Determine the null space of the following matrix












1












$begingroup$


Determine the null space of the following matrix:



$$begin{bmatrix}
1 & 2 & -3& -1 \
-2& -4 &6 &3
end{bmatrix}$$



For this question, I reduced the row echelon form into $$begin{bmatrix}
1 & 2 & -3& -1 \
0& 0 &0 &1
end{bmatrix},$$ but then I have no idea how to determine the null space, because there's no relationship between $x_1, x_2, x_3, x_4$.






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    you are encouraged to include your attempt.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 6:44










  • $begingroup$
    For this question, I reduced the row echelon form into ( 1 2 -3 -1 0 0 0 1 ), but then I have no idea how to determine the null space, because there's no relationship between x1, x2, x3, x4
    $endgroup$
    – Shadow Z
    Jan 6 at 6:45










  • $begingroup$
    Include your attempt in the original post directly.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 6:46










  • $begingroup$
    You require $$begin{bmatrix} 1 & 2 & -3& -1 \ 0& 0 &0 &1 end{bmatrix}begin{bmatrix} x_1\x_2\x_3\x_4 end{bmatrix}=begin{bmatrix} 0\ 0\0\0 end{bmatrix}$$ Now just find the relations between $x_1,x_2,x_3,x_4$ as in the answer below.
    $endgroup$
    – Shubham Johri
    Jan 6 at 6:57










  • $begingroup$
    See math.stackexchange.com/a/1521354/265466 for how to read a basis for the null space directly from the RREF.
    $endgroup$
    – amd
    Jan 6 at 8:13
















1












$begingroup$


Determine the null space of the following matrix:



$$begin{bmatrix}
1 & 2 & -3& -1 \
-2& -4 &6 &3
end{bmatrix}$$



For this question, I reduced the row echelon form into $$begin{bmatrix}
1 & 2 & -3& -1 \
0& 0 &0 &1
end{bmatrix},$$ but then I have no idea how to determine the null space, because there's no relationship between $x_1, x_2, x_3, x_4$.






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    you are encouraged to include your attempt.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 6:44










  • $begingroup$
    For this question, I reduced the row echelon form into ( 1 2 -3 -1 0 0 0 1 ), but then I have no idea how to determine the null space, because there's no relationship between x1, x2, x3, x4
    $endgroup$
    – Shadow Z
    Jan 6 at 6:45










  • $begingroup$
    Include your attempt in the original post directly.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 6:46










  • $begingroup$
    You require $$begin{bmatrix} 1 & 2 & -3& -1 \ 0& 0 &0 &1 end{bmatrix}begin{bmatrix} x_1\x_2\x_3\x_4 end{bmatrix}=begin{bmatrix} 0\ 0\0\0 end{bmatrix}$$ Now just find the relations between $x_1,x_2,x_3,x_4$ as in the answer below.
    $endgroup$
    – Shubham Johri
    Jan 6 at 6:57










  • $begingroup$
    See math.stackexchange.com/a/1521354/265466 for how to read a basis for the null space directly from the RREF.
    $endgroup$
    – amd
    Jan 6 at 8:13














1












1








1





$begingroup$


Determine the null space of the following matrix:



$$begin{bmatrix}
1 & 2 & -3& -1 \
-2& -4 &6 &3
end{bmatrix}$$



For this question, I reduced the row echelon form into $$begin{bmatrix}
1 & 2 & -3& -1 \
0& 0 &0 &1
end{bmatrix},$$ but then I have no idea how to determine the null space, because there's no relationship between $x_1, x_2, x_3, x_4$.






linear-algebra vector-spaces







share|cite|improve this question











$endgroup$




Determine the null space of the following matrix:



$$begin{bmatrix}
1 & 2 & -3& -1 \
-2& -4 &6 &3
end{bmatrix}$$



For this question, I reduced the row echelon form into $$begin{bmatrix}
1 & 2 & -3& -1 \
0& 0 &0 &1
end{bmatrix},$$ but then I have no idea how to determine the null space, because there's no relationship between $x_1, x_2, x_3, x_4$.






