can you confirm my solution for the basis












0












$begingroup$


If S is the subspace spanned by $$
pmatrix{1 \ 2\1} pmatrix{-1 \ 1\-1} $$ then find a basis for the orthogonal complement of S ?



the solution is:



since the null space of any matrix is the orthogonal compliment of the row apace then we solve for the null space $Ax=0$



$$
pmatrix{1&2&1 \ -1&1&-1}X=0 $$



and tha basis will be $$
pmatrix{-1 \ 0\1} $$



is this the right answer ? and thank you






linear-algebra matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can check to see that you have the right answer by confirming that the vector in your answer is in fact orthogonal to both of the basis vectors in your problem. It is, and therefore your answer is indeed correct.
    $endgroup$
    – JMoravitz
    Dec 12 '18 at 17:28










  • $begingroup$
    @JMoravitz thank you very much for confirming and for that info
    $endgroup$
    – The Beard
    Dec 12 '18 at 17:32
















0












$begingroup$


If S is the subspace spanned by $$
pmatrix{1 \ 2\1} pmatrix{-1 \ 1\-1} $$ then find a basis for the orthogonal complement of S ?



the solution is:



since the null space of any matrix is the orthogonal compliment of the row apace then we solve for the null space $Ax=0$



$$
pmatrix{1&2&1 \ -1&1&-1}X=0 $$



and tha basis will be $$
pmatrix{-1 \ 0\1} $$



is this the right answer ? and thank you






linear-algebra matrices







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can check to see that you have the right answer by confirming that the vector in your answer is in fact orthogonal to both of the basis vectors in your problem. It is, and therefore your answer is indeed correct.
    $endgroup$
    – JMoravitz
    Dec 12 '18 at 17:28










  • $begingroup$
    @JMoravitz thank you very much for confirming and for that info
    $endgroup$
    – The Beard
    Dec 12 '18 at 17:32














0












0








0





$begingroup$


If S is the subspace spanned by $$
pmatrix{1 \ 2\1} pmatrix{-1 \ 1\-1} $$ then find a basis for the orthogonal complement of S ?



the solution is:



since the null space of any matrix is the orthogonal compliment of the row apace then we solve for the null space $Ax=0$



$$
pmatrix{1&2&1 \ -1&1&-1}X=0 $$



and tha basis will be $$
pmatrix{-1 \ 0\1} $$



is this the right answer ? and thank you






linear-algebra matrices







share|cite|improve this question











$endgroup$




If S is the subspace spanned by $$
pmatrix{1 \ 2\1} pmatrix{-1 \ 1\-1} $$ then find a basis for the orthogonal complement of S ?



the solution is:



since the null space of any matrix is the orthogonal compliment of the row apace then we solve for the null space $Ax=0$



$$
pmatrix{1&2&1 \ -1&1&-1}X=0 $$



and tha basis will be $$
pmatrix{-1 \ 0\1} $$



is this the right answer ? and thank you






linear-algebra matrices




linear-algebra matrices






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 17:43









Key Flex

7,98461233




7,98461233










asked Dec 12 '18 at 17:18









The BeardThe Beard

54




54












  • $begingroup$
    You can check to see that you have the right answer by confirming that the vector in your answer is in fact orthogonal to both of the basis vectors in your problem. It is, and therefore your answer is indeed correct.
    $endgroup$
    – JMoravitz
    Dec 12 '18 at 17:28










  • $begingroup$
    @JMoravitz thank you very much for confirming and for that info
    $endgroup$
    – The Beard
    Dec 12 '18 at 17:32


















  • $begingroup$
    You can check to see that you have the right answer by confirming that the vector in your answer is in fact orthogonal to both of the basis vectors in your problem. It is, and therefore your answer is indeed correct.
    $endgroup$
    – JMoravitz
    Dec 12 '18 at 17:28










  • $begingroup$
    @JMoravitz thank you very much for confirming and for that info
    $endgroup$
    – The Beard
    Dec 12 '18 at 17:32
















$begingroup$
You can check to see that you have the right answer by confirming that the vector in your answer is in fact orthogonal to both of the basis vectors in your problem. It is, and therefore your answer is indeed correct.
$endgroup$
– JMoravitz
Dec 12 '18 at 17:28




$begingroup$
You can check to see that you have the right answer by confirming that the vector in your answer is in fact orthogonal to both of the basis vectors in your problem. It is, and therefore your answer is indeed correct.
$endgroup$
– JMoravitz
Dec 12 '18 at 17:28












$begingroup$
@JMoravitz thank you very much for confirming and for that info
$endgroup$
– The Beard
Dec 12 '18 at 17:32




$begingroup$
@JMoravitz thank you very much for confirming and for that info
$endgroup$
– The Beard
Dec 12 '18 at 17:32










2 Answers
2






active

oldest

votes


















1












$begingroup$

Consider the matrix form as $$A=begin{bmatrix} 1 & 2 & 1 \ -1 & 1 & -1 end{bmatrix}$$
By construction, the row space of $A$ is equal to $S$. Therefore, since the nullspace of any matrix is the orthogonal complement of the row space, it must be the case that $S^{⊥}=$ nul$(A)$.



