Is Cross Product of two vectors a linear transformation? (Linear Algebra)
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- Background Information:
I am studying linear algebra. For this question, I understand the definition of a vector in $$R^3 => v =(x,y,z)$$, and I know that A linear transformation between two vector spaces V and W is a map
$$T: V->W $$such that the following hold:
$$T(v1+v2)=T(v1) + T(v2)$$ for any vectors v1 and v2 in V, and
$$T(av) = a T(v)$$ for any scalar alpha a.
I also know how to calculate the cross product between two vectors.
- Question:
Let a be a fixed vector in R3. Does T(x) = a × x define a linear transformation?
- My thoughts:
I don't understand how to show T(a + x) = T(x) + T(a) and T(ax) = aT(x) for a cross product. The fact that numbers are not given makes it confusing as well. How can I approach this problem and prove the cross product is a transformation?
vectors linear-transformations cross-product
asked Aug 11 at 3:51
Kourosh
381113
add a comment |Â
up vote
3
down vote
favorite
- Background Information:
I am studying linear algebra. For this question, I understand the definition of a vector in $$R^3 => v =(x,y,z)$$, and I know that A linear transformation between two vector spaces V and W is a map
$$T: V->W $$such that the following hold:
$$T(v1+v2)=T(v1) + T(v2)$$ for any vectors v1 and v2 in V, and
$$T(av) = a T(v)$$ for any scalar alpha a.
I also know how to calculate the cross product between two vectors.
- Question:
Let a be a fixed vector in R3. Does T(x) = a × x define a linear transformation?
- My thoughts:
I don't understand how to show T(a + x) = T(x) + T(a) and T(ax) = aT(x) for a cross product. The fact that numbers are not given makes it confusing as well. How can I approach this problem and prove the cross product is a transformation?
vectors linear-transformations cross-product
asked Aug 11 at 3:51
Kourosh
381113
1
Hint: what is $a times ( x+y)$? Use the properties!
– Sean Roberson
Aug 11 at 3:54
@SeanRoberson, thanks for the hint, It will be (a X x ) + (a X y), so what would the value of the cross product represent? Also, how can we do this without numbers?
– Kourosh
Aug 11 at 4:02
1
First, I recommend changing your scalar from "a" to $alpha$ since you also define "a" to be a fixed vector. You can see that $T(alpha x) = atimes alpha x$ and that $alpha T(x) = alpha (atimes x)$. If you can show that $ atimes alpha x = alpha (atimes x)$ you are done with that part. Now fill in that blank and write a suitable expression for $T(v_1+v_2)$, $T(v_1)$, and $T(v_2)$ then proceed again.
– JessicaK
Aug 11 at 4:15
@JessicaK, thanks for your help, so having T(αx)=a×αx => αT(x)=α(a×x), and then a×αx=α(a×x) will satisfy the homogeneity property? Or should I add more to it?
– Kourosh
Aug 11 at 5:58
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
- Background Information:
I am studying linear algebra. For this question, I understand the definition of a vector in $$R^3 => v =(x,y,z)$$, and I know that A linear transformation between two vector spaces V and W is a map
$$T: V->W $$such that the following hold:
$$T(v1+v2)=T(v1) + T(v2)$$ for any vectors v1 and v2 in V, and
$$T(av) = a T(v)$$ for any scalar alpha a.
I also know how to calculate the cross product between two vectors.
- Question:
Let a be a fixed vector in R3. Does T(x) = a × x define a linear transformation?
- My thoughts:
I don't understand how to show T(a + x) = T(x) + T(a) and T(ax) = aT(x) for a cross product. The fact that numbers are not given makes it confusing as well. How can I approach this problem and prove the cross product is a transformation?
vectors linear-transformations cross-product
asked Aug 11 at 3:51
Kourosh
381113
- Background Information:
I am studying linear algebra. For this question, I understand the definition of a vector in $$R^3 => v =(x,y,z)$$, and I know that A linear transformation between two vector spaces V and W is a map
$$T: V->W $$such that the following hold:
$$T(v1+v2)=T(v1) + T(v2)$$ for any vectors v1 and v2 in V, and
$$T(av) = a T(v)$$ for any scalar alpha a.
I also know how to calculate the cross product between two vectors.
- Question:
Let a be a fixed vector in R3. Does T(x) = a × x define a linear transformation?
- My thoughts:
I don't understand how to show T(a + x) = T(x) + T(a) and T(ax) = aT(x) for a cross product. The fact that numbers are not given makes it confusing as well. How can I approach this problem and prove the cross product is a transformation?
vectors linear-transformations cross-product
vectors linear-transformations cross-product
asked Aug 11 at 3:51
Kourosh
381113
asked Aug 11 at 3:51
Kourosh
381113
edited Aug 11 at 4:05
asked Aug 11 at 3:51
Kourosh
381113
asked Aug 11 at 3:51
Kourosh
381113
asked Aug 11 at 3:51
Kourosh
381113
381113
1
Hint: what is $a times ( x+y)$? Use the properties!
