Is $U$ a subspace of $Bbb R^3$?
$begingroup$
$U={(2r,-s^2,t) mid r, s text{ and} t in Bbb R}$
So I know I have to make sure of the following:
(a) The zero vector is in $U$ (replace $r,s,t$ by $0$ and you get $0$)
(b) It is closed under addition (the resultant of adding other vectors, like $(r,s,t)$ to $(2r,-s^2,t)$, is still in $U$)
(c) It is closed under scalar multiplication (any scalar times $U$ is still in $U$)
I found that they all work. But apparently, $U$ is not a subset. Why? Is it because there are three variables, $r,s,t$?
linear-algebra
edited Dec 2 '18 at 5:00
Tianlalu
3,08121038
asked Dec 2 '18 at 4:41
Sami JrSami Jr
72
$endgroup$
add a comment |
$begingroup$
$U={(2r,-s^2,t) mid r, s text{ and} t in Bbb R}$
So I know I have to make sure of the following:
(a) The zero vector is in $U$ (replace $r,s,t$ by $0$ and you get $0$)
(b) It is closed under addition (the resultant of adding other vectors, like $(r,s,t)$ to $(2r,-s^2,t)$, is still in $U$)
(c) It is closed under scalar multiplication (any scalar times $U$ is still in $U$)
I found that they all work. But apparently, $U$ is not a subset. Why? Is it because there are three variables, $r,s,t$?
linear-algebra
edited Dec 2 '18 at 5:00
Tianlalu
3,08121038
asked Dec 2 '18 at 4:41
Sami JrSami Jr
72
$endgroup$
add a comment |
$begingroup$
$U={(2r,-s^2,t) mid r, s text{ and} t in Bbb R}$
So I know I have to make sure of the following:
(a) The zero vector is in $U$ (replace $r,s,t$ by $0$ and you get $0$)
(b) It is closed under addition (the resultant of adding other vectors, like $(r,s,t)$ to $(2r,-s^2,t)$, is still in $U$)
(c) It is closed under scalar multiplication (any scalar times $U$ is still in $U$)
I found that they all work. But apparently, $U$ is not a subset. Why? Is it because there are three variables, $r,s,t$?
linear-algebra
edited Dec 2 '18 at 5:00
Tianlalu
3,08121038
asked Dec 2 '18 at 4:41
Sami JrSami Jr
72
$endgroup$
$U={(2r,-s^2,t) mid r, s text{ and} t in Bbb R}$
So I know I have to make sure of the following:
(a) The zero vector is in $U$ (replace $r,s,t$ by $0$ and you get $0$)
(b) It is closed under addition (the resultant of adding other vectors, like $(r,s,t)$ to $(2r,-s^2,t)$, is still in $U$)
(c) It is closed under scalar multiplication (any scalar times $U$ is still in $U$)
I found that they all work. But apparently, $U$ is not a subset. Why? Is it because there are three variables, $r,s,t$?
linear-algebra
linear-algebra
edited Dec 2 '18 at 5:00
Tianlalu
3,08121038
asked Dec 2 '18 at 4:41
Sami JrSami Jr
72
edited Dec 2 '18 at 5:00
Tianlalu
3,08121038
asked Dec 2 '18 at 4:41
Sami JrSami Jr
72
edited Dec 2 '18 at 5:00
Tianlalu
3,08121038
edited Dec 2 '18 at 5:00
Tianlalu
3,08121038
edited Dec 2 '18 at 5:00
Tianlalu
3,08121038
3,08121038
asked Dec 2 '18 at 4:41
Sami JrSami Jr
72
asked Dec 2 '18 at 4:41
Sami JrSami Jr
72
asked Dec 2 '18 at 4:41
Sami JrSami Jr
72
72
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$(1,-1,1)$ is in $U$. Multiply by $-1$, which is a scalar. The vector $(-1,1,-1)$ is not in $U$. (Notice $-s^2$ can only be a negative number. In my experience such condition is a red flag!!)
edited Dec 2 '18 at 5:01
Tianlalu
3,08121038
answered Dec 2 '18 at 4:45
Ko ByeongminKo Byeongmin
1326
$endgroup$
1
$begingroup$
But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
$endgroup$
– Sami Jr
Dec 2 '18 at 4:47
$begingroup$
As you pointed out, a subspace must be closed under scalar multiplication. If one entry of a vector inside a subset of interest can only be a negative number, multiplying it by a negative number would give us a vector outside that set (because the negative entry now becomes positive,) making the subset not closed under scalar multiplication.
