Finding the basis of the intersection of a subspace and span
$begingroup$
I need help with determining the basis of $U_1 cap U_2$ in the following problem:
Let $V=mathbb{R}^4$. ${U_1} = left{ {left( {begin{array}{*{20}{c}}
{{x_1}} \
{{x_2}} \
{{x_3}} \
{{x_4}}
end{array}} right)left| {{x_1} - {x_2} + {x_3} - 3{x_4} = 0} right.} right}$ and $U_2=leftlangle {left( {begin{array}{*{20}{c}}
1 \
1 \
0 \
3
end{array}} right),left( {begin{array}{*{20}{c}}
0 \
{ - 1} \
0 \
1
end{array}} right)} rightrangle$.
If $U_1$ is a subspace of $V$, determine a basis of $U_1 cap U_2$.
My attempt:
I know that ${U_2} = left{ {left( {begin{array}{*{20}{c}}
lambda \
{lambda - mu } \
0 \
{3lambda + mu }
end{array}} right)left| {lambda ,mu in mathbb{R}} right.} right}$, and that the next step is that I should choose an element in $U_1$ and in $U_2$, e.g. Let $w in {U_1}$ and let $w in {U_2}$. Then we know that $w$ is of the form $w = left( {begin{array}{*{20}{c}}
lambda \
{lambda - mu } \
0 \
{3lambda + mu }
end{array}} right)$, but I'm not sure what the procedure is from there.
linear-algebra
$endgroup$
add a comment |
$begingroup$
I need help with determining the basis of $U_1 cap U_2$ in the following problem:
Let $V=mathbb{R}^4$. ${U_1} = left{ {left( {begin{array}{*{20}{c}}
{{x_1}} \
{{x_2}} \
{{x_3}} \
{{x_4}}
end{array}} right)left| {{x_1} - {x_2} + {x_3} - 3{x_4} = 0} right.} right}$ and $U_2=leftlangle {left( {begin{array}{*{20}{c}}
1 \
1 \
0 \
3
end{array}} right),left( {begin{array}{*{20}{c}}
0 \
{ - 1} \
0 \
1
end{array}} right)} rightrangle$.
If $U_1$ is a subspace of $V$, determine a basis of $U_1 cap U_2$.
My attempt:
I know that ${U_2} = left{ {left( {begin{array}{*{20}{c}}
lambda \
{lambda - mu } \
0 \
{3lambda + mu }
end{array}} right)left| {lambda ,mu in mathbb{R}} right.} right}$, and that the next step is that I should choose an element in $U_1$ and in $U_2$, e.g. Let $w in {U_1}$ and let $w in {U_2}$. Then we know that $w$ is of the form $w = left( {begin{array}{*{20}{c}}
lambda \
{lambda - mu } \
0 \
{3lambda + mu }
end{array}} right)$, but I'm not sure what the procedure is from there.
linear-algebra
$endgroup$
$begingroup$
I think the easiest way to solve this without being clever is to rewrite $U_2$ as the solution set to an appropriate system of equations. Having found that system of equations you are merely looking for points where both the equations for $U_1$ and $U_2$ hold true. That is a known calculation, then you can find the basis for that. Moreover, this method generalizes to other similar problems. In short, span bad, equation good (for intersections).
$endgroup$
– James S. Cook
4 hours ago
add a comment |
$begingroup$
I need help with determining the basis of $U_1 cap U_2$ in the following problem:
Let $V=mathbb{R}^4$. ${U_1} = left{ {left( {begin{array}{*{20}{c}}
{{x_1}} \
{{x_2}} \
{{x_3}} \
{{x_4}}
end{array}} right)left| {{x_1} - {x_2} + {x_3} - 3{x_4} = 0} right.} right}$ and $U_2=leftlangle {left( {begin{array}{*{20}{c}}
1 \
1 \
0 \
3
end{array}} right),left( {begin{array}{*{20}{c}}
0 \
{ - 1} \
0 \
1
end{array}} right)} rightrangle$.
If $U_1$ is a subspace of $V$, determine a basis of $U_1 cap U_2$.
