How do determine whether a subset is a subspace?
$begingroup$
So I have
begin{align*}
A&= {(x_1,x_2)'in mathbb{R}^2: max (|x_1|,|x_2|) leq 1} \
B &= {(x_1,x_2,x_3,x_4)'in mathbb{R}^4:x_1⋅x_2=0} \
C&={(x_1,x_2,x_3)' in mathbb{R}^3 :x_2-x_3-1=0 }
end{align*}
And i want to determine whether these subsets of $mathbb{R}^n$ are vector subspaces or not.
I know that $V subset mathbb{R}^n$ is a subspace when : the $0$ vector is in $V$, $V$ is closed under addition and $V$ is closed under multiplication .
I also learned that if $V$ is the set of solutions to homogenoeus linear equations , then $V$ is a subspace, and if $V$ is the span of some vectors , then $V$ is a subspace.
Since I have dyscalculia, I really do not know how to apply this information in order to check whether they are subspaces or not... They don't make sense to me unfortunately . Moroever, i don't know which of these concepts i should apply to which subset.
If anybody could help me and show me how these sets are subspaces or not , in the clearest and most simple way possible, that would be great ! Thank you !
matrices vector-spaces vectors matrix-equations
edited Dec 10 '18 at 16:16
Digitalis
530216
asked Dec 10 '18 at 15:43
BM97BM97
758
$endgroup$
add a comment |
$begingroup$
So I have
begin{align*}
A&= {(x_1,x_2)'in mathbb{R}^2: max (|x_1|,|x_2|) leq 1} \
B &= {(x_1,x_2,x_3,x_4)'in mathbb{R}^4:x_1⋅x_2=0} \
C&={(x_1,x_2,x_3)' in mathbb{R}^3 :x_2-x_3-1=0 }
end{align*}
And i want to determine whether these subsets of $mathbb{R}^n$ are vector subspaces or not.
I know that $V subset mathbb{R}^n$ is a subspace when : the $0$ vector is in $V$, $V$ is closed under addition and $V$ is closed under multiplication .
I also learned that if $V$ is the set of solutions to homogenoeus linear equations , then $V$ is a subspace, and if $V$ is the span of some vectors , then $V$ is a subspace.
Since I have dyscalculia, I really do not know how to apply this information in order to check whether they are subspaces or not... They don't make sense to me unfortunately . Moroever, i don't know which of these concepts i should apply to which subset.
If anybody could help me and show me how these sets are subspaces or not , in the clearest and most simple way possible, that would be great ! Thank you !
matrices vector-spaces vectors matrix-equations
edited Dec 10 '18 at 16:16
Digitalis
530216
asked Dec 10 '18 at 15:43
BM97BM97
758
$endgroup$
$begingroup$
"These are column vectors but i'm not sure how to put them in the correct format" You can do this with mathjax. You can find a basic tutorial math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Digitalis
Dec 10 '18 at 16:01
1
$begingroup$
All of these subsets are in fact are not subspace. So you need to find a counter example which shows that one of the requirement is not satisfied. For example for set $A$ notice that $(1,1)in A$ (do you see why?). But if we scalar multiply this vector with say 5 we have $5(1,1)=(5,5)$ and this vector is clearly not in $A$. Hence $A$ is not a subspace.
$endgroup$
– user9077
Dec 10 '18 at 16:22
1
$begingroup$
I think "dyscalculia" is simply an excuse, and you will be able to improve in mathematics if you drop it. True mathematics is not something that only talented people can do. It always involves hard work and many failed attempts before one successful attempt. When you think a sentence of the logical form "For every x in S, something is true about x" is false, then try as many simple values of x in S that you can think of, instead of saying "I do not know how to apply this information".
$endgroup$
– user21820
Dec 24 '18 at 7:04
$begingroup$
And it occurs to me that you may not even get how what I said is relevant to your question. Closure of $V$ under addition is defined as "For every vectors $x,y$ in $V$, $x+y$ is also in $V$." If you are not familiar with this way of expressing mathematical statements, then you really need to learn basic (first-order) logic first before anything else.