The reduced row echelon form of the above matrix is $$A=begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 0 end{bmatrix}$$ The matrix $A$ is now in reduced echelon form, so we can see that the homogeneous equation $Ax=0$ is equivalent to $$x_1=-x_3$$
$$x_2=0$$
So, the following is a basis for nul$(A) =S^⊥$
$$begin{bmatrix} -1 \ 0 \ 1 end{bmatrix}$$



So, your answer is correct.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much for confirming and for explaining that
    $endgroup$
    – The Beard
    Dec 12 '18 at 17:35



















0












$begingroup$

You could also take the cross product of the two vectors to get an orthogonal vector: $begin{vmatrix} i&j&k\1&2&1\-1&1&-1end{vmatrix}=begin{vmatrix}2&1\1&-1end{vmatrix}i-begin{vmatrix}1&1\-1&-1end{vmatrix}j+begin{vmatrix}1&2\-1&1end{vmatrix}k=-3i+3k$. Thus we get $begin{pmatrix}-3\0\3end{pmatrix}$.



You could also pull a rabbit out of a hat and guess it, since both have $x$ and $z$ coordinates the same.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks for explaining other method to solve the problem
    $endgroup$
    – The Beard
    Dec 13 '18 at 21:03











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Consider the matrix form as $$A=begin{bmatrix} 1 & 2 & 1 \ -1 & 1 & -1 end{bmatrix}$$
By construction, the row space of $A$ is equal to $S$. Therefore, since the nullspace of any matrix is the orthogonal complement of the row space, it must be the case that $S^{⊥}=$ nul$(A)$.



The reduced row echelon form of the above matrix is $$A=begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 0 end{bmatrix}$$ The matrix $A$ is now in reduced echelon form, so we can see that the homogeneous equation $Ax=0$ is equivalent to $$x_1=-x_3$$
$$x_2=0$$
So, the following is a basis for nul$(A) =S^⊥$
$$begin{bmatrix} -1 \ 0 \ 1 end{bmatrix}$$



So, your answer is correct.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much for confirming and for explaining that
    $endgroup$
    – The Beard
    Dec 12 '18 at 17:35
















1












$begingroup$

Consider the matrix form as $$A=begin{bmatrix} 1 & 2 & 1 \ -1 & 1 & -1 end{bmatrix}$$
By construction, the row space of $A$ is equal to $S$. Therefore, since the nullspace of any matrix is the orthogonal complement of the row space, it must be the case that $S^{⊥}=$ nul$(A)$.



The reduced row echelon form of the above matrix is $$A=begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 0 end{bmatrix}$$ The matrix $A$ is now in reduced echelon form, so we can see that the homogeneous equation $Ax=0$ is equivalent to $$x_1=-x_3$$
$$x_2=0$$
So, the following is a basis for nul$(A) =S^⊥$
$$begin{bmatrix} -1 \ 0 \ 1 end{bmatrix}$$



So, your answer is correct.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much for confirming and for explaining that
    $endgroup$
    – The Beard
    Dec 12 '18 at 17:35














1












1








1





$begingroup$

Consider the matrix form as $$A=begin{bmatrix} 1 & 2 & 1 \ -1 & 1 & -1 end{bmatrix}$$
By construction, the row space of $A$ is equal to $S$. Therefore, since the nullspace of any matrix is the orthogonal complement of the row space, it must be the case that $S^{⊥}=$ nul$(A)$.



The reduced row echelon form of the above matrix is $$A=begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 0 end{bmatrix}$$ The matrix $A$ is now in reduced echelon form, so we can see that the homogeneous equation $Ax=0$ is equivalent to $$x_1=-x_3$$
$$x_2=0$$
So, the following is a basis for nul$(A) =S^⊥$
$$begin{bmatrix} -1 \ 0 \ 1 end{bmatrix}$$



So, your answer is correct.






share|cite|improve this answer









$endgroup$



Consider the matrix form as $$A=begin{bmatrix} 1 & 2 & 1 \ -1 & 1 & -1 end{bmatrix}$$
By construction, the row space of $A$ is equal to $S$. Therefore, since the nullspace of any matrix is the orthogonal complement of the row space, it must be the case that $S^{⊥}=$ nul$(A)$.



