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Finding which sets are subspaces of R3
Clash Royale CLAN TAG#URR8PPP
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https://i.stack.imgur.com/Bpl28.png
Hello. I have attached an image of the question I am having trouble with. I thought that it was 1,2 and 6 that were subspaces of $mathbb R^3$.
Here is my working:
1) Rearranged equation ---> $x+y-z=0$. Is a subspace since it is the set of solutions to a homogeneous linear equation.
2) $0$ is in the set if $x=y=0$. Is a subspace. (I know that to be a subspace, it must be closed under scalar multiplication and vector addition, but there was no equation linking the variables, so I just jumped into thinking it would be a subspace...)
3) Rearranged equation ---> $xy - xz=0$. $0$ is in the set if $x=0$ and $y=z$. I said that $(1,2,3)$ element of $R^3$ since $x,y,z$ are all real numbers, but when putting this into the rearranged equation, there was a contradiction. So, not a subspace.
4) = space $,(1,0,0),(0,0,1),$. set is not a subspace (no zero vector)
5) Similar to above.
6) $0$ is in the set if $m=0$. Again, I was not sure how to check if it is closed under vector addition and multiplication. From seeing that $0$ is in the set, I claimed it was a subspace.
I appreciate any help. Thank you.
linear-algebra vector-spaces vectors invariant-subspace
edited Aug 6 at 7:26
Babelfish
629114
asked Aug 6 at 4:20
Michel
84
add a comment |Â
up vote
1
down vote
favorite
https://i.stack.imgur.com/Bpl28.png
Hello. I have attached an image of the question I am having trouble with. I thought that it was 1,2 and 6 that were subspaces of $mathbb R^3$.
Here is my working:
1) Rearranged equation ---> $x+y-z=0$. Is a subspace since it is the set of solutions to a homogeneous linear equation.
2) $0$ is in the set if $x=y=0$. Is a subspace. (I know that to be a subspace, it must be closed under scalar multiplication and vector addition, but there was no equation linking the variables, so I just jumped into thinking it would be a subspace...)
3) Rearranged equation ---> $xy - xz=0$. $0$ is in the set if $x=0$ and $y=z$. I said that $(1,2,3)$ element of $R^3$ since $x,y,z$ are all real numbers, but when putting this into the rearranged equation, there was a contradiction. So, not a subspace.
4) = space $,(1,0,0),(0,0,1),$. set is not a subspace (no zero vector)
5) Similar to above.
6) $0$ is in the set if $m=0$. Again, I was not sure how to check if it is closed under vector addition and multiplication. From seeing that $0$ is in the set, I claimed it was a subspace.
I appreciate any help. Thank you.
linear-algebra vector-spaces vectors invariant-subspace
edited Aug 6 at 7:26
Babelfish
629114
asked Aug 6 at 4:20
Michel
84
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
https://i.stack.imgur.com/Bpl28.png
Hello. I have attached an image of the question I am having trouble with. I thought that it was 1,2 and 6 that were subspaces of $mathbb R^3$.
Here is my working:
1) Rearranged equation ---> $x+y-z=0$. Is a subspace since it is the set of solutions to a homogeneous linear equation.
2) $0$ is in the set if $x=y=0$. Is a subspace. (I know that to be a subspace, it must be closed under scalar multiplication and vector addition, but there was no equation linking the variables, so I just jumped into thinking it would be a subspace...)
3) Rearranged equation ---> $xy - xz=0$. $0$ is in the set if $x=0$ and $y=z$. I said that $(1,2,3)$ element of $R^3$ since $x,y,z$ are all real numbers, but when putting this into the rearranged equation, there was a contradiction. So, not a subspace.
4) = space $,(1,0,0),(0,0,1),$. set is not a subspace (no zero vector)
5) Similar to above.
6) $0$ is in the set if $m=0$. Again, I was not sure how to check if it is closed under vector addition and multiplication. From seeing that $0$ is in the set, I claimed it was a subspace.
I appreciate any help. Thank you.
linear-algebra vector-spaces vectors invariant-subspace
edited Aug 6 at 7:26
Babelfish
629114
asked Aug 6 at 4:20
Michel
84
https://i.stack.imgur.com/Bpl28.png
Hello. I have attached an image of the question I am having trouble with. I thought that it was 1,2 and 6 that were subspaces of $mathbb R^3$.