$endgroup$
– user21820
Dec 24 '18 at 7:13
$begingroup$
I have dyscalculia. I would also consider myself a working mathematician (I am nearly done with my PhD and have a paper out from my masters work, and should have a couple more papers out by the end of next year). Dyscalculia is cannot be an excuse. It just means that you have a neurological processing disorder that forces you to slow down, take your time, and double check your work. Write things out very carefully. Keep the appropriate definitions in sight. Work slowly and double check frequently. Find ways to compensate for your condition.
$endgroup$
– Xander Henderson
Dec 24 '18 at 16:48
add a comment |
$begingroup$
So I have
begin{align*}
A&= {(x_1,x_2)'in mathbb{R}^2: max (|x_1|,|x_2|) leq 1} \
B &= {(x_1,x_2,x_3,x_4)'in mathbb{R}^4:x_1⋅x_2=0} \
C&={(x_1,x_2,x_3)' in mathbb{R}^3 :x_2-x_3-1=0 }
end{align*}
And i want to determine whether these subsets of $mathbb{R}^n$ are vector subspaces or not.
I know that $V subset mathbb{R}^n$ is a subspace when : the $0$ vector is in $V$, $V$ is closed under addition and $V$ is closed under multiplication .
I also learned that if $V$ is the set of solutions to homogenoeus linear equations , then $V$ is a subspace, and if $V$ is the span of some vectors , then $V$ is a subspace.
Since I have dyscalculia, I really do not know how to apply this information in order to check whether they are subspaces or not... They don't make sense to me unfortunately . Moroever, i don't know which of these concepts i should apply to which subset.
If anybody could help me and show me how these sets are subspaces or not , in the clearest and most simple way possible, that would be great ! Thank you !
matrices vector-spaces vectors matrix-equations
edited Dec 10 '18 at 16:16
Digitalis
530216
asked Dec 10 '18 at 15:43
BM97BM97
758
$endgroup$
So I have
begin{align*}
A&= {(x_1,x_2)'in mathbb{R}^2: max (|x_1|,|x_2|) leq 1} \
B &= {(x_1,x_2,x_3,x_4)'in mathbb{R}^4:x_1⋅x_2=0} \
C&={(x_1,x_2,x_3)' in mathbb{R}^3 :x_2-x_3-1=0 }
end{align*}
And i want to determine whether these subsets of $mathbb{R}^n$ are vector subspaces or not.
I know that $V subset mathbb{R}^n$ is a subspace when : the $0$ vector is in $V$, $V$ is closed under addition and $V$ is closed under multiplication .
I also learned that if $V$ is the set of solutions to homogenoeus linear equations , then $V$ is a subspace, and if $V$ is the span of some vectors , then $V$ is a subspace.
Since I have dyscalculia, I really do not know how to apply this information in order to check whether they are subspaces or not... They don't make sense to me unfortunately . Moroever, i don't know which of these concepts i should apply to which subset.
If anybody could help me and show me how these sets are subspaces or not , in the clearest and most simple way possible, that would be great ! Thank you !
matrices vector-spaces vectors matrix-equations
matrices vector-spaces vectors matrix-equations
edited Dec 10 '18 at 16:16
Digitalis
530216
asked Dec 10 '18 at 15:43
BM97BM97
758
edited Dec 10 '18 at 16:16
Digitalis
530216
asked Dec 10 '18 at 15:43
BM97BM97
758
edited Dec 10 '18 at 16:16
Digitalis
530216
edited Dec 10 '18 at 16:16
Digitalis
530216
edited Dec 10 '18 at 16:16
Digitalis
530216
530216
asked Dec 10 '18 at 15:43
BM97BM97
758
asked Dec 10 '18 at 15:43
BM97BM97
758
asked Dec 10 '18 at 15:43
BM97BM97
758
758
$begingroup$
"These are column vectors but i'm not sure how to put them in the correct format" You can do this with mathjax. You can find a basic tutorial math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Digitalis
Dec 10 '18 at 16:01
1
$begingroup$
All of these subsets are in fact are not subspace. So you need to find a counter example which shows that one of the requirement is not satisfied. For example for set $A$ notice that $(1,1)in A$ (do you see why?). But if we scalar multiply this vector with say 5 we have $5(1,1)=(5,5)$ and this vector is clearly not in $A$. Hence $A$ is not a subspace.
