Is $U$ a subspace of $Bbb R^3$?
$U={(2r,-s^2,t) mid r, s text{ and} t in Bbb R}$
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
3,09621038
asked Dec 2 '18 at 4:41
Sami Jr
72
add a comment |
$U={(2r,-s^2,t) mid r, s text{ and} t in Bbb R}$
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
3,09621038
asked Dec 2 '18 at 4:41
Sami Jr
72
add a comment |
$U={(2r,-s^2,t) mid r, s text{ and} t in Bbb R}$
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
3,09621038
asked Dec 2 '18 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 '18 at 5:00
Tianlalu
3,09621038
asked Dec 2 '18 at 4:41
Sami Jr
72
edited Dec 2 '18 at 5:00
Tianlalu
3,09621038
asked Dec 2 '18 at 4:41
Sami Jr
72
edited Dec 2 '18 at 5:00
Tianlalu
3,09621038
edited Dec 2 '18 at 5:00
Tianlalu
3,09621038
edited Dec 2 '18 at 5:00
Tianlalu
3,09621038
3,09621038
asked Dec 2 '18 at 4:41
Sami Jr
72
asked Dec 2 '18 at 4:41
Sami Jr
72
asked Dec 2 '18 at 4:41
Sami Jr
72
72
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$(1,-1,1)$ 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
3,09621038
answered Dec 2 '18 at 4:45
Ko Byeongmin
1326
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 '18 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 '18 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 '18 at 4:53
1
Ohhh thank you so much for the added explanation!!
– Sami Jr
Dec 2 '18 at 4:57
add a comment |
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$(1,-1,1)$ 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
3,09621038
answered Dec 2 '18 at 4:45
Ko Byeongmin
1326
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 '18 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 '18 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 '18 at 4:53
1
Ohhh thank you so much for the added explanation!!
– Sami Jr
Dec 2 '18 at 4:57
add a comment |
$(1,-1,1)$ 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
3,09621038
answered Dec 2 '18 at 4:45
Ko Byeongmin
1326
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 '18 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 '18 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 '18 at 4:53
1
Ohhh thank you so much for the added explanation!!
– Sami Jr
Dec 2 '18 at 4:57
add a comment |
$(1,-1,1)$ 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
3,09621038
answered Dec 2 '18 at 4:45
Ko Byeongmin
1326
$(1,-1,1)$ 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
3,09621038
answered Dec 2 '18 at 4:45
Ko Byeongmin
1326
edited Dec 2 '18 at 5:01
Tianlalu
3,09621038
edited Dec 2 '18 at 5:01
Tianlalu
3,09621038
edited Dec 2 '18 at 5:01
Tianlalu
3,09621038
3,09621038
answered Dec 2 '18 at 4:45
Ko Byeongmin
1326
answered Dec 2 '18 at 4:45
Ko Byeongmin
1326
answered Dec 2 '18 at 4:45
Ko Byeongmin
1326
1326
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 '18 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 '18 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 '18 at 4:53
1
Ohhh thank you so much for the added explanation!!
– Sami Jr
Dec 2 '18 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 '18 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 '18 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 '18 at 4:53
1
Ohhh thank you so much for the added explanation!!
– Sami Jr
Dec 2 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 4:53
1
1
Ohhh thank you so much for the added explanation!!
– Sami Jr
Dec 2 '18 at 4:57
Ohhh thank you so much for the added explanation!!
– Sami Jr
Dec 2 '18 at 4:57
add a comment |
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