Here is my working:
1) Rearranged equation ---> $x+y-z=0$. Is a subspace since it is the set of solutions to a homogeneous linear equation.
2) $0$ is in the set if $x=y=0$. Is a subspace. (I know that to be a subspace, it must be closed under scalar multiplication and vector addition, but there was no equation linking the variables, so I just jumped into thinking it would be a subspace...)
3) Rearranged equation ---> $xy - xz=0$. $0$ is in the set if $x=0$ and $y=z$. I said that $(1,2,3)$ element of $R^3$ since $x,y,z$ are all real numbers, but when putting this into the rearranged equation, there was a contradiction. So, not a subspace.
4) = space $,(1,0,0),(0,0,1),$. set is not a subspace (no zero vector)
5) Similar to above.
6) $0$ is in the set if $m=0$. Again, I was not sure how to check if it is closed under vector addition and multiplication. From seeing that $0$ is in the set, I claimed it was a subspace.
I appreciate any help. Thank you.
linear-algebra vector-spaces vectors invariant-subspace
edited Aug 6 at 7:26
Babelfish
629114
asked Aug 6 at 4:20
Michel
84
edited Aug 6 at 7:26
Babelfish
629114
edited Aug 6 at 7:26
Babelfish
629114
edited Aug 6 at 7:26
Babelfish
629114
629114
asked Aug 6 at 4:20
Michel
84
asked Aug 6 at 4:20
Michel
84
asked Aug 6 at 4:20
Michel
84
84
add a comment |Â
add a comment |Â
2 Answers
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up vote
1
down vote
Here are the definitions I think you are missing:
Closure under vector addition:
A subset $S$ of $mathbbR^3$ is closed under vector addition if the sum of any two vectors in $S$ is also in $S$. In other words, if $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are in the subspace, then so is $(x_1+x_2,y_1+y_2,z_1+z_2)$. For example, if we were to check this definition against problem 2, we would be asking whether it is true that, for any $x_1,y_1,x_2,y_2inmathbbR$, the vector $(x_1,y_2,x_1y_1)+(x_2,y_2,x_2y_2)=(x_1+x_2,y_1+y_2,x_1x_2+y_1y_2)$ is in the subset.
Closure under scalar multiplication:
A subset $S$ of $mathbbR^3$ is closed under scalar multiplication if any real multiple of any vector in $S$ is also in $S$. In other words, if $r$ is any real number and $(x_1,y_1,z_1)$ is in the subspace, then so is $(rx_1,ry_1,rz_1)$. For example, if we were to check this definition against problem 2, we would be asking whether it is true that, for any $r,x_1,y_1inmathbbR$, the vector $(rx_1,ry_2,rx_1y_1)$ is in the subset.
The set $s(1,0,0)+t(0,0,1)$ from problem 4 is the set of vectors that can be expressed in the form $s(1,0,0)+t(0,0,1)$ for some pair of real numbers $s,tinmathbbR$. A similar definition holds for problem 5.
(Also I don't follow your reasoning at all for 3.)
answered Aug 6 at 4:42
alphacapture
1,806420
add a comment |Â
up vote
1
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Guide:
For part $4$, note that
$U_4=operatornameSpan (1,0,0), (0,0,1)$, it is written in the form of span of elements of $mathbbR^3$ which is closed under addition and scalar multiplication. Hence it is a subspace. I will leave part $5$ as an exercise.
Alternatively, let me prove $U_4$ is a subspace by verifying it is closed under additon and scalar multiplicaiton explicitly.
Closed under addition:
Let $x in U_4$, $exists s_x, t_x$ such that $x=s_x(1,0,0)+t_x(0,0,1)$ . Let $y in U_4$, $exists s_y, t_y$ such that $y=s_y(1,0,0)+t_y(0,0,1)$, then $x+y = (s_x+s_y)(1,0,0)+(s_y+t_y)(0,0,1)$ but we have $s_x+s_y, t_x+t_y in mathbbR$, hence $x+y in U_4$.
Closed under scalar multiplication, let $c in mathbbR$, $cx = (cs_x)(1,0,0)+(ct_x)(0,0,1)$ but we have $cs_x, ct_x in mathbbR$, hence $cx in U_4$.