$endgroup$
– user9077
Dec 10 '18 at 16:22
1
$begingroup$
I think "dyscalculia" is simply an excuse, and you will be able to improve in mathematics if you drop it. True mathematics is not something that only talented people can do. It always involves hard work and many failed attempts before one successful attempt. When you think a sentence of the logical form "For every x in S, something is true about x" is false, then try as many simple values of x in S that you can think of, instead of saying "I do not know how to apply this information".
$endgroup$
– user21820
Dec 24 '18 at 7:04
$begingroup$
And it occurs to me that you may not even get how what I said is relevant to your question. Closure of $V$ under addition is defined as "For every vectors $x,y$ in $V$, $x+y$ is also in $V$." If you are not familiar with this way of expressing mathematical statements, then you really need to learn basic (first-order) logic first before anything else.
$endgroup$
– user21820
Dec 24 '18 at 7:13
$begingroup$
I have dyscalculia. I would also consider myself a working mathematician (I am nearly done with my PhD and have a paper out from my masters work, and should have a couple more papers out by the end of next year). Dyscalculia is cannot be an excuse. It just means that you have a neurological processing disorder that forces you to slow down, take your time, and double check your work. Write things out very carefully. Keep the appropriate definitions in sight. Work slowly and double check frequently. Find ways to compensate for your condition.
$endgroup$
– Xander Henderson
Dec 24 '18 at 16:48
add a comment |
$begingroup$
"These are column vectors but i'm not sure how to put them in the correct format" You can do this with mathjax. You can find a basic tutorial math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Digitalis
Dec 10 '18 at 16:01
1
$begingroup$
All of these subsets are in fact are not subspace. So you need to find a counter example which shows that one of the requirement is not satisfied. For example for set $A$ notice that $(1,1)in A$ (do you see why?). But if we scalar multiply this vector with say 5 we have $5(1,1)=(5,5)$ and this vector is clearly not in $A$. Hence $A$ is not a subspace.
$endgroup$
– user9077
Dec 10 '18 at 16:22
1
$begingroup$
I think "dyscalculia" is simply an excuse, and you will be able to improve in mathematics if you drop it. True mathematics is not something that only talented people can do. It always involves hard work and many failed attempts before one successful attempt. When you think a sentence of the logical form "For every x in S, something is true about x" is false, then try as many simple values of x in S that you can think of, instead of saying "I do not know how to apply this information".
$endgroup$
– user21820
Dec 24 '18 at 7:04
$begingroup$
And it occurs to me that you may not even get how what I said is relevant to your question. Closure of $V$ under addition is defined as "For every vectors $x,y$ in $V$, $x+y$ is also in $V$." If you are not familiar with this way of expressing mathematical statements, then you really need to learn basic (first-order) logic first before anything else.
$endgroup$
– user21820
Dec 24 '18 at 7:13
$begingroup$
I have dyscalculia. I would also consider myself a working mathematician (I am nearly done with my PhD and have a paper out from my masters work, and should have a couple more papers out by the end of next year). Dyscalculia is cannot be an excuse. It just means that you have a neurological processing disorder that forces you to slow down, take your time, and double check your work. Write things out very carefully. Keep the appropriate definitions in sight. Work slowly and double check frequently. Find ways to compensate for your condition.
$endgroup$
– Xander Henderson
Dec 24 '18 at 16:48
$begingroup$
"These are column vectors but i'm not sure how to put them in the correct format" You can do this with mathjax. You can find a basic tutorial math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Digitalis
Dec 10 '18 at 16:01
$begingroup$
"These are column vectors but i'm not sure how to put them in the correct format" You can do this with mathjax. You can find a basic tutorial math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Digitalis
Dec 10 '18 at 16:01
1
1
$begingroup$
All of these subsets are in fact are not subspace. So you need to find a counter example which shows that one of the requirement is not satisfied. For example for set $A$ notice that $(1,1)in A$ (do you see why?). But if we scalar multiply this vector with say 5 we have $5(1,1)=(5,5)$ and this vector is clearly not in $A$. Hence $A$ is not a subspace.
$endgroup$
– user9077
Dec 10 '18 at 16:22
$begingroup$
All of these subsets are in fact are not subspace. So you need to find a counter example which shows that one of the requirement is not satisfied. For example for set $A$ notice that $(1,1)in A$ (do you see why?). But if we scalar multiply this vector with say 5 we have $5(1,1)=(5,5)$ and this vector is clearly not in $A$. Hence $A$ is not a subspace.