My attempt:
I know that ${U_2} = left{ {left( {begin{array}{*{20}{c}}
lambda \
{lambda - mu } \
0 \
{3lambda + mu }
end{array}} right)left| {lambda ,mu in mathbb{R}} right.} right}$, and that the next step is that I should choose an element in $U_1$ and in $U_2$, e.g. Let $w in {U_1}$ and let $w in {U_2}$. Then we know that $w$ is of the form $w = left( {begin{array}{*{20}{c}}
lambda \
{lambda - mu } \
0 \
{3lambda + mu }
end{array}} right)$, but I'm not sure what the procedure is from there.
linear-algebra
$endgroup$
I need help with determining the basis of $U_1 cap U_2$ in the following problem:
Let $V=mathbb{R}^4$. ${U_1} = left{ {left( {begin{array}{*{20}{c}}
{{x_1}} \
{{x_2}} \
{{x_3}} \
{{x_4}}
end{array}} right)left| {{x_1} - {x_2} + {x_3} - 3{x_4} = 0} right.} right}$ and $U_2=leftlangle {left( {begin{array}{*{20}{c}}
1 \
1 \
0 \
3
end{array}} right),left( {begin{array}{*{20}{c}}
0 \
{ - 1} \
0 \
1
end{array}} right)} rightrangle$.
If $U_1$ is a subspace of $V$, determine a basis of $U_1 cap U_2$.
My attempt:
I know that ${U_2} = left{ {left( {begin{array}{*{20}{c}}
lambda \
{lambda - mu } \
0 \
{3lambda + mu }
end{array}} right)left| {lambda ,mu in mathbb{R}} right.} right}$, and that the next step is that I should choose an element in $U_1$ and in $U_2$, e.g. Let $w in {U_1}$ and let $w in {U_2}$. Then we know that $w$ is of the form $w = left( {begin{array}{*{20}{c}}
lambda \
{lambda - mu } \
0 \
{3lambda + mu }
end{array}} right)$, but I'm not sure what the procedure is from there.
linear-algebra
linear-algebra
edited 4 hours ago
bb411
asked 5 hours ago
bb411bb411
15119
15119
$begingroup$
I think the easiest way to solve this without being clever is to rewrite $U_2$ as the solution set to an appropriate system of equations. Having found that system of equations you are merely looking for points where both the equations for $U_1$ and $U_2$ hold true. That is a known calculation, then you can find the basis for that. Moreover, this method generalizes to other similar problems. In short, span bad, equation good (for intersections).
$endgroup$
– James S. Cook
4 hours ago
add a comment |
$begingroup$
I think the easiest way to solve this without being clever is to rewrite $U_2$ as the solution set to an appropriate system of equations. Having found that system of equations you are merely looking for points where both the equations for $U_1$ and $U_2$ hold true. That is a known calculation, then you can find the basis for that. Moreover, this method generalizes to other similar problems. In short, span bad, equation good (for intersections).
$endgroup$
– James S. Cook
4 hours ago
$begingroup$
I think the easiest way to solve this without being clever is to rewrite $U_2$ as the solution set to an appropriate system of equations. Having found that system of equations you are merely looking for points where both the equations for $U_1$ and $U_2$ hold true. That is a known calculation, then you can find the basis for that. Moreover, this method generalizes to other similar problems. In short, span bad, equation good (for intersections).
$endgroup$
– James S. Cook
4 hours ago
$begingroup$
I think the easiest way to solve this without being clever is to rewrite $U_2$ as the solution set to an appropriate system of equations. Having found that system of equations you are merely looking for points where both the equations for $U_1$ and $U_2$ hold true. That is a known calculation, then you can find the basis for that. Moreover, this method generalizes to other similar problems. In short, span bad, equation good (for intersections).