– Sean Roberson
Aug 11 at 3:54
@SeanRoberson, thanks for the hint, It will be (a X x ) + (a X y), so what would the value of the cross product represent? Also, how can we do this without numbers?
– Kourosh
Aug 11 at 4:02
1
First, I recommend changing your scalar from "a" to $alpha$ since you also define "a" to be a fixed vector. You can see that $T(alpha x) = atimes alpha x$ and that $alpha T(x) = alpha (atimes x)$. If you can show that $ atimes alpha x = alpha (atimes x)$ you are done with that part. Now fill in that blank and write a suitable expression for $T(v_1+v_2)$, $T(v_1)$, and $T(v_2)$ then proceed again.
– JessicaK
Aug 11 at 4:15
@JessicaK, thanks for your help, so having T(αx)=a×αx => αT(x)=α(a×x), and then a×αx=α(a×x) will satisfy the homogeneity property? Or should I add more to it?
– Kourosh
Aug 11 at 5:58
add a comment |Â
1
Hint: what is $a times ( x+y)$? Use the properties!
– Sean Roberson
Aug 11 at 3:54
@SeanRoberson, thanks for the hint, It will be (a X x ) + (a X y), so what would the value of the cross product represent? Also, how can we do this without numbers?
– Kourosh
Aug 11 at 4:02
1
First, I recommend changing your scalar from "a" to $alpha$ since you also define "a" to be a fixed vector. You can see that $T(alpha x) = atimes alpha x$ and that $alpha T(x) = alpha (atimes x)$. If you can show that $ atimes alpha x = alpha (atimes x)$ you are done with that part. Now fill in that blank and write a suitable expression for $T(v_1+v_2)$, $T(v_1)$, and $T(v_2)$ then proceed again.
– JessicaK
Aug 11 at 4:15
@JessicaK, thanks for your help, so having T(αx)=a×αx => αT(x)=α(a×x), and then a×αx=α(a×x) will satisfy the homogeneity property? Or should I add more to it?
– Kourosh
Aug 11 at 5:58
1
1
Hint: what is $a times ( x+y)$? Use the properties!
– Sean Roberson
Aug 11 at 3:54
Hint: what is $a times ( x+y)$? Use the properties!
– Sean Roberson
Aug 11 at 3:54
@SeanRoberson, thanks for the hint, It will be (a X x ) + (a X y), so what would the value of the cross product represent? Also, how can we do this without numbers?
– Kourosh
Aug 11 at 4:02
@SeanRoberson, thanks for the hint, It will be (a X x ) + (a X y), so what would the value of the cross product represent? Also, how can we do this without numbers?
– Kourosh
Aug 11 at 4:02
1
1
First, I recommend changing your scalar from "a" to $alpha$ since you also define "a" to be a fixed vector. You can see that $T(alpha x) = atimes alpha x$ and that $alpha T(x) = alpha (atimes x)$. If you can show that $ atimes alpha x = alpha (atimes x)$ you are done with that part. Now fill in that blank and write a suitable expression for $T(v_1+v_2)$, $T(v_1)$, and $T(v_2)$ then proceed again.
– JessicaK
Aug 11 at 4:15
First, I recommend changing your scalar from "a" to $alpha$ since you also define "a" to be a fixed vector. You can see that $T(alpha x) = atimes alpha x$ and that $alpha T(x) = alpha (atimes x)$. If you can show that $ atimes alpha x = alpha (atimes x)$ you are done with that part. Now fill in that blank and write a suitable expression for $T(v_1+v_2)$, $T(v_1)$, and $T(v_2)$ then proceed again.
– JessicaK
Aug 11 at 4:15
@JessicaK, thanks for your help, so having T(αx)=a×αx => αT(x)=α(a×x), and then a×αx=α(a×x) will satisfy the homogeneity property? Or should I add more to it?
– Kourosh
Aug 11 at 5:58
@JessicaK, thanks for your help, so having T(αx)=a×αx => αT(x)=α(a×x), and then a×αx=α(a×x) will satisfy the homogeneity property? Or should I add more to it?
– Kourosh
Aug 11 at 5:58
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
Believe it or not, the cross product is linear! Let $T(x) = a times x$ for fixed $a$. Now, I'll show both conditions at once. Choose $x, y in mathbbR^3$. Now:
beginalign*
T(kx + y) &= a times (kx + y) \
&= a times (kx) + a times y \
&= k(a times x) + a times y \
&= kT(x) + T(y)
endalign*
Done! So this map is linear!
answered Aug 11 at 5:14
Sean Roberson
5,90931226
Oh wow, it is this easy. So the additivity property is done. Would it be enough to show "T(αx)=a×αx => αT(x)=α(a×x), and then a×αx=α(a×x)" for the homogeneity property?
