Is $U$ a subspace of $Bbb R^3$?
up vote
0
down vote
favorite
$U={(2r,-s^2,t) mid r, s text{ and} t in Bbb R}$
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 at 5:00
Tianlalu
2,9451936
asked Dec 2 at 4:41
Sami Jr
72
add a comment |
up vote
0
down vote
favorite
$U={(2r,-s^2,t) mid r, s text{ and} t in Bbb R}$
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 at 5:00
Tianlalu
2,9451936
asked Dec 2 at 4:41
Sami Jr
72
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$U={(2r,-s^2,t) mid r, s text{ and} t in Bbb R}$
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 at 5:00
Tianlalu
2,9451936
asked Dec 2 at 4:41
Sami Jr
72
$U={(2r,-s^2,t) mid r, s text{ and} t in Bbb R}$
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 at 5:00
Tianlalu
2,9451936
asked Dec 2 at 4:41
Sami Jr
72
edited Dec 2 at 5:00
Tianlalu
2,9451936
asked Dec 2 at 4:41
Sami Jr
72
edited Dec 2 at 5:00
Tianlalu
2,9451936
edited Dec 2 at 5:00
Tianlalu
2,9451936
edited Dec 2 at 5:00
Tianlalu
2,9451936
2,9451936
asked Dec 2 at 4:41
Sami Jr
72
asked Dec 2 at 4:41
Sami Jr
72
asked Dec 2 at 4:41
Sami Jr
72
72
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
$(1,-1,1)$ 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 at 5:01
Tianlalu
2,9451936
answered Dec 2 at 4:45
Ko Byeongmin
1406
1
But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
– Sami Jr
Dec 2 at 4:47
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.
– Ko Byeongmin
Dec 2 at 4:49
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.
– Ko Byeongmin
Dec 2 at 4:53
1
Ohhh thank you so much for the added explanation!!
– Sami Jr
Dec 2 at 4:57
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$(1,-1,1)$ 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 at 5:01
Tianlalu
2,9451936
answered Dec 2 at 4:45
Ko Byeongmin
1406
1
But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
– Sami Jr
Dec 2 at 4:47
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.
– Ko Byeongmin
Dec 2 at 4:49
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.
– Ko Byeongmin
Dec 2 at 4:53
1
Ohhh thank you so much for the added explanation!!
– Sami Jr
Dec 2 at 4:57
add a comment |
up vote
1
down vote
accepted
$(1,-1,1)$ 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 at 5:01
Tianlalu
2,9451936
answered Dec 2 at 4:45
Ko Byeongmin
1406
1
But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
– Sami Jr
Dec 2 at 4:47
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.
– Ko Byeongmin
Dec 2 at 4:49
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.
– Ko Byeongmin
Dec 2 at 4:53
1
Ohhh thank you so much for the added explanation!!
– Sami Jr
Dec 2 at 4:57
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$(1,-1,1)$ 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 at 5:01
Tianlalu
2,9451936
answered Dec 2 at 4:45
Ko Byeongmin
1406
$(1,-1,1)$ 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 at 5:01
Tianlalu
2,9451936
answered Dec 2 at 4:45
Ko Byeongmin
1406
edited Dec 2 at 5:01
Tianlalu
2,9451936
edited Dec 2 at 5:01
Tianlalu
2,9451936
edited Dec 2 at 5:01
Tianlalu
2,9451936
2,9451936
answered Dec 2 at 4:45
Ko Byeongmin
1406
answered Dec 2 at 4:45
Ko Byeongmin
1406
answered Dec 2 at 4:45
Ko Byeongmin
1406
1406
1
But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
– Sami Jr
Dec 2 at 4:47
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.
– Ko Byeongmin
Dec 2 at 4:49
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.
– Ko Byeongmin
Dec 2 at 4:53
1
Ohhh thank you so much for the added explanation!!
– Sami Jr
Dec 2 at 4:57
add a comment |
1
But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
– Sami Jr
Dec 2 at 4:47
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.
– Ko Byeongmin
Dec 2 at 4:49
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.
– Ko Byeongmin
Dec 2 at 4:53
1
Ohhh thank you so much for the added explanation!!
– Sami Jr
Dec 2 at 4:57
1
1
But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
– Sami Jr
Dec 2 at 4:47
But what's wrong with $-s^2$ only being a negative number when the condition on $s$ is that it must be in R?
– Sami Jr
Dec 2 at 4:47
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.
– Ko Byeongmin
Dec 2 at 4:49
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.
– Ko Byeongmin
Dec 2 at 4:49
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.
– Ko Byeongmin
Dec 2 at 4:53
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.
– Ko Byeongmin
Dec 2 at 4:53
1
1
Ohhh thank you so much for the added explanation!!
– Sami Jr
Dec 2 at 4:57
Ohhh thank you so much for the added explanation!!
– Sami Jr
Dec 2 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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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
