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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063554%2fdetermine-the-null-space-of-the-following-matrix%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063554%2fdetermine-the-null-space-of-the-following-matrix%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more