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 Jan 6 at 6:50









Siong Thye Goh

103k1468119




103k1468119










asked Jan 6 at 6:41









Shadow ZShadow Z

396




396












  • $begingroup$
    you are encouraged to include your attempt.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 6:44










  • $begingroup$
    For this question, I reduced the row echelon form into ( 1 2 -3 -1 0 0 0 1 ), but then I have no idea how to determine the null space, because there's no relationship between x1, x2, x3, x4
    $endgroup$
    – Shadow Z
    Jan 6 at 6:45










  • $begingroup$
    Include your attempt in the original post directly.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 6:46










  • $begingroup$
    You require $$begin{bmatrix} 1 & 2 & -3& -1 \ 0& 0 &0 &1 end{bmatrix}begin{bmatrix} x_1\x_2\x_3\x_4 end{bmatrix}=begin{bmatrix} 0\ 0\0\0 end{bmatrix}$$ Now just find the relations between $x_1,x_2,x_3,x_4$ as in the answer below.
    $endgroup$
    – Shubham Johri
    Jan 6 at 6:57










  • $begingroup$
    See math.stackexchange.com/a/1521354/265466 for how to read a basis for the null space directly from the RREF.
    $endgroup$
    – amd
    Jan 6 at 8:13


















  • $begingroup$
    you are encouraged to include your attempt.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 6:44










  • $begingroup$
    For this question, I reduced the row echelon form into ( 1 2 -3 -1 0 0 0 1 ), but then I have no idea how to determine the null space, because there's no relationship between x1, x2, x3, x4
    $endgroup$
    – Shadow Z
    Jan 6 at 6:45










  • $begingroup$
    Include your attempt in the original post directly.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 6:46










  • $begingroup$
    You require $$begin{bmatrix} 1 & 2 & -3& -1 \ 0& 0 &0 &1 end{bmatrix}begin{bmatrix} x_1\x_2\x_3\x_4 end{bmatrix}=begin{bmatrix} 0\ 0\0\0 end{bmatrix}$$ Now just find the relations between $x_1,x_2,x_3,x_4$ as in the answer below.
    $endgroup$
    – Shubham Johri
    Jan 6 at 6:57










  • $begingroup$
    See math.stackexchange.com/a/1521354/265466 for how to read a basis for the null space directly from the RREF.
    $endgroup$
    – amd
    Jan 6 at 8:13
















$begingroup$
you are encouraged to include your attempt.
$endgroup$
– Siong Thye Goh
Jan 6 at 6:44




$begingroup$
you are encouraged to include your attempt.
$endgroup$
– Siong Thye Goh
Jan 6 at 6:44












$begingroup$
For this question, I reduced the row echelon form into ( 1 2 -3 -1 0 0 0 1 ), but then I have no idea how to determine the null space, because there's no relationship between x1, x2, x3, x4
$endgroup$
– Shadow Z
Jan 6 at 6:45




$begingroup$
For this question, I reduced the row echelon form into ( 1 2 -3 -1 0 0 0 1 ), but then I have no idea how to determine the null space, because there's no relationship between x1, x2, x3, x4
$endgroup$
– Shadow Z
Jan 6 at 6:45












$begingroup$
Include your attempt in the original post directly.
$endgroup$
– Siong Thye Goh
Jan 6 at 6:46




$begingroup$
Include your attempt in the original post directly.
$endgroup$
– Siong Thye Goh
Jan 6 at 6:46












$begingroup$
You require $$begin{bmatrix} 1 & 2 & -3& -1 \ 0& 0 &0 &1 end{bmatrix}begin{bmatrix} x_1\x_2\x_3\x_4 end{bmatrix}=begin{bmatrix} 0\ 0\0\0 end{bmatrix}$$ Now just find the relations between $x_1,x_2,x_3,x_4$ as in the answer below.
$endgroup$
– Shubham Johri
Jan 6 at 6:57




$begingroup$
You require $$begin{bmatrix} 1 & 2 & -3& -1 \ 0& 0 &0 &1 end{bmatrix}begin{bmatrix} x_1\x_2\x_3\x_4 end{bmatrix}=begin{bmatrix} 0\ 0\0\0 end{bmatrix}$$ Now just find the relations between $x_1,x_2,x_3,x_4$ as in the answer below.
$endgroup$
– Shubham Johri
Jan 6 at 6:57












$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a basis for the null space directly from the RREF.
$endgroup$
– amd
Jan 6 at 8:13




$begingroup$
See math.stackexchange.com/a/1521354/265466 for how to read a basis for the null space directly from the RREF.
$endgroup$
– amd
Jan 6 at 8:13










1 Answer
1






active

oldest

votes


















1












$begingroup$

Great that you have found a row echelon form.



From the second row, we can conclude that $x_4=0$.



Also, from there, and the first row of the row echelon form, we have



$$x_1+2x_2-3x_3=0$$



Now you have a relationship between the variables. Hopefully you can take it from here.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Then I set x2=α, x3= β, then x1 = 3β-2α. So the null space is spanned by α(-2 1 0 0)^T + β(3 0 1 0)^T, is that true?
    $endgroup$
    – Shadow Z
    Jan 6 at 6:56










  • $begingroup$
    @ShadowZ Yes, that is correct
    $endgroup$
    – Shubham Johri
    Jan 6 at 6:59










  • $begingroup$
    Congratulations, good job. try to pick up mathjax, it makes maths prettier and more importantly makes communication smoother.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 6:59











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Great that you have found a row echelon form.