The reduced row echelon form of the above matrix is $$A=begin{bmatrix} 1 & 0 & 1 \ 0 & 1 & 0 end{bmatrix}$$ The matrix $A$ is now in reduced echelon form, so we can see that the homogeneous equation $Ax=0$ is equivalent to $$x_1=-x_3$$
$$x_2=0$$
So, the following is a basis for nul$(A) =S^⊥$
$$begin{bmatrix} -1 \ 0 \ 1 end{bmatrix}$$



So, your answer is correct.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 17:32









Key FlexKey Flex

7,98461233




7,98461233












  • $begingroup$
    thank you very much for confirming and for explaining that
    $endgroup$
    – The Beard
    Dec 12 '18 at 17:35


















  • $begingroup$
    thank you very much for confirming and for explaining that
    $endgroup$
    – The Beard
    Dec 12 '18 at 17:35
















$begingroup$
thank you very much for confirming and for explaining that
$endgroup$
– The Beard
Dec 12 '18 at 17:35




$begingroup$
thank you very much for confirming and for explaining that
$endgroup$
– The Beard
Dec 12 '18 at 17:35











0












$begingroup$

You could also take the cross product of the two vectors to get an orthogonal vector: $begin{vmatrix} i&j&k\1&2&1\-1&1&-1end{vmatrix}=begin{vmatrix}2&1\1&-1end{vmatrix}i-begin{vmatrix}1&1\-1&-1end{vmatrix}j+begin{vmatrix}1&2\-1&1end{vmatrix}k=-3i+3k$. Thus we get $begin{pmatrix}-3\0\3end{pmatrix}$.



You could also pull a rabbit out of a hat and guess it, since both have $x$ and $z$ coordinates the same.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks for explaining other method to solve the problem
    $endgroup$
    – The Beard
    Dec 13 '18 at 21:03
















0












$begingroup$

You could also take the cross product of the two vectors to get an orthogonal vector: $begin{vmatrix} i&j&k\1&2&1\-1&1&-1end{vmatrix}=begin{vmatrix}2&1\1&-1end{vmatrix}i-begin{vmatrix}1&1\-1&-1end{vmatrix}j+begin{vmatrix}1&2\-1&1end{vmatrix}k=-3i+3k$. Thus we get $begin{pmatrix}-3\0\3end{pmatrix}$.



You could also pull a rabbit out of a hat and guess it, since both have $x$ and $z$ coordinates the same.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks for explaining other method to solve the problem
    $endgroup$
    – The Beard
    Dec 13 '18 at 21:03














0












0








0





$begingroup$

You could also take the cross product of the two vectors to get an orthogonal vector: $begin{vmatrix} i&j&k\1&2&1\-1&1&-1end{vmatrix}=begin{vmatrix}2&1\1&-1end{vmatrix}i-begin{vmatrix}1&1\-1&-1end{vmatrix}j+begin{vmatrix}1&2\-1&1end{vmatrix}k=-3i+3k$. Thus we get $begin{pmatrix}-3\0\3end{pmatrix}$.



You could also pull a rabbit out of a hat and guess it, since both have $x$ and $z$ coordinates the same.






share|cite|improve this answer









$endgroup$



You could also take the cross product of the two vectors to get an orthogonal vector: $begin{vmatrix} i&j&k\1&2&1\-1&1&-1end{vmatrix}=begin{vmatrix}2&1\1&-1end{vmatrix}i-begin{vmatrix}1&1\-1&-1end{vmatrix}j+begin{vmatrix}1&2\-1&1end{vmatrix}k=-3i+3k$. Thus we get $begin{pmatrix}-3\0\3end{pmatrix}$.



You could also pull a rabbit out of a hat and guess it, since both have $x$ and $z$ coordinates the same.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 18:11









Chris CusterChris Custer

12.8k3825




12.8k3825












  • $begingroup$
    thanks for explaining other method to solve the problem
    $endgroup$
    – The Beard
    Dec 13 '18 at 21:03


















  • $begingroup$
    thanks for explaining other method to solve the problem
    $endgroup$
    – The Beard
    Dec 13 '18 at 21:03
















$begingroup$
thanks for explaining other method to solve the problem
$endgroup$
– The Beard
Dec 13 '18 at 21:03




$begingroup$
thanks for explaining other method to solve the problem
$endgroup$
– The Beard
Dec 13 '18 at 21:03


















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