Proving a set of vectors are a basis
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Question Let A $subseteq {mathbb R}^{3}$ be the plane ${(x,y,z) in mathbb R^3 : 2x+3y-4z = 0}$. Prove that $B = {(-6,4,0),(5,-2,1)}$ is a basis of the real vector space $A$.
So I have set $u = (a,b,c) in A$
such that $a = frac{4c-2b}{2}$, then using this
$u$ is now $(frac{4c-2b}{2},b,c)$
$$ u = c (2,0,1)+ b(frac{-3}{2},1,0)= c (2,0,1)+4b(-6,4,0)$$
I'm uncertain how to show the combination of both vectors span $A$.
linear-algebra
edited Dec 21 '18 at 19:04
hardmath
28.8k95296
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
465
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add a comment |
$begingroup$
Question Let A $subseteq {mathbb R}^{3}$ be the plane ${(x,y,z) in mathbb R^3 : 2x+3y-4z = 0}$. Prove that $B = {(-6,4,0),(5,-2,1)}$ is a basis of the real vector space $A$.
So I have set $u = (a,b,c) in A$
such that $a = frac{4c-2b}{2}$, then using this
$u$ is now $(frac{4c-2b}{2},b,c)$
$$ u = c (2,0,1)+ b(frac{-3}{2},1,0)= c (2,0,1)+4b(-6,4,0)$$
I'm uncertain how to show the combination of both vectors span $A$.
linear-algebra
edited Dec 21 '18 at 19:04
hardmath
28.8k95296
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
465
$endgroup$
add a comment |
$begingroup$
Question Let A $subseteq {mathbb R}^{3}$ be the plane ${(x,y,z) in mathbb R^3 : 2x+3y-4z = 0}$. Prove that $B = {(-6,4,0),(5,-2,1)}$ is a basis of the real vector space $A$.
So I have set $u = (a,b,c) in A$
such that $a = frac{4c-2b}{2}$, then using this
$u$ is now $(frac{4c-2b}{2},b,c)$
$$ u = c (2,0,1)+ b(frac{-3}{2},1,0)= c (2,0,1)+4b(-6,4,0)$$
I'm uncertain how to show the combination of both vectors span $A$.
linear-algebra
edited Dec 21 '18 at 19:04
hardmath
28.8k95296
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
465
$endgroup$
Question Let A $subseteq {mathbb R}^{3}$ be the plane ${(x,y,z) in mathbb R^3 : 2x+3y-4z = 0}$. Prove that $B = {(-6,4,0),(5,-2,1)}$ is a basis of the real vector space $A$.
So I have set $u = (a,b,c) in A$
such that $a = frac{4c-2b}{2}$, then using this
$u$ is now $(frac{4c-2b}{2},b,c)$
$$ u = c (2,0,1)+ b(frac{-3}{2},1,0)= c (2,0,1)+4b(-6,4,0)$$
I'm uncertain how to show the combination of both vectors span $A$.
linear-algebra
linear-algebra
edited Dec 21 '18 at 19:04
hardmath
28.8k95296
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
465
edited Dec 21 '18 at 19:04
hardmath
28.8k95296
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
465
edited Dec 21 '18 at 19:04
hardmath
28.8k95296
edited Dec 21 '18 at 19:04
hardmath
28.8k95296
edited Dec 21 '18 at 19:04
hardmath
28.8k95296
28.8k95296
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
465
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
465
asked Dec 15 '18 at 15:52
Rito LoweRito Lowe
465
465
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free.
To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent.
answered Dec 15 '18 at 15:56
user3482749user3482749
3,714417
$endgroup$
add a comment |
$begingroup$
You mean $a = frac{4c-3b}2$.
If you let $b=4, c=0$, then $a=frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.
If you let $b=-2, c=1$, then $a=frac{4(1)-3(-2)}2=5$, that is how you get the second vector.
Since they are not multiple of each other and you know the dimension is $2$, it is a basis.
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5168
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free.
To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent.
answered Dec 15 '18 at 15:56
user3482749user3482749
3,714417
$endgroup$
add a comment |
$begingroup$
First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free.
To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent.
answered Dec 15 '18 at 15:56
user3482749user3482749
3,714417
$endgroup$
add a comment |
$begingroup$
First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free.
To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent.
answered Dec 15 '18 at 15:56
user3482749user3482749
3,714417
$endgroup$
First, note that both vectors are in $A$. Second, note that $A$ has codimension 1, so dimension 2, so any two linearly independent vectors in $A$ form a basis of it (if that made no sense to you, simply note that it's not all of $mathbb{R}^3$, so has dimension at most 2). It's then reasonably easy to check that the vectors in question are linearly independent, so the basis theorem gives that they span $A$ for free.
To actually do that check: if $a(-6,4,0) + b(5,-2,1) = (0,0,0)$, then $(5b-6a,4a-2b,b)=(0,0,0)$. The last coordinate then gives us that $b = 0$, and either of the first two gives us that $a = 0$. Thus, $B$ is linearly independent.
answered Dec 15 '18 at 15:56
user3482749user3482749
3,714417
answered Dec 15 '18 at 15:56
user3482749user3482749
3,714417
answered Dec 15 '18 at 15:56
user3482749user3482749
3,714417
answered Dec 15 '18 at 15:56
user3482749user3482749
3,714417
3,714417
add a comment |
add a comment |
$begingroup$
You mean $a = frac{4c-3b}2$.
If you let $b=4, c=0$, then $a=frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.
If you let $b=-2, c=1$, then $a=frac{4(1)-3(-2)}2=5$, that is how you get the second vector.
Since they are not multiple of each other and you know the dimension is $2$, it is a basis.
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
You mean $a = frac{4c-3b}2$.
If you let $b=4, c=0$, then $a=frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.
If you let $b=-2, c=1$, then $a=frac{4(1)-3(-2)}2=5$, that is how you get the second vector.
Since they are not multiple of each other and you know the dimension is $2$, it is a basis.
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
You mean $a = frac{4c-3b}2$.
If you let $b=4, c=0$, then $a=frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.
If you let $b=-2, c=1$, then $a=frac{4(1)-3(-2)}2=5$, that is how you get the second vector.
Since they are not multiple of each other and you know the dimension is $2$, it is a basis.
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
You mean $a = frac{4c-3b}2$.
If you let $b=4, c=0$, then $a=frac{4(0)-3(4)}2=-6$, hence you obtain the first vector.
If you let $b=-2, c=1$, then $a=frac{4(1)-3(-2)}2=5$, that is how you get the second vector.
Since they are not multiple of each other and you know the dimension is $2$, it is a basis.
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
100k1465117
answered Dec 15 '18 at 15:57
Siong Thye GohSiong Thye Goh
100k1465117
100k1465117
add a comment |
add a comment |
$begingroup$
Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5168
$endgroup$
add a comment |
$begingroup$
Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5168
$endgroup$
add a comment |
$begingroup$
Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5168
$endgroup$
Both the vectors would span $A$ if they are perpendicular to its normal. The normal is $(2,3,-4)$ and the dot products $(-6,4,0).(2,3,-4)$ and $(5,−2,1).(2,3,-4)$ are $0$. Hence, they are two non-parallel vectors in the plane $A$ any combination of which in the form $a(-6,4,0)+b(5,−2,1)$ can get one to any point in the plane $A$.
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5168
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5168
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5168
answered Dec 15 '18 at 16:01
Sameer BahetiSameer Baheti
5168
5168
add a comment |
add a comment |
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