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:



  1. $$T(v1+v2)=T(v1) + T(v2)$$ for any vectors v1 and v2 in V, and


  2. $$T(av) = a T(v)$$ 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?










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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














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:



  1. $$T(v1+v2)=T(v1) + T(v2)$$ for any vectors v1 and v2 in V, and


  2. $$T(av) = a T(v)$$ 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?










share|cite|improve this question



















  • 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












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:



  1. $$T(v1+v2)=T(v1) + T(v2)$$ for any vectors v1 and v2 in V, and


  2. $$T(av) = a T(v)$$ 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?










share|cite|improve this question















  • 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:



  1. $$T(v1+v2)=T(v1) + T(v2)$$ for any vectors v1 and v2 in V, and


  2. $$T(av) = a T(v)$$ 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






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share|cite|improve this question













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edited Aug 11 at 4:05

























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












  • 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










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!






share|cite|improve this answer




















  • 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

















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.






share|cite|improve this answer




















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    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!






    share|cite|improve this answer




















    • 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














    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!






    share|cite|improve this answer




















    • 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












    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!






    share|cite|improve this answer












    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!







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    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
















    • 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










    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.






    share|cite|improve this answer
























      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.






      share|cite|improve this answer






















        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 11 at 5:24









        onurcanbektas

        3,1221834




        3,1221834



























             

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