathforjob
2 days ago












You have to "escape" the braces, by typing {, since they are usually used for something else.
– Chris Custer
2 days ago






You have to "escape" the braces, by typing {, since they are usually used for something else.
– Chris Custer
2 days ago














Thanks for the hint
– Mathforjob
yesterday




Thanks for the hint
– Mathforjob
yesterday










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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





















  • Thank you for the help
    – Mathforjob
    yesterday











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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





















  • Thank you for the help
    – Mathforjob
    yesterday
















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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





















  • Thank you for the help
    – Mathforjob
    yesterday














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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












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 yesterday









Chris Custer

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  • Thank you for the help
    – Mathforjob
    yesterday


















  • Thank you for the help
    – Mathforjob
    yesterday
















Thank you for the help
– Mathforjob
yesterday




Thank you for the help
– Mathforjob
yesterday










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