Checking whether the zero vector is in is not sufficient.
Try to exhibit counter examples for part $2,3,6$ to prove that they are either not closed under addition or scalar multiplication. For example, for part $2$, $(1,1,1) in U_2$, what about $frac12 (1,1,1)$, is it in $U_2$?
answered Aug 6 at 4:41
Siong Thye Goh
78k134997
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Here are the definitions I think you are missing:
Closure under vector addition:
A subset $S$ of $mathbbR^3$ is closed under vector addition if the sum of any two vectors in $S$ is also in $S$. In other words, if $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are in the subspace, then so is $(x_1+x_2,y_1+y_2,z_1+z_2)$. For example, if we were to check this definition against problem 2, we would be asking whether it is true that, for any $x_1,y_1,x_2,y_2inmathbbR$, the vector $(x_1,y_2,x_1y_1)+(x_2,y_2,x_2y_2)=(x_1+x_2,y_1+y_2,x_1x_2+y_1y_2)$ is in the subset.
Closure under scalar multiplication:
A subset $S$ of $mathbbR^3$ is closed under scalar multiplication if any real multiple of any vector in $S$ is also in $S$. In other words, if $r$ is any real number and $(x_1,y_1,z_1)$ is in the subspace, then so is $(rx_1,ry_1,rz_1)$. For example, if we were to check this definition against problem 2, we would be asking whether it is true that, for any $r,x_1,y_1inmathbbR$, the vector $(rx_1,ry_2,rx_1y_1)$ is in the subset.
The set $s(1,0,0)+t(0,0,1)$ from problem 4 is the set of vectors that can be expressed in the form $s(1,0,0)+t(0,0,1)$ for some pair of real numbers $s,tinmathbbR$. A similar definition holds for problem 5.
(Also I don't follow your reasoning at all for 3.)
answered Aug 6 at 4:42
alphacapture
1,806420
add a comment |Â
up vote
1
down vote
Here are the definitions I think you are missing:
Closure under vector addition:
A subset $S$ of $mathbbR^3$ is closed under vector addition if the sum of any two vectors in $S$ is also in $S$. In other words, if $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are in the subspace, then so is $(x_1+x_2,y_1+y_2,z_1+z_2)$. For example, if we were to check this definition against problem 2, we would be asking whether it is true that, for any $x_1,y_1,x_2,y_2inmathbbR$, the vector $(x_1,y_2,x_1y_1)+(x_2,y_2,x_2y_2)=(x_1+x_2,y_1+y_2,x_1x_2+y_1y_2)$ is in the subset.
Closure under scalar multiplication:
A subset $S$ of $mathbbR^3$ is closed under scalar multiplication if any real multiple of any vector in $S$ is also in $S$. In other words, if $r$ is any real number and $(x_1,y_1,z_1)$ is in the subspace, then so is $(rx_1,ry_1,rz_1)$. For example, if we were to check this definition against problem 2, we would be asking whether it is true that, for any $r,x_1,y_1inmathbbR$, the vector $(rx_1,ry_2,rx_1y_1)$ is in the subset.
The set $s(1,0,0)+t(0,0,1)$ from problem 4 is the set of vectors that can be expressed in the form $s(1,0,0)+t(0,0,1)$ for some pair of real numbers $s,tinmathbbR$. A similar definition holds for problem 5.
(Also I don't follow your reasoning at all for 3.)
answered Aug 6 at 4:42
alphacapture
1,806420
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here are the definitions I think you are missing:
Closure under vector addition:
A subset $S$ of $mathbbR^3$ is closed under vector addition if the sum of any two vectors in $S$ is also in $S$. In other words, if $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are in the subspace, then so is $(x_1+x_2,y_1+y_2,z_1+z_2)$. For example, if we were to check this definition against problem 2, we would be asking whether it is true that, for any $x_1,y_1,x_2,y_2inmathbbR$, the vector $(x_1,y_2,x_1y_1)+(x_2,y_2,x_2y_2)=(x_1+x_2,y_1+y_2,x_1x_2+y_1y_2)$ is in the subset.