From the second row, we can conclude that $x_4=0$.



Also, from there, and the first row of the row echelon form, we have



$$x_1+2x_2-3x_3=0$$



Now you have a relationship between the variables. Hopefully you can take it from here.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Then I set x2=α, x3= β, then x1 = 3β-2α. So the null space is spanned by α(-2 1 0 0)^T + β(3 0 1 0)^T, is that true?
    $endgroup$
    – Shadow Z
    Jan 6 at 6:56










  • $begingroup$
    @ShadowZ Yes, that is correct
    $endgroup$
    – Shubham Johri
    Jan 6 at 6:59










  • $begingroup$
    Congratulations, good job. try to pick up mathjax, it makes maths prettier and more importantly makes communication smoother.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 6:59
















1












$begingroup$

Great that you have found a row echelon form.



From the second row, we can conclude that $x_4=0$.



Also, from there, and the first row of the row echelon form, we have



$$x_1+2x_2-3x_3=0$$



Now you have a relationship between the variables. Hopefully you can take it from here.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! Then I set x2=α, x3= β, then x1 = 3β-2α. So the null space is spanned by α(-2 1 0 0)^T + β(3 0 1 0)^T, is that true?
    $endgroup$
    – Shadow Z
    Jan 6 at 6:56










  • $begingroup$
    @ShadowZ Yes, that is correct
    $endgroup$
    – Shubham Johri
    Jan 6 at 6:59










  • $begingroup$
    Congratulations, good job. try to pick up mathjax, it makes maths prettier and more importantly makes communication smoother.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 6:59














1












1








1





$begingroup$

Great that you have found a row echelon form.



From the second row, we can conclude that $x_4=0$.



Also, from there, and the first row of the row echelon form, we have



$$x_1+2x_2-3x_3=0$$



Now you have a relationship between the variables. Hopefully you can take it from here.






share|cite|improve this answer









$endgroup$



Great that you have found a row echelon form.



From the second row, we can conclude that $x_4=0$.



Also, from there, and the first row of the row echelon form, we have



$$x_1+2x_2-3x_3=0$$



Now you have a relationship between the variables. Hopefully you can take it from here.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 6 at 6:48









Siong Thye GohSiong Thye Goh

103k1468119




103k1468119












  • $begingroup$
    Thanks! Then I set x2=α, x3= β, then x1 = 3β-2α. So the null space is spanned by α(-2 1 0 0)^T + β(3 0 1 0)^T, is that true?
    $endgroup$
    – Shadow Z
    Jan 6 at 6:56










  • $begingroup$
    @ShadowZ Yes, that is correct
    $endgroup$
    – Shubham Johri
    Jan 6 at 6:59










  • $begingroup$
    Congratulations, good job. try to pick up mathjax, it makes maths prettier and more importantly makes communication smoother.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 6:59


















  • $begingroup$
    Thanks! Then I set x2=α, x3= β, then x1 = 3β-2α. So the null space is spanned by α(-2 1 0 0)^T + β(3 0 1 0)^T, is that true?
    $endgroup$
    – Shadow Z
    Jan 6 at 6:56










  • $begingroup$
    @ShadowZ Yes, that is correct
    $endgroup$
    – Shubham Johri
    Jan 6 at 6:59










  • $begingroup$
    Congratulations, good job. try to pick up mathjax, it makes maths prettier and more importantly makes communication smoother.
    $endgroup$
    – Siong Thye Goh
    Jan 6 at 6:59
















$begingroup$
Thanks! Then I set x2=α, x3= β, then x1 = 3β-2α. So the null space is spanned by α(-2 1 0 0)^T + β(3 0 1 0)^T, is that true?
$endgroup$
– Shadow Z
Jan 6 at 6:56




$begingroup$
Thanks! Then I set x2=α, x3= β, then x1 = 3β-2α. So the null space is spanned by α(-2 1 0 0)^T + β(3 0 1 0)^T, is that true?
$endgroup$
– Shadow Z
Jan 6 at 6:56












$begingroup$
@ShadowZ Yes, that is correct
$endgroup$
– Shubham Johri
Jan 6 at 6:59




$begingroup$
@ShadowZ Yes, that is correct
$endgroup$
– Shubham Johri
Jan 6 at 6:59












$begingroup$
Congratulations, good job. try to pick up mathjax, it makes maths prettier and more importantly makes communication smoother.
$endgroup$
– Siong Thye Goh
Jan 6 at 6:59




$begingroup$
Congratulations, good job. try to pick up mathjax, it makes maths prettier and more importantly makes communication smoother.
$endgroup$
– Siong Thye Goh
Jan 6 at 6:59


















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