$endgroup$
– Ko Byeongmin
Dec 2 '18 at 4:49
$begingroup$
Similar example: consider our usual xy-plane. The first quadrant is not a subspace of $R^2$, because multiply any nonzero vector inside the first quadrant by a negative number and the multiplied vector is outside the first quadrant.
$endgroup$
– Ko Byeongmin
Dec 2 '18 at 4:53
1
$begingroup$
Ohhh thank you so much for the added explanation!!
$endgroup$
– Sami Jr
Dec 2 '18 at 4:57
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%2f3022241%2fis-u-a-subspace-of-bbb-r3%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
$begingroup$
$(1,-1,1)$ is in $U$. Multiply by $-1$, which is a scalar. The vector $(-1,1,-1)$ is not in $U$. (Notice $-s^2$ can only be a negative number. In my experience such condition is a red flag!!)
edited Dec 2 '18 at 5:01
Tianlalu
3,08121038
answered Dec 2 '18 at 4:45
Ko ByeongminKo Byeongmin
1326
$endgroup$
1
$begingroup$
But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
$endgroup$
– Sami Jr
Dec 2 '18 at 4:47
$begingroup$
As you pointed out, a subspace must be closed under scalar multiplication. If one entry of a vector inside a subset of interest can only be a negative number, multiplying it by a negative number would give us a vector outside that set (because the negative entry now becomes positive,) making the subset not closed under scalar multiplication.
$endgroup$
– Ko Byeongmin
Dec 2 '18 at 4:49
$begingroup$
Similar example: consider our usual xy-plane. The first quadrant is not a subspace of $R^2$, because multiply any nonzero vector inside the first quadrant by a negative number and the multiplied vector is outside the first quadrant.
$endgroup$
– Ko Byeongmin
Dec 2 '18 at 4:53
1
$begingroup$
Ohhh thank you so much for the added explanation!!
$endgroup$
– Sami Jr
Dec 2 '18 at 4:57
add a comment |
$begingroup$
$(1,-1,1)$ is in $U$. Multiply by $-1$, which is a scalar. The vector $(-1,1,-1)$ is not in $U$. (Notice $-s^2$ can only be a negative number. In my experience such condition is a red flag!!)
edited Dec 2 '18 at 5:01
Tianlalu
3,08121038
answered Dec 2 '18 at 4:45
Ko ByeongminKo Byeongmin
1326
$endgroup$
1
$begingroup$
But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
$endgroup$
– Sami Jr
Dec 2 '18 at 4:47
$begingroup$
As you pointed out, a subspace must be closed under scalar multiplication. If one entry of a vector inside a subset of interest can only be a negative number, multiplying it by a negative number would give us a vector outside that set (because the negative entry now becomes positive,) making the subset not closed under scalar multiplication.
$endgroup$
– Ko Byeongmin
Dec 2 '18 at 4:49
$begingroup$
Similar example: consider our usual xy-plane. The first quadrant is not a subspace of $R^2$, because multiply any nonzero vector inside the first quadrant by a negative number and the multiplied vector is outside the first quadrant.
$endgroup$
– Ko Byeongmin
Dec 2 '18 at 4:53
1
$begingroup$
Ohhh thank you so much for the added explanation!!
$endgroup$
– Sami Jr
Dec 2 '18 at 4:57
add a comment |
$begingroup$
$(1,-1,1)$ is in $U$. Multiply by $-1$, which is a scalar. The vector $(-1,1,-1)$ is not in $U$. (Notice $-s^2$ can only be a negative number. In my experience such condition is a red flag!!)
edited Dec 2 '18 at 5:01
Tianlalu
3,08121038
answered Dec 2 '18 at 4:45
Ko ByeongminKo Byeongmin
1326
$endgroup$
$(1,-1,1)$ is in $U$. Multiply by $-1$, which is a scalar. The vector $(-1,1,-1)$ is not in $U$. (Notice $-s^2$ can only be a negative number. In my experience such condition is a red flag!!)