Closure under scalar multiplication:
A subset $S$ of $mathbbR^3$ is closed under scalar multiplication if any real multiple of any vector in $S$ is also in $S$. In other words, if $r$ is any real number and $(x_1,y_1,z_1)$ is in the subspace, then so is $(rx_1,ry_1,rz_1)$. For example, if we were to check this definition against problem 2, we would be asking whether it is true that, for any $r,x_1,y_1inmathbbR$, the vector $(rx_1,ry_2,rx_1y_1)$ is in the subset.
The set $s(1,0,0)+t(0,0,1)$ from problem 4 is the set of vectors that can be expressed in the form $s(1,0,0)+t(0,0,1)$ for some pair of real numbers $s,tinmathbbR$. A similar definition holds for problem 5.
(Also I don't follow your reasoning at all for 3.)
answered Aug 6 at 4:42
alphacapture
1,806420
Here are the definitions I think you are missing:
Closure under vector addition:
A subset $S$ of $mathbbR^3$ is closed under vector addition if the sum of any two vectors in $S$ is also in $S$. In other words, if $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are in the subspace, then so is $(x_1+x_2,y_1+y_2,z_1+z_2)$. For example, if we were to check this definition against problem 2, we would be asking whether it is true that, for any $x_1,y_1,x_2,y_2inmathbbR$, the vector $(x_1,y_2,x_1y_1)+(x_2,y_2,x_2y_2)=(x_1+x_2,y_1+y_2,x_1x_2+y_1y_2)$ is in the subset.
Closure under scalar multiplication:
A subset $S$ of $mathbbR^3$ is closed under scalar multiplication if any real multiple of any vector in $S$ is also in $S$. In other words, if $r$ is any real number and $(x_1,y_1,z_1)$ is in the subspace, then so is $(rx_1,ry_1,rz_1)$. For example, if we were to check this definition against problem 2, we would be asking whether it is true that, for any $r,x_1,y_1inmathbbR$, the vector $(rx_1,ry_2,rx_1y_1)$ is in the subset.
The set $s(1,0,0)+t(0,0,1)$ from problem 4 is the set of vectors that can be expressed in the form $s(1,0,0)+t(0,0,1)$ for some pair of real numbers $s,tinmathbbR$. A similar definition holds for problem 5.
(Also I don't follow your reasoning at all for 3.)
answered Aug 6 at 4:42
alphacapture
1,806420
answered Aug 6 at 4:42
alphacapture
1,806420
answered Aug 6 at 4:42
alphacapture
1,806420
answered Aug 6 at 4:42
alphacapture
1,806420
1,806420
add a comment |Â
add a comment |Â
up vote
1
down vote
Guide:
For part $4$, note that
$U_4=operatornameSpan (1,0,0), (0,0,1)$, it is written in the form of span of elements of $mathbbR^3$ which is closed under addition and scalar multiplication. Hence it is a subspace. I will leave part $5$ as an exercise.
Alternatively, let me prove $U_4$ is a subspace by verifying it is closed under additon and scalar multiplicaiton explicitly.
Closed under addition:
Let $x in U_4$, $exists s_x, t_x$ such that $x=s_x(1,0,0)+t_x(0,0,1)$ . Let $y in U_4$, $exists s_y, t_y$ such that $y=s_y(1,0,0)+t_y(0,0,1)$, then $x+y = (s_x+s_y)(1,0,0)+(s_y+t_y)(0,0,1)$ but we have $s_x+s_y, t_x+t_y in mathbbR$, hence $x+y in U_4$.
Closed under scalar multiplication, let $c in mathbbR$, $cx = (cs_x)(1,0,0)+(ct_x)(0,0,1)$ but we have $cs_x, ct_x in mathbbR$, hence $cx in U_4$.
Checking whether the zero vector is in is not sufficient.
Try to exhibit counter examples for part $2,3,6$ to prove that they are either not closed under addition or scalar multiplication. For example, for part $2$, $(1,1,1) in U_2$, what about $frac12 (1,1,1)$, is it in $U_2$?
answered Aug 6 at 4:41
Siong Thye Goh
78k134997
add a comment |Â
up vote
1
down vote
Guide:
For part $4$, note that
$U_4=operatornameSpan (1,0,0), (0,0,1)$, it is written in the form of span of elements of $mathbbR^3$ which is closed under addition and scalar multiplication. Hence it is a subspace. I will leave part $5$ as an exercise.
Alternatively, let me prove $U_4$ is a subspace by verifying it is closed under additon and scalar multiplicaiton explicitly.