$endgroup$
– James S. Cook
4 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The next step is to note thatbegin{align}U_1cap U_2&=left{begin{pmatrix}lambda\lambda-mu\0\3lambda+muend{pmatrix},middle|,lambda-(lambda-mu)-3(3lambda+mu)=0right}\&=left{begin{pmatrix}lambda\lambda-mu\0\3lambda+muend{pmatrix},middle|,9lambda+2mu=0right}\&=left{begin{pmatrix}lambda\frac{11}2lambda\0\-frac32lambdaend{pmatrix},middle|,lambdainmathbb{R}right}.end{align}Can you take it from here?
$endgroup$
$begingroup$
How should I proceed with finding a basis of ${U_1} cap {U_2} = leftlangle {left( {begin{array}{*{20}{c}} 1 \ {{raise0.5exhbox{$scriptstyle {11}$} kern-0.1em/kern-0.15em lower0.25exhbox{$scriptstyle 2$}}} \ 0 \ {{raise0.5exhbox{$scriptstyle { - 3}$} kern-0.1em/kern-0.15em lower0.25exhbox{$scriptstyle 2$}}} end{array}} right)} rightrangle $?
$endgroup$
– bb411
3 hours ago
1
$begingroup$
What about taking$$left{begin{pmatrix}1\frac{11}2\0\-frac32end{pmatrix}right}?$$
$endgroup$
– José Carlos Santos
3 hours ago
add a comment |
$begingroup$
$U_1$ is the set of vector such that $begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} x_1 \x_2 \x_3 \x_4end{bmatrix} = 0$
$begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} 1 \1 \0 \3end{bmatrix} = -9\
begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} 0 \-1 \0 \1end{bmatrix} = -2$
$2begin{bmatrix} 1 \1 \0 \3end{bmatrix} - 9begin{bmatrix} 0 \-1 \0 \1end{bmatrix} = begin{bmatrix} 2 \11 \0 \-3end{bmatrix}$
$endgroup$
add a comment |
$begingroup$
If $(1,1,0,3)$ and $(0,-1,0,1)$ span $U_2$ then if $(x_1,x_2,x_3,x_4) in U_2$ there exist $a,b$ such that:
$$ a(1,1,0,3)+b(0,-1,0,1)=(x_1,x_2,x_3,x_4)$$
We face,
$$ a=x_1, a-b = x_2, x_3=0, 3a+b=x_4 $$
Ok, so,
$$ b = x_1-x_2 qquad & qquad b = x_4-3x_1 Rightarrow x_1-x_2 = x_4-3x_1$$
Ok, in summary, $(x_1,x_2,x_3,x_4) in U_2$ if we have
$$ 4x_1-x_2-x_4 = 0 & x_3=0. $$
If $(x_1,x_2,x_3,x_4) in U_1$ then we know $x_1-x_2+x_3-3x_4 = 0$. Consequently, to find $(x_1,x_2,x_3,x_4) in U_1 cap U_2$ we need to solve equations for both subspaces simultaneously:
$$ left[ begin{array}{cccc|c} 1 & -1 & 1 & -3 & 0 \ 4 & -1 & 0 & -1 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] sim
left[ begin{array}{cccc|c} 1 & -1 & 0 & -9/3 & 0 \ 0 & 1 & 0 & 11/3 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] sim left[ begin{array}{cccc|c} 1 & 0 & 0 & 2/3 & 0 \ 0 & 1 & 0 & 11/3 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] $$
Thus $x_1 = -2x_4/3$ and $x_2 = -11x_4/3$ and $x_3=0$ with $x_4$ free. In short,
$$ U_1 cap U_2 = text{span}(-2,-11,0,3). $$
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3130546%2ffinding-the-basis-of-the-intersection-of-a-subspace-and-span%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The next step is to note thatbegin{align}U_1cap U_2&=left{begin{pmatrix}lambda\lambda-mu\0\3lambda+muend{pmatrix},middle|,lambda-(lambda-mu)-3(3lambda+mu)=0right}\&=left{begin{pmatrix}lambda\lambda-mu\0\3lambda+muend{pmatrix},middle|,9lambda+2mu=0right}\&=left{begin{pmatrix}lambda\frac{11}2lambda\0\-frac32lambdaend{pmatrix},middle|,lambdainmathbb{R}right}.end{align}Can you take it from here?