– Kourosh
Aug 11 at 5:53
Notice I also showed homogeneity. "...I"ll show both conditions at once."
– Sean Roberson
Aug 11 at 6:21
My bad, I missed that, but you are the real MVP, I appreciate your help :)
– Kourosh
Aug 11 at 6:35
This proof is kind-of tautologic. The algebraic properties that you use are, in effect, equivalent to the linearity of the cross product.
– Federico Poloni
Aug 11 at 8:45
add a comment |Â
up vote
2
down vote
Even if you consider $$T(p,q) = ptimes q,$$
this function is a 2-linear map, i.e when you fix one argument, the function is linear wrt to the other argument.
Therefore, the cross product is more than just a linear transformation, but it is a 2-linear transformation.
answered Aug 11 at 5:24
onurcanbektas
3,1221834
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Believe it or not, the cross product is linear! Let $T(x) = a times x$ for fixed $a$. Now, I'll show both conditions at once. Choose $x, y in mathbbR^3$. Now:
beginalign*
T(kx + y) &= a times (kx + y) \
&= a times (kx) + a times y \
&= k(a times x) + a times y \
&= kT(x) + T(y)
endalign*
Done! So this map is linear!
answered Aug 11 at 5:14
Sean Roberson
5,90931226
Oh wow, it is this easy. So the additivity property is done. Would it be enough to show "T(αx)=a×αx => αT(x)=α(a×x), and then a×αx=α(a×x)" for the homogeneity property?
– Kourosh
Aug 11 at 5:53
Notice I also showed homogeneity. "...I"ll show both conditions at once."
– Sean Roberson
Aug 11 at 6:21
My bad, I missed that, but you are the real MVP, I appreciate your help :)
– Kourosh
Aug 11 at 6:35
This proof is kind-of tautologic. The algebraic properties that you use are, in effect, equivalent to the linearity of the cross product.
– Federico Poloni
Aug 11 at 8:45
add a comment |Â
up vote
5
down vote
accepted
Believe it or not, the cross product is linear! Let $T(x) = a times x$ for fixed $a$. Now, I'll show both conditions at once. Choose $x, y in mathbbR^3$. Now:
beginalign*
T(kx + y) &= a times (kx + y) \
&= a times (kx) + a times y \
&= k(a times x) + a times y \
&= kT(x) + T(y)
endalign*
Done! So this map is linear!
answered Aug 11 at 5:14
Sean Roberson
5,90931226
Oh wow, it is this easy. So the additivity property is done. Would it be enough to show "T(αx)=a×αx => αT(x)=α(a×x), and then a×αx=α(a×x)" for the homogeneity property?
– Kourosh
Aug 11 at 5:53
Notice I also showed homogeneity. "...I"ll show both conditions at once."
– Sean Roberson
Aug 11 at 6:21
My bad, I missed that, but you are the real MVP, I appreciate your help :)
– Kourosh
Aug 11 at 6:35
This proof is kind-of tautologic. The algebraic properties that you use are, in effect, equivalent to the linearity of the cross product.
– Federico Poloni
Aug 11 at 8:45
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Believe it or not, the cross product is linear! Let $T(x) = a times x$ for fixed $a$. Now, I'll show both conditions at once. Choose $x, y in mathbbR^3$. Now:
beginalign*
T(kx + y) &= a times (kx + y) \
&= a times (kx) + a times y \
&= k(a times x) + a times y \
&= kT(x) + T(y)
endalign*
Done! So this map is linear!
answered Aug 11 at 5:14
Sean Roberson
5,90931226
Believe it or not, the cross product is linear! Let $T(x) = a times x$ for fixed $a$. Now, I'll show both conditions at once. Choose $x, y in mathbbR^3$. Now:
beginalign*
T(kx + y) &= a times (kx + y) \
&= a times (kx) + a times y \
&= k(a times x) + a times y \
&= kT(x) + T(y)
endalign*
Done! So this map is linear!
answered Aug 11 at 5:14
Sean Roberson
5,90931226
answered Aug 11 at 5:14
Sean Roberson
5,90931226
answered Aug 11 at 5:14
Sean Roberson
5,90931226
answered Aug 11 at 5:14
Sean Roberson
5,90931226
5,90931226
Oh wow, it is this easy. So the additivity property is done. Would it be enough to show "T(αx)=a×αx => αT(x)=α(a×x), and then a×αx=α(a×x)" for the homogeneity property?
– Kourosh
Aug 11 at 5:53
Notice I also showed homogeneity. "...I"ll show both conditions at once."