$endgroup$
– user9077
Dec 10 '18 at 16:22
1
1
$begingroup$
I think "dyscalculia" is simply an excuse, and you will be able to improve in mathematics if you drop it. True mathematics is not something that only talented people can do. It always involves hard work and many failed attempts before one successful attempt. When you think a sentence of the logical form "For every x in S, something is true about x" is false, then try as many simple values of x in S that you can think of, instead of saying "I do not know how to apply this information".
$endgroup$
– user21820
Dec 24 '18 at 7:04
$begingroup$
I think "dyscalculia" is simply an excuse, and you will be able to improve in mathematics if you drop it. True mathematics is not something that only talented people can do. It always involves hard work and many failed attempts before one successful attempt. When you think a sentence of the logical form "For every x in S, something is true about x" is false, then try as many simple values of x in S that you can think of, instead of saying "I do not know how to apply this information".
$endgroup$
– user21820
Dec 24 '18 at 7:04
$begingroup$
And it occurs to me that you may not even get how what I said is relevant to your question. Closure of $V$ under addition is defined as "For every vectors $x,y$ in $V$, $x+y$ is also in $V$." If you are not familiar with this way of expressing mathematical statements, then you really need to learn basic (first-order) logic first before anything else.
$endgroup$
– user21820
Dec 24 '18 at 7:13
$begingroup$
And it occurs to me that you may not even get how what I said is relevant to your question. Closure of $V$ under addition is defined as "For every vectors $x,y$ in $V$, $x+y$ is also in $V$." If you are not familiar with this way of expressing mathematical statements, then you really need to learn basic (first-order) logic first before anything else.
$endgroup$
– user21820
Dec 24 '18 at 7:13
$begingroup$
I have dyscalculia. I would also consider myself a working mathematician (I am nearly done with my PhD and have a paper out from my masters work, and should have a couple more papers out by the end of next year). Dyscalculia is cannot be an excuse. It just means that you have a neurological processing disorder that forces you to slow down, take your time, and double check your work. Write things out very carefully. Keep the appropriate definitions in sight. Work slowly and double check frequently. Find ways to compensate for your condition.
$endgroup$
– Xander Henderson
Dec 24 '18 at 16:48
$begingroup$
I have dyscalculia. I would also consider myself a working mathematician (I am nearly done with my PhD and have a paper out from my masters work, and should have a couple more papers out by the end of next year). Dyscalculia is cannot be an excuse. It just means that you have a neurological processing disorder that forces you to slow down, take your time, and double check your work. Write things out very carefully. Keep the appropriate definitions in sight. Work slowly and double check frequently. Find ways to compensate for your condition.
$endgroup$
– Xander Henderson
Dec 24 '18 at 16:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$A$ fails to be a subspace because it is not closed under scalar multiplication. For example, $(1,1) in A$ but $2(1,1) =(2,2) notin A$.
$B$ fails to be a subspace because it is not closed under addition. For example, $(0,1,0,0) in B$ and $(1,0,0,0) in B$ but $(0,1,0,0) + (1,0,0,0) = (1,1,0,0) notin B$.
$C$ fails to be a subspace because $(0,0,0) notin C$.
answered Dec 10 '18 at 16:39
gandalf61gandalf61
8,761725
$endgroup$
$begingroup$
For B , could i use the two vectors (1,0,0,0) and (1,0,0,0) ? meaning, can i use whichver vector as long as x1 times x2 is 0?
$endgroup$
– BM97
Dec 10 '18 at 16:43
$begingroup$
@BM97 Yes, but $(1,0,0,0)+(1,0,0,0) = (2,0,0,0) in B$ so this doesn't work as a counter example.
$endgroup$
– gandalf61
Dec 10 '18 at 16:58
add a comment |
$begingroup$
$A$ is clearly not a subspace since $(1,0) in A $ but $(1,0) + (1,0) = (2,0) notin A.$
$C$ is not a subspace since $(x_1,x_2,x_3) = (0,0,0)$ does not verify $x_2 - x_3 - 1 = 0.$
For $B$ we must check
$0 in B$ ? Yes ! $(x_1,x_2,x_3) = (0,0,0)$ does verify $x_1x_2 = 0.$
The sum of two vectors of $B$ is in $B$ ? $forall x_1,x_2,x_3,x_4, x_1',x_2',x_3',x_4' in mathbb{R}:$
$$(x_1,x_2,x_3,x_4) + (x_1',x_2',x_3',x_4') = (x_1+ x_1',x_2+ x_2',x_3+ x_3',x_4') $$
$(x_1+ x_1')(x_2+ x_2') = x_1x_2 + x_1'x_2 + x_1x_2' + x_1'x_2' = 0 + x_1'x_2 + x_1x_2' + 0 = x_1'x_2 + x_1x_2' neq 0$
So $B$ is not a subspace.