$endgroup$
$begingroup$
How should I proceed with finding a basis of ${U_1} cap {U_2} = leftlangle {left( {begin{array}{*{20}{c}} 1 \ {{raise0.5exhbox{$scriptstyle {11}$} kern-0.1em/kern-0.15em lower0.25exhbox{$scriptstyle 2$}}} \ 0 \ {{raise0.5exhbox{$scriptstyle { - 3}$} kern-0.1em/kern-0.15em lower0.25exhbox{$scriptstyle 2$}}} end{array}} right)} rightrangle $?
$endgroup$
– bb411
3 hours ago
1
$begingroup$
What about taking$$left{begin{pmatrix}1\frac{11}2\0\-frac32end{pmatrix}right}?$$
$endgroup$
– José Carlos Santos
3 hours ago
add a comment |
$begingroup$
The next step is to note thatbegin{align}U_1cap U_2&=left{begin{pmatrix}lambda\lambda-mu\0\3lambda+muend{pmatrix},middle|,lambda-(lambda-mu)-3(3lambda+mu)=0right}\&=left{begin{pmatrix}lambda\lambda-mu\0\3lambda+muend{pmatrix},middle|,9lambda+2mu=0right}\&=left{begin{pmatrix}lambda\frac{11}2lambda\0\-frac32lambdaend{pmatrix},middle|,lambdainmathbb{R}right}.end{align}Can you take it from here?
$endgroup$
$begingroup$
How should I proceed with finding a basis of ${U_1} cap {U_2} = leftlangle {left( {begin{array}{*{20}{c}} 1 \ {{raise0.5exhbox{$scriptstyle {11}$} kern-0.1em/kern-0.15em lower0.25exhbox{$scriptstyle 2$}}} \ 0 \ {{raise0.5exhbox{$scriptstyle { - 3}$} kern-0.1em/kern-0.15em lower0.25exhbox{$scriptstyle 2$}}} end{array}} right)} rightrangle $?
$endgroup$
– bb411
3 hours ago
1
$begingroup$
What about taking$$left{begin{pmatrix}1\frac{11}2\0\-frac32end{pmatrix}right}?$$
$endgroup$
– José Carlos Santos
3 hours ago
add a comment |
$begingroup$
The next step is to note thatbegin{align}U_1cap U_2&=left{begin{pmatrix}lambda\lambda-mu\0\3lambda+muend{pmatrix},middle|,lambda-(lambda-mu)-3(3lambda+mu)=0right}\&=left{begin{pmatrix}lambda\lambda-mu\0\3lambda+muend{pmatrix},middle|,9lambda+2mu=0right}\&=left{begin{pmatrix}lambda\frac{11}2lambda\0\-frac32lambdaend{pmatrix},middle|,lambdainmathbb{R}right}.end{align}Can you take it from here?
$endgroup$
The next step is to note thatbegin{align}U_1cap U_2&=left{begin{pmatrix}lambda\lambda-mu\0\3lambda+muend{pmatrix},middle|,lambda-(lambda-mu)-3(3lambda+mu)=0right}\&=left{begin{pmatrix}lambda\lambda-mu\0\3lambda+muend{pmatrix},middle|,9lambda+2mu=0right}\&=left{begin{pmatrix}lambda\frac{11}2lambda\0\-frac32lambdaend{pmatrix},middle|,lambdainmathbb{R}right}.end{align}Can you take it from here?
answered 4 hours ago
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
$begingroup$
How should I proceed with finding a basis of ${U_1} cap {U_2} = leftlangle {left( {begin{array}{*{20}{c}} 1 \ {{raise0.5exhbox{$scriptstyle {11}$} kern-0.1em/kern-0.15em lower0.25exhbox{$scriptstyle 2$}}} \ 0 \ {{raise0.5exhbox{$scriptstyle { - 3}$} kern-0.1em/kern-0.15em lower0.25exhbox{$scriptstyle 2$}}} end{array}} right)} rightrangle $?