– Sean Roberson
Aug 11 at 6:21
My bad, I missed that, but you are the real MVP, I appreciate your help :)
– Kourosh
Aug 11 at 6:35
This proof is kind-of tautologic. The algebraic properties that you use are, in effect, equivalent to the linearity of the cross product.
– Federico Poloni
Aug 11 at 8:45
add a comment |Â
Oh wow, it is this easy. So the additivity property is done. Would it be enough to show "T(αx)=a×αx => αT(x)=α(a×x), and then a×αx=α(a×x)" for the homogeneity property?
– Kourosh
Aug 11 at 5:53
Notice I also showed homogeneity. "...I"ll show both conditions at once."
– Sean Roberson
Aug 11 at 6:21
My bad, I missed that, but you are the real MVP, I appreciate your help :)
– Kourosh
Aug 11 at 6:35
This proof is kind-of tautologic. The algebraic properties that you use are, in effect, equivalent to the linearity of the cross product.
– Federico Poloni
Aug 11 at 8:45
Oh wow, it is this easy. So the additivity property is done. Would it be enough to show "T(αx)=a×αx => αT(x)=α(a×x), and then a×αx=α(a×x)" for the homogeneity property?
– Kourosh
Aug 11 at 5:53
Oh wow, it is this easy. So the additivity property is done. Would it be enough to show "T(αx)=a×αx => αT(x)=α(a×x), and then a×αx=α(a×x)" for the homogeneity property?
– Kourosh
Aug 11 at 5:53
Notice I also showed homogeneity. "...I"ll show both conditions at once."
– Sean Roberson
Aug 11 at 6:21
Notice I also showed homogeneity. "...I"ll show both conditions at once."
– Sean Roberson
Aug 11 at 6:21
My bad, I missed that, but you are the real MVP, I appreciate your help :)
– Kourosh
Aug 11 at 6:35
My bad, I missed that, but you are the real MVP, I appreciate your help :)
– Kourosh
Aug 11 at 6:35
This proof is kind-of tautologic. The algebraic properties that you use are, in effect, equivalent to the linearity of the cross product.
– Federico Poloni
Aug 11 at 8:45
This proof is kind-of tautologic. The algebraic properties that you use are, in effect, equivalent to the linearity of the cross product.
– Federico Poloni
Aug 11 at 8:45
add a comment |Â
up vote
2
down vote
Even if you consider $$T(p,q) = ptimes q,$$
this function is a 2-linear map, i.e when you fix one argument, the function is linear wrt to the other argument.
Therefore, the cross product is more than just a linear transformation, but it is a 2-linear transformation.
answered Aug 11 at 5:24
onurcanbektas
3,1221834
add a comment |Â
up vote
2
down vote
Even if you consider $$T(p,q) = ptimes q,$$
this function is a 2-linear map, i.e when you fix one argument, the function is linear wrt to the other argument.
Therefore, the cross product is more than just a linear transformation, but it is a 2-linear transformation.
answered Aug 11 at 5:24
onurcanbektas
3,1221834
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Even if you consider $$T(p,q) = ptimes q,$$
this function is a 2-linear map, i.e when you fix one argument, the function is linear wrt to the other argument.
Therefore, the cross product is more than just a linear transformation, but it is a 2-linear transformation.
answered Aug 11 at 5:24
onurcanbektas
3,1221834
Even if you consider $$T(p,q) = ptimes q,$$
this function is a 2-linear map, i.e when you fix one argument, the function is linear wrt to the other argument.
Therefore, the cross product is more than just a linear transformation, but it is a 2-linear transformation.
answered Aug 11 at 5:24
onurcanbektas
3,1221834
answered Aug 11 at 5:24
onurcanbektas
3,1221834
answered Aug 11 at 5:24
onurcanbektas
3,1221834
answered Aug 11 at 5:24
onurcanbektas
3,1221834
3,1221834
add a comment |Â
add a comment |Â
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1
Hint: what is $a times ( x+y)$? Use the properties!
– Sean Roberson
Aug 11 at 3:54
@SeanRoberson, thanks for the hint, It will be (a X x ) + (a X y), so what would the value of the cross product represent? Also, how can we do this without numbers?
– Kourosh
Aug 11 at 4:02
1
First, I recommend changing your scalar from "a" to $alpha$ since you also define "a" to be a fixed vector. You can see that $T(alpha x) = atimes alpha x$ and that $alpha T(x) = alpha (atimes x)$. If you can show that $ atimes alpha x = alpha (atimes x)$ you are done with that part. Now fill in that blank and write a suitable expression for $T(v_1+v_2)$, $T(v_1)$, and $T(v_2)$ then proceed again.
– JessicaK
Aug 11 at 4:15
@JessicaK, thanks for your help, so having T(αx)=a×αx => αT(x)=α(a×x), and then a×αx=α(a×x) will satisfy the homogeneity property? Or should I add more to it?
– Kourosh
Aug 11 at 5:58