answered Dec 10 '18 at 16:31
DigitalisDigitalis
530216
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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$begingroup$
$A$ fails to be a subspace because it is not closed under scalar multiplication. For example, $(1,1) in A$ but $2(1,1) =(2,2) notin A$.
$B$ fails to be a subspace because it is not closed under addition. For example, $(0,1,0,0) in B$ and $(1,0,0,0) in B$ but $(0,1,0,0) + (1,0,0,0) = (1,1,0,0) notin B$.
$C$ fails to be a subspace because $(0,0,0) notin C$.
answered Dec 10 '18 at 16:39
gandalf61gandalf61
8,761725
$endgroup$
$begingroup$
For B , could i use the two vectors (1,0,0,0) and (1,0,0,0) ? meaning, can i use whichver vector as long as x1 times x2 is 0?
$endgroup$
– BM97
Dec 10 '18 at 16:43
$begingroup$
@BM97 Yes, but $(1,0,0,0)+(1,0,0,0) = (2,0,0,0) in B$ so this doesn't work as a counter example.
$endgroup$
– gandalf61
Dec 10 '18 at 16:58
add a comment |
$begingroup$
$A$ fails to be a subspace because it is not closed under scalar multiplication. For example, $(1,1) in A$ but $2(1,1) =(2,2) notin A$.
$B$ fails to be a subspace because it is not closed under addition. For example, $(0,1,0,0) in B$ and $(1,0,0,0) in B$ but $(0,1,0,0) + (1,0,0,0) = (1,1,0,0) notin B$.
$C$ fails to be a subspace because $(0,0,0) notin C$.
answered Dec 10 '18 at 16:39
gandalf61gandalf61
8,761725
$endgroup$
$begingroup$
For B , could i use the two vectors (1,0,0,0) and (1,0,0,0) ? meaning, can i use whichver vector as long as x1 times x2 is 0?
$endgroup$
– BM97
Dec 10 '18 at 16:43
$begingroup$
@BM97 Yes, but $(1,0,0,0)+(1,0,0,0) = (2,0,0,0) in B$ so this doesn't work as a counter example.
$endgroup$
– gandalf61
Dec 10 '18 at 16:58
add a comment |
$begingroup$
$A$ fails to be a subspace because it is not closed under scalar multiplication. For example, $(1,1) in A$ but $2(1,1) =(2,2) notin A$.
$B$ fails to be a subspace because it is not closed under addition. For example, $(0,1,0,0) in B$ and $(1,0,0,0) in B$ but $(0,1,0,0) + (1,0,0,0) = (1,1,0,0) notin B$.
$C$ fails to be a subspace because $(0,0,0) notin C$.
answered Dec 10 '18 at 16:39
gandalf61gandalf61
8,761725
$endgroup$
$A$ fails to be a subspace because it is not closed under scalar multiplication. For example, $(1,1) in A$ but $2(1,1) =(2,2) notin A$.
$B$ fails to be a subspace because it is not closed under addition. For example, $(0,1,0,0) in B$ and $(1,0,0,0) in B$ but $(0,1,0,0) + (1,0,0,0) = (1,1,0,0) notin B$.
$C$ fails to be a subspace because $(0,0,0) notin C$.
answered Dec 10 '18 at 16:39
gandalf61gandalf61
8,761725
answered Dec 10 '18 at 16:39
gandalf61gandalf61
8,761725
answered Dec 10 '18 at 16:39
gandalf61gandalf61
8,761725
answered Dec 10 '18 at 16:39
gandalf61gandalf61
8,761725
8,761725
$begingroup$
For B , could i use the two vectors (1,0,0,0) and (1,0,0,0) ? meaning, can i use whichver vector as long as x1 times x2 is 0?