Closed under addition:
Let $x in U_4$, $exists s_x, t_x$ such that $x=s_x(1,0,0)+t_x(0,0,1)$ . Let $y in U_4$, $exists s_y, t_y$ such that $y=s_y(1,0,0)+t_y(0,0,1)$, then $x+y = (s_x+s_y)(1,0,0)+(s_y+t_y)(0,0,1)$ but we have $s_x+s_y, t_x+t_y in mathbbR$, hence $x+y in U_4$.
Closed under scalar multiplication, let $c in mathbbR$, $cx = (cs_x)(1,0,0)+(ct_x)(0,0,1)$ but we have $cs_x, ct_x in mathbbR$, hence $cx in U_4$.
Checking whether the zero vector is in is not sufficient.
Try to exhibit counter examples for part $2,3,6$ to prove that they are either not closed under addition or scalar multiplication. For example, for part $2$, $(1,1,1) in U_2$, what about $frac12 (1,1,1)$, is it in $U_2$?
answered Aug 6 at 4:41
Siong Thye Goh
78k134997
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Guide:
For part $4$, note that
$U_4=operatornameSpan (1,0,0), (0,0,1)$, it is written in the form of span of elements of $mathbbR^3$ which is closed under addition and scalar multiplication. Hence it is a subspace. I will leave part $5$ as an exercise.
Alternatively, let me prove $U_4$ is a subspace by verifying it is closed under additon and scalar multiplicaiton explicitly.
Closed under addition:
Let $x in U_4$, $exists s_x, t_x$ such that $x=s_x(1,0,0)+t_x(0,0,1)$ . Let $y in U_4$, $exists s_y, t_y$ such that $y=s_y(1,0,0)+t_y(0,0,1)$, then $x+y = (s_x+s_y)(1,0,0)+(s_y+t_y)(0,0,1)$ but we have $s_x+s_y, t_x+t_y in mathbbR$, hence $x+y in U_4$.
Closed under scalar multiplication, let $c in mathbbR$, $cx = (cs_x)(1,0,0)+(ct_x)(0,0,1)$ but we have $cs_x, ct_x in mathbbR$, hence $cx in U_4$.
Checking whether the zero vector is in is not sufficient.
Try to exhibit counter examples for part $2,3,6$ to prove that they are either not closed under addition or scalar multiplication. For example, for part $2$, $(1,1,1) in U_2$, what about $frac12 (1,1,1)$, is it in $U_2$?
answered Aug 6 at 4:41
Siong Thye Goh
78k134997
Guide:
For part $4$, note that
$U_4=operatornameSpan (1,0,0), (0,0,1)$, it is written in the form of span of elements of $mathbbR^3$ which is closed under addition and scalar multiplication. Hence it is a subspace. I will leave part $5$ as an exercise.
Alternatively, let me prove $U_4$ is a subspace by verifying it is closed under additon and scalar multiplicaiton explicitly.
Closed under addition:
Let $x in U_4$, $exists s_x, t_x$ such that $x=s_x(1,0,0)+t_x(0,0,1)$ . Let $y in U_4$, $exists s_y, t_y$ such that $y=s_y(1,0,0)+t_y(0,0,1)$, then $x+y = (s_x+s_y)(1,0,0)+(s_y+t_y)(0,0,1)$ but we have $s_x+s_y, t_x+t_y in mathbbR$, hence $x+y in U_4$.
Closed under scalar multiplication, let $c in mathbbR$, $cx = (cs_x)(1,0,0)+(ct_x)(0,0,1)$ but we have $cs_x, ct_x in mathbbR$, hence $cx in U_4$.
Checking whether the zero vector is in is not sufficient.
Try to exhibit counter examples for part $2,3,6$ to prove that they are either not closed under addition or scalar multiplication. For example, for part $2$, $(1,1,1) in U_2$, what about $frac12 (1,1,1)$, is it in $U_2$?
answered Aug 6 at 4:41
Siong Thye Goh
78k134997
edited Aug 6 at 4:52
answered Aug 6 at 4:41
Siong Thye Goh
78k134997
answered Aug 6 at 4:41
Siong Thye Goh
78k134997
answered Aug 6 at 4:41
Siong Thye Goh
78k134997
78k134997
add a comment |Â
add a comment |Â
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