edited Dec 2 '18 at 5:01
Tianlalu
3,08121038
answered Dec 2 '18 at 4:45
Ko ByeongminKo Byeongmin
1326
edited Dec 2 '18 at 5:01
Tianlalu
3,08121038
edited Dec 2 '18 at 5:01
Tianlalu
3,08121038
edited Dec 2 '18 at 5:01
Tianlalu
3,08121038
3,08121038
answered Dec 2 '18 at 4:45
Ko ByeongminKo Byeongmin
1326
answered Dec 2 '18 at 4:45
Ko ByeongminKo Byeongmin
1326
answered Dec 2 '18 at 4:45
Ko ByeongminKo Byeongmin
1326
1326
1
$begingroup$
But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
$endgroup$
– Sami Jr
Dec 2 '18 at 4:47
$begingroup$
As you pointed out, a subspace must be closed under scalar multiplication. If one entry of a vector inside a subset of interest can only be a negative number, multiplying it by a negative number would give us a vector outside that set (because the negative entry now becomes positive,) making the subset not closed under scalar multiplication.
$endgroup$
– Ko Byeongmin
Dec 2 '18 at 4:49
$begingroup$
Similar example: consider our usual xy-plane. The first quadrant is not a subspace of $R^2$, because multiply any nonzero vector inside the first quadrant by a negative number and the multiplied vector is outside the first quadrant.
$endgroup$
– Ko Byeongmin
Dec 2 '18 at 4:53
1
$begingroup$
Ohhh thank you so much for the added explanation!!
$endgroup$
– Sami Jr
Dec 2 '18 at 4:57
add a comment |
1
$begingroup$
But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
$endgroup$
– Sami Jr
Dec 2 '18 at 4:47
$begingroup$
As you pointed out, a subspace must be closed under scalar multiplication. If one entry of a vector inside a subset of interest can only be a negative number, multiplying it by a negative number would give us a vector outside that set (because the negative entry now becomes positive,) making the subset not closed under scalar multiplication.
$endgroup$
– Ko Byeongmin
Dec 2 '18 at 4:49
$begingroup$
Similar example: consider our usual xy-plane. The first quadrant is not a subspace of $R^2$, because multiply any nonzero vector inside the first quadrant by a negative number and the multiplied vector is outside the first quadrant.
$endgroup$
– Ko Byeongmin
Dec 2 '18 at 4:53
1
$begingroup$
Ohhh thank you so much for the added explanation!!
$endgroup$
– Sami Jr
Dec 2 '18 at 4:57
1
1
$begingroup$
But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
$endgroup$
– Sami Jr
Dec 2 '18 at 4:47
$begingroup$
But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
$endgroup$
– Sami Jr
Dec 2 '18 at 4:47
$begingroup$
As you pointed out, a subspace must be closed under scalar multiplication. If one entry of a vector inside a subset of interest can only be a negative number, multiplying it by a negative number would give us a vector outside that set (because the negative entry now becomes positive,) making the subset not closed under scalar multiplication.
$endgroup$
– Ko Byeongmin
Dec 2 '18 at 4:49
$begingroup$
As you pointed out, a subspace must be closed under scalar multiplication. If one entry of a vector inside a subset of interest can only be a negative number, multiplying it by a negative number would give us a vector outside that set (because the negative entry now becomes positive,) making the subset not closed under scalar multiplication.
$endgroup$
– Ko Byeongmin
Dec 2 '18 at 4:49
$begingroup$
Similar example: consider our usual xy-plane. The first quadrant is not a subspace of $R^2$, because multiply any nonzero vector inside the first quadrant by a negative number and the multiplied vector is outside the first quadrant.
$endgroup$
– Ko Byeongmin
Dec 2 '18 at 4:53
$begingroup$
Similar example: consider our usual xy-plane. The first quadrant is not a subspace of $R^2$, because multiply any nonzero vector inside the first quadrant by a negative number and the multiplied vector is outside the first quadrant.
$endgroup$
– Ko Byeongmin
Dec 2 '18 at 4:53
1
1
$begingroup$
Ohhh thank you so much for the added explanation!!
$endgroup$
– Sami Jr
Dec 2 '18 at 4:57
$begingroup$
Ohhh thank you so much for the added explanation!!
$endgroup$
– Sami Jr
Dec 2 '18 at 4:57
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%2f3022241%2fis-u-a-subspace-of-bbb-r3%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