$endgroup$
– bb411
3 hours ago
1
$begingroup$
What about taking$$left{begin{pmatrix}1\frac{11}2\0\-frac32end{pmatrix}right}?$$
$endgroup$
– José Carlos Santos
3 hours ago
add a comment |
$begingroup$
How should I proceed with finding a basis of ${U_1} cap {U_2} = leftlangle {left( {begin{array}{*{20}{c}} 1 \ {{raise0.5exhbox{$scriptstyle {11}$} kern-0.1em/kern-0.15em lower0.25exhbox{$scriptstyle 2$}}} \ 0 \ {{raise0.5exhbox{$scriptstyle { - 3}$} kern-0.1em/kern-0.15em lower0.25exhbox{$scriptstyle 2$}}} end{array}} right)} rightrangle $?
$endgroup$
– bb411
3 hours ago
1
$begingroup$
What about taking$$left{begin{pmatrix}1\frac{11}2\0\-frac32end{pmatrix}right}?$$
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
How should I proceed with finding a basis of ${U_1} cap {U_2} = leftlangle {left( {begin{array}{*{20}{c}} 1 \ {{raise0.5exhbox{$scriptstyle {11}$} kern-0.1em/kern-0.15em lower0.25exhbox{$scriptstyle 2$}}} \ 0 \ {{raise0.5exhbox{$scriptstyle { - 3}$} kern-0.1em/kern-0.15em lower0.25exhbox{$scriptstyle 2$}}} end{array}} right)} rightrangle $?
$endgroup$
– bb411
3 hours ago
$begingroup$
How should I proceed with finding a basis of ${U_1} cap {U_2} = leftlangle {left( {begin{array}{*{20}{c}} 1 \ {{raise0.5exhbox{$scriptstyle {11}$} kern-0.1em/kern-0.15em lower0.25exhbox{$scriptstyle 2$}}} \ 0 \ {{raise0.5exhbox{$scriptstyle { - 3}$} kern-0.1em/kern-0.15em lower0.25exhbox{$scriptstyle 2$}}} end{array}} right)} rightrangle $?
$endgroup$
– bb411
3 hours ago
1
1
$begingroup$
What about taking$$left{begin{pmatrix}1\frac{11}2\0\-frac32end{pmatrix}right}?$$
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
What about taking$$left{begin{pmatrix}1\frac{11}2\0\-frac32end{pmatrix}right}?$$
$endgroup$
– José Carlos Santos
3 hours ago
add a comment |
$begingroup$
$U_1$ is the set of vector such that $begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} x_1 \x_2 \x_3 \x_4end{bmatrix} = 0$
$begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} 1 \1 \0 \3end{bmatrix} = -9\
begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} 0 \-1 \0 \1end{bmatrix} = -2$
$2begin{bmatrix} 1 \1 \0 \3end{bmatrix} - 9begin{bmatrix} 0 \-1 \0 \1end{bmatrix} = begin{bmatrix} 2 \11 \0 \-3end{bmatrix}$
$endgroup$
add a comment |
$begingroup$
$U_1$ is the set of vector such that $begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} x_1 \x_2 \x_3 \x_4end{bmatrix} = 0$
$begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} 1 \1 \0 \3end{bmatrix} = -9\
begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} 0 \-1 \0 \1end{bmatrix} = -2$
$2begin{bmatrix} 1 \1 \0 \3end{bmatrix} - 9begin{bmatrix} 0 \-1 \0 \1end{bmatrix} = begin{bmatrix} 2 \11 \0 \-3end{bmatrix}$
$endgroup$
add a comment |
$begingroup$
$U_1$ is the set of vector such that $begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} x_1 \x_2 \x_3 \x_4end{bmatrix} = 0$
$begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} 1 \1 \0 \3end{bmatrix} = -9\
begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} 0 \-1 \0 \1end{bmatrix} = -2$
$2begin{bmatrix} 1 \1 \0 \3end{bmatrix} - 9begin{bmatrix} 0 \-1 \0 \1end{bmatrix} = begin{bmatrix} 2 \11 \0 \-3end{bmatrix}$
$endgroup$
$U_1$ is the set of vector such that $begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} x_1 \x_2 \x_3 \x_4end{bmatrix} = 0$
$begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} 1 \1 \0 \3end{bmatrix} = -9\
begin{bmatrix} 1 &-1 &1 &-3end{bmatrix}begin{bmatrix} 0 \-1 \0 \1end{bmatrix} = -2$
$2begin{bmatrix} 1 \1 \0 \3end{bmatrix} - 9begin{bmatrix} 0 \-1 \0 \1end{bmatrix} = begin{bmatrix} 2 \11 \0 \-3end{bmatrix}$
answered 4 hours ago
Doug MDoug M
45.3k31954
45.3k31954
add a comment |
add a comment |
$begingroup$
If $(1,1,0,3)$ and $(0,-1,0,1)$ span $U_2$ then if $(x_1,x_2,x_3,x_4) in U_2$ there exist $a,b$ such that:
$$ a(1,1,0,3)+b(0,-1,0,1)=(x_1,x_2,x_3,x_4)$$
We face,
$$ a=x_1, a-b = x_2, x_3=0, 3a+b=x_4 $$
Ok, so,
$$ b = x_1-x_2 qquad & qquad b = x_4-3x_1 Rightarrow x_1-x_2 = x_4-3x_1$$
Ok, in summary, $(x_1,x_2,x_3,x_4) in U_2$ if we have
$$ 4x_1-x_2-x_4 = 0 & x_3=0. $$
If $(x_1,x_2,x_3,x_4) in U_1$ then we know $x_1-x_2+x_3-3x_4 = 0$. Consequently, to find $(x_1,x_2,x_3,x_4) in U_1 cap U_2$ we need to solve equations for both subspaces simultaneously:
$$ left[ begin{array}{cccc|c} 1 & -1 & 1 & -3 & 0 \ 4 & -1 & 0 & -1 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] sim
left[ begin{array}{cccc|c} 1 & -1 & 0 & -9/3 & 0 \ 0 & 1 & 0 & 11/3 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] sim left[ begin{array}{cccc|c} 1 & 0 & 0 & 2/3 & 0 \ 0 & 1 & 0 & 11/3 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] $$
Thus $x_1 = -2x_4/3$ and $x_2 = -11x_4/3$ and $x_3=0$ with $x_4$ free. In short,
$$ U_1 cap U_2 = text{span}(-2,-11,0,3). $$
$endgroup$
add a comment |
$begingroup$
If $(1,1,0,3)$ and $(0,-1,0,1)$ span $U_2$ then if $(x_1,x_2,x_3,x_4) in U_2$ there exist $a,b$ such that:
$$ a(1,1,0,3)+b(0,-1,0,1)=(x_1,x_2,x_3,x_4)$$
We face,
$$ a=x_1, a-b = x_2, x_3=0, 3a+b=x_4 $$
Ok, so,
$$ b = x_1-x_2 qquad & qquad b = x_4-3x_1 Rightarrow x_1-x_2 = x_4-3x_1$$
Ok, in summary, $(x_1,x_2,x_3,x_4) in U_2$ if we have
$$ 4x_1-x_2-x_4 = 0 & x_3=0. $$
If $(x_1,x_2,x_3,x_4) in U_1$ then we know $x_1-x_2+x_3-3x_4 = 0$. Consequently, to find $(x_1,x_2,x_3,x_4) in U_1 cap U_2$ we need to solve equations for both subspaces simultaneously:
$$ left[ begin{array}{cccc|c} 1 & -1 & 1 & -3 & 0 \ 4 & -1 & 0 & -1 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] sim
left[ begin{array}{cccc|c} 1 & -1 & 0 & -9/3 & 0 \ 0 & 1 & 0 & 11/3 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] sim left[ begin{array}{cccc|c} 1 & 0 & 0 & 2/3 & 0 \ 0 & 1 & 0 & 11/3 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] $$
Thus $x_1 = -2x_4/3$ and $x_2 = -11x_4/3$ and $x_3=0$ with $x_4$ free. In short,