$endgroup$
– BM97
Dec 10 '18 at 16:43
$begingroup$
@BM97 Yes, but $(1,0,0,0)+(1,0,0,0) = (2,0,0,0) in B$ so this doesn't work as a counter example.
$endgroup$
– gandalf61
Dec 10 '18 at 16:58
add a comment |
$begingroup$
For B , could i use the two vectors (1,0,0,0) and (1,0,0,0) ? meaning, can i use whichver vector as long as x1 times x2 is 0?
$endgroup$
– BM97
Dec 10 '18 at 16:43
$begingroup$
@BM97 Yes, but $(1,0,0,0)+(1,0,0,0) = (2,0,0,0) in B$ so this doesn't work as a counter example.
$endgroup$
– gandalf61
Dec 10 '18 at 16:58
$begingroup$
For B , could i use the two vectors (1,0,0,0) and (1,0,0,0) ? meaning, can i use whichver vector as long as x1 times x2 is 0?
$endgroup$
– BM97
Dec 10 '18 at 16:43
$begingroup$
For B , could i use the two vectors (1,0,0,0) and (1,0,0,0) ? meaning, can i use whichver vector as long as x1 times x2 is 0?
$endgroup$
– BM97
Dec 10 '18 at 16:43
$begingroup$
@BM97 Yes, but $(1,0,0,0)+(1,0,0,0) = (2,0,0,0) in B$ so this doesn't work as a counter example.
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– gandalf61
Dec 10 '18 at 16:58
$begingroup$
@BM97 Yes, but $(1,0,0,0)+(1,0,0,0) = (2,0,0,0) in B$ so this doesn't work as a counter example.
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– gandalf61
Dec 10 '18 at 16:58
add a comment |
$begingroup$
$A$ is clearly not a subspace since $(1,0) in A $ but $(1,0) + (1,0) = (2,0) notin A.$
$C$ is not a subspace since $(x_1,x_2,x_3) = (0,0,0)$ does not verify $x_2 - x_3 - 1 = 0.$
For $B$ we must check
$0 in B$ ? Yes ! $(x_1,x_2,x_3) = (0,0,0)$ does verify $x_1x_2 = 0.$
The sum of two vectors of $B$ is in $B$ ? $forall x_1,x_2,x_3,x_4, x_1',x_2',x_3',x_4' in mathbb{R}:$
$$(x_1,x_2,x_3,x_4) + (x_1',x_2',x_3',x_4') = (x_1+ x_1',x_2+ x_2',x_3+ x_3',x_4') $$
$(x_1+ x_1')(x_2+ x_2') = x_1x_2 + x_1'x_2 + x_1x_2' + x_1'x_2' = 0 + x_1'x_2 + x_1x_2' + 0 = x_1'x_2 + x_1x_2' neq 0$
So $B$ is not a subspace.
answered Dec 10 '18 at 16:31
DigitalisDigitalis
530216
$endgroup$
add a comment |
$begingroup$
$A$ is clearly not a subspace since $(1,0) in A $ but $(1,0) + (1,0) = (2,0) notin A.$
$C$ is not a subspace since $(x_1,x_2,x_3) = (0,0,0)$ does not verify $x_2 - x_3 - 1 = 0.$
For $B$ we must check
$0 in B$ ? Yes ! $(x_1,x_2,x_3) = (0,0,0)$ does verify $x_1x_2 = 0.$
The sum of two vectors of $B$ is in $B$ ? $forall x_1,x_2,x_3,x_4, x_1',x_2',x_3',x_4' in mathbb{R}:$
$$(x_1,x_2,x_3,x_4) + (x_1',x_2',x_3',x_4') = (x_1+ x_1',x_2+ x_2',x_3+ x_3',x_4') $$
$(x_1+ x_1')(x_2+ x_2') = x_1x_2 + x_1'x_2 + x_1x_2' + x_1'x_2' = 0 + x_1'x_2 + x_1x_2' + 0 = x_1'x_2 + x_1x_2' neq 0$
So $B$ is not a subspace.
answered Dec 10 '18 at 16:31
DigitalisDigitalis
530216
$endgroup$
add a comment |
$begingroup$
$A$ is clearly not a subspace since $(1,0) in A $ but $(1,0) + (1,0) = (2,0) notin A.$
$C$ is not a subspace since $(x_1,x_2,x_3) = (0,0,0)$ does not verify $x_2 - x_3 - 1 = 0.$
For $B$ we must check
$0 in B$ ? Yes ! $(x_1,x_2,x_3) = (0,0,0)$ does verify $x_1x_2 = 0.$
The sum of two vectors of $B$ is in $B$ ? $forall x_1,x_2,x_3,x_4, x_1',x_2',x_3',x_4' in mathbb{R}:$
$$(x_1,x_2,x_3,x_4) + (x_1',x_2',x_3',x_4') = (x_1+ x_1',x_2+ x_2',x_3+ x_3',x_4') $$
$(x_1+ x_1')(x_2+ x_2') = x_1x_2 + x_1'x_2 + x_1x_2' + x_1'x_2' = 0 + x_1'x_2 + x_1x_2' + 0 = x_1'x_2 + x_1x_2' neq 0$
So $B$ is not a subspace.
answered Dec 10 '18 at 16:31
DigitalisDigitalis
530216
$endgroup$
$A$ is clearly not a subspace since $(1,0) in A $ but $(1,0) + (1,0) = (2,0) notin A.$
$C$ is not a subspace since $(x_1,x_2,x_3) = (0,0,0)$ does not verify $x_2 - x_3 - 1 = 0.$
For $B$ we must check
$0 in B$ ? Yes ! $(x_1,x_2,x_3) = (0,0,0)$ does verify $x_1x_2 = 0.$
The sum of two vectors of $B$ is in $B$ ? $forall x_1,x_2,x_3,x_4, x_1',x_2',x_3',x_4' in mathbb{R}:$
$$(x_1,x_2,x_3,x_4) + (x_1',x_2',x_3',x_4') = (x_1+ x_1',x_2+ x_2',x_3+ x_3',x_4') $$
$(x_1+ x_1')(x_2+ x_2') = x_1x_2 + x_1'x_2 + x_1x_2' + x_1'x_2' = 0 + x_1'x_2 + x_1x_2' + 0 = x_1'x_2 + x_1x_2' neq 0$
So $B$ is not a subspace.
answered Dec 10 '18 at 16:31
DigitalisDigitalis
530216
edited Dec 10 '18 at 16:41
answered Dec 10 '18 at 16:31
DigitalisDigitalis
530216
answered Dec 10 '18 at 16:31
DigitalisDigitalis
530216
answered Dec 10 '18 at 16:31
DigitalisDigitalis
530216
530216
add a comment |
add a comment |
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"These are column vectors but i'm not sure how to put them in the correct format" You can do this with mathjax. You can find a basic tutorial math.meta.stackexchange.com/questions/5020/…
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– Digitalis
Dec 10 '18 at 16:01
1
$begingroup$
All of these subsets are in fact are not subspace. So you need to find a counter example which shows that one of the requirement is not satisfied. For example for set $A$ notice that $(1,1)in A$ (do you see why?). But if we scalar multiply this vector with say 5 we have $5(1,1)=(5,5)$ and this vector is clearly not in $A$. Hence $A$ is not a subspace.
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– user9077
Dec 10 '18 at 16:22
1
$begingroup$
I think "dyscalculia" is simply an excuse, and you will be able to improve in mathematics if you drop it. True mathematics is not something that only talented people can do. It always involves hard work and many failed attempts before one successful attempt. When you think a sentence of the logical form "For every x in S, something is true about x" is false, then try as many simple values of x in S that you can think of, instead of saying "I do not know how to apply this information".
$endgroup$
– user21820
Dec 24 '18 at 7:04
$begingroup$
And it occurs to me that you may not even get how what I said is relevant to your question. Closure of $V$ under addition is defined as "For every vectors $x,y$ in $V$, $x+y$ is also in $V$." If you are not familiar with this way of expressing mathematical statements, then you really need to learn basic (first-order) logic first before anything else.
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– user21820
Dec 24 '18 at 7:13
$begingroup$
I have dyscalculia. I would also consider myself a working mathematician (I am nearly done with my PhD and have a paper out from my masters work, and should have a couple more papers out by the end of next year). Dyscalculia is cannot be an excuse. It just means that you have a neurological processing disorder that forces you to slow down, take your time, and double check your work. Write things out very carefully. Keep the appropriate definitions in sight. Work slowly and double check frequently. Find ways to compensate for your condition.
$endgroup$
– Xander Henderson
Dec 24 '18 at 16:48