$$ U_1 cap U_2 = text{span}(-2,-11,0,3). $$
$endgroup$
add a comment |
$begingroup$
If $(1,1,0,3)$ and $(0,-1,0,1)$ span $U_2$ then if $(x_1,x_2,x_3,x_4) in U_2$ there exist $a,b$ such that:
$$ a(1,1,0,3)+b(0,-1,0,1)=(x_1,x_2,x_3,x_4)$$
We face,
$$ a=x_1, a-b = x_2, x_3=0, 3a+b=x_4 $$
Ok, so,
$$ b = x_1-x_2 qquad & qquad b = x_4-3x_1 Rightarrow x_1-x_2 = x_4-3x_1$$
Ok, in summary, $(x_1,x_2,x_3,x_4) in U_2$ if we have
$$ 4x_1-x_2-x_4 = 0 & x_3=0. $$
If $(x_1,x_2,x_3,x_4) in U_1$ then we know $x_1-x_2+x_3-3x_4 = 0$. Consequently, to find $(x_1,x_2,x_3,x_4) in U_1 cap U_2$ we need to solve equations for both subspaces simultaneously:
$$ left[ begin{array}{cccc|c} 1 & -1 & 1 & -3 & 0 \ 4 & -1 & 0 & -1 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] sim
left[ begin{array}{cccc|c} 1 & -1 & 0 & -9/3 & 0 \ 0 & 1 & 0 & 11/3 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] sim left[ begin{array}{cccc|c} 1 & 0 & 0 & 2/3 & 0 \ 0 & 1 & 0 & 11/3 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] $$
Thus $x_1 = -2x_4/3$ and $x_2 = -11x_4/3$ and $x_3=0$ with $x_4$ free. In short,
$$ U_1 cap U_2 = text{span}(-2,-11,0,3). $$
$endgroup$
If $(1,1,0,3)$ and $(0,-1,0,1)$ span $U_2$ then if $(x_1,x_2,x_3,x_4) in U_2$ there exist $a,b$ such that:
$$ a(1,1,0,3)+b(0,-1,0,1)=(x_1,x_2,x_3,x_4)$$
We face,
$$ a=x_1, a-b = x_2, x_3=0, 3a+b=x_4 $$
Ok, so,
$$ b = x_1-x_2 qquad & qquad b = x_4-3x_1 Rightarrow x_1-x_2 = x_4-3x_1$$
Ok, in summary, $(x_1,x_2,x_3,x_4) in U_2$ if we have
$$ 4x_1-x_2-x_4 = 0 & x_3=0. $$
If $(x_1,x_2,x_3,x_4) in U_1$ then we know $x_1-x_2+x_3-3x_4 = 0$. Consequently, to find $(x_1,x_2,x_3,x_4) in U_1 cap U_2$ we need to solve equations for both subspaces simultaneously:
$$ left[ begin{array}{cccc|c} 1 & -1 & 1 & -3 & 0 \ 4 & -1 & 0 & -1 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] sim
left[ begin{array}{cccc|c} 1 & -1 & 0 & -9/3 & 0 \ 0 & 1 & 0 & 11/3 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] sim left[ begin{array}{cccc|c} 1 & 0 & 0 & 2/3 & 0 \ 0 & 1 & 0 & 11/3 & 0 \ 0 & 0 & 1 & 0 & 0 end{array}right] $$
Thus $x_1 = -2x_4/3$ and $x_2 = -11x_4/3$ and $x_3=0$ with $x_4$ free. In short,
$$ U_1 cap U_2 = text{span}(-2,-11,0,3). $$
answered 3 hours ago
James S. CookJames S. Cook
13.2k22872
13.2k22872
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3130546%2ffinding-the-basis-of-the-intersection-of-a-subspace-and-span%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
I think the easiest way to solve this without being clever is to rewrite $U_2$ as the solution set to an appropriate system of equations. Having found that system of equations you are merely looking for points where both the equations for $U_1$ and $U_2$ hold true. That is a known calculation, then you can find the basis for that. Moreover, this method generalizes to other similar problems. In short, span bad, equation good (for intersections).
$endgroup$
– James S. Cook
4 hours ago