Concept: Multiplying a vector by a scalar quantity

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A scalar is a quantity which has the only magnitude $|m|$ in contrast to a vector which has both direction and magnitude $|m|angletheta$.
Intuitively, multiplying a vector by a scalar scales a vector by the respective magnitude. Moreover, various texts claim, that multiplying a vector by scalar only changes the magnitude but direction remains unaffected.



An unexpected result is noticed when we multiply a negative scalar by a vector. Notably, a vector $vec v = (x, y)$, when its multiplied by a negative scalar $-|q|$, it not only scales the magnitude by a quantity equals to $|q|$, but also flips the vector $-|q|vec v = (-|q|x, -|q|y) = |q(x^2+y^2)|angleleft(pi+arctanfracyxright)$.



Many notable texts, claim the behavior to be valid. But, I find a few contradictions



  1. If by multiplying a vector $vec v$ scales a vector, what is the intuition of negative scaling?

  2. If the magnitude is absolute, and scalar has the only magnitude, how can scalar be negative?

  3. If multiplying by a scalar does not change the direction, why is multiplying a vector by a negative quantity and thus flipping it, a valid behavior?

The way, I am trying to understand is, multiplying by a negative scalar is actually two operations.



  1. Multiplying the vector by (-1) which flips a vector. This generates the negative inverse of a vector $because vec v + (-1)times vec v = 0$

  2. Scale the resultant vector by the appropriate scalar quantity.

So, scalar -q is actually $|q| times -1$ and thus multiplying with the vector $vec v = |v|angletheta$



$$Rightarrow -q times vec v = |q| times -1 vec v = |q| times |v|angleleft(pi + thetaright)$$



But, I cannot find a source to substantiate my reasoning and need the community support to help me with my understanding.







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  • I think when they say that multiplying by a scalar does not change the direction, what they mean is the span of the vector remains unchanged.
    – gd1035
    Jul 26 at 20:01










  • Multiplying by a negative does "flip" and scale the vector, but the span of the vector remains unchanged.
    – gd1035
    Jul 26 at 20:02










  • Multiplying by a negative number changes the magnitude and direction, as you say. I wouldn't worry too much about the words the text used to describe this.
    – saulspatz
    Jul 26 at 20:02











  • When they say a scalar is "a quantity that has magnitude but not direction", they are speaking roughly. The term "magnitude" in this context is not meant to imply that a scalar cannot be negative. A scalar can be negative, and that is fine. When they say that multiplying by a scalar "does not change direction", they are also speaking roughly, because multiplying by a negative scalar does reverse the direction.
    – littleO
    Jul 26 at 20:04










  • For a vector $v$: Magnitude: $|v|$, Direction: $O=rv: rtext scalar$, Orientation: Fixing a vector $win O$, with $wneq0$, $v$ is positively oriented with respect to $w$ if there is $r>0$ such that $rv=w$, negatively oriented with respect to $w$ if there is $r<0$ such that $rv=w$. With these definitions, multiplying $v$ by $-2$, changes magnitude, and orientation with respect to a fixed $w$, and not direction.
    – user577471
    Jul 26 at 20:11














up vote
0
down vote

favorite












A scalar is a quantity which has the only magnitude $|m|$ in contrast to a vector which has both direction and magnitude $|m|angletheta$.
Intuitively, multiplying a vector by a scalar scales a vector by the respective magnitude. Moreover, various texts claim, that multiplying a vector by scalar only changes the magnitude but direction remains unaffected.



An unexpected result is noticed when we multiply a negative scalar by a vector. Notably, a vector $vec v = (x, y)$, when its multiplied by a negative scalar $-|q|$, it not only scales the magnitude by a quantity equals to $|q|$, but also flips the vector $-|q|vec v = (-|q|x, -|q|y) = |q(x^2+y^2)|angleleft(pi+arctanfracyxright)$.



Many notable texts, claim the behavior to be valid. But, I find a few contradictions



  1. If by multiplying a vector $vec v$ scales a vector, what is the intuition of negative scaling?

  2. If the magnitude is absolute, and scalar has the only magnitude, how can scalar be negative?

  3. If multiplying by a scalar does not change the direction, why is multiplying a vector by a negative quantity and thus flipping it, a valid behavior?

The way, I am trying to understand is, multiplying by a negative scalar is actually two operations.



  1. Multiplying the vector by (-1) which flips a vector. This generates the negative inverse of a vector $because vec v + (-1)times vec v = 0$

  2. Scale the resultant vector by the appropriate scalar quantity.

So, scalar -q is actually $|q| times -1$ and thus multiplying with the vector $vec v = |v|angletheta$



$$Rightarrow -q times vec v = |q| times -1 vec v = |q| times |v|angleleft(pi + thetaright)$$



But, I cannot find a source to substantiate my reasoning and need the community support to help me with my understanding.







share|cite|improve this question





















  • I think when they say that multiplying by a scalar does not change the direction, what they mean is the span of the vector remains unchanged.
    – gd1035
    Jul 26 at 20:01










  • Multiplying by a negative does "flip" and scale the vector, but the span of the vector remains unchanged.
    – gd1035
    Jul 26 at 20:02










  • Multiplying by a negative number changes the magnitude and direction, as you say. I wouldn't worry too much about the words the text used to describe this.
    – saulspatz
    Jul 26 at 20:02











  • When they say a scalar is "a quantity that has magnitude but not direction", they are speaking roughly. The term "magnitude" in this context is not meant to imply that a scalar cannot be negative. A scalar can be negative, and that is fine. When they say that multiplying by a scalar "does not change direction", they are also speaking roughly, because multiplying by a negative scalar does reverse the direction.
    – littleO
    Jul 26 at 20:04










  • For a vector $v$: Magnitude: $|v|$, Direction: $O=rv: rtext scalar$, Orientation: Fixing a vector $win O$, with $wneq0$, $v$ is positively oriented with respect to $w$ if there is $r>0$ such that $rv=w$, negatively oriented with respect to $w$ if there is $r<0$ such that $rv=w$. With these definitions, multiplying $v$ by $-2$, changes magnitude, and orientation with respect to a fixed $w$, and not direction.
    – user577471
    Jul 26 at 20:11












up vote
0
down vote

favorite









up vote
0
down vote

favorite











A scalar is a quantity which has the only magnitude $|m|$ in contrast to a vector which has both direction and magnitude $|m|angletheta$.
Intuitively, multiplying a vector by a scalar scales a vector by the respective magnitude. Moreover, various texts claim, that multiplying a vector by scalar only changes the magnitude but direction remains unaffected.



An unexpected result is noticed when we multiply a negative scalar by a vector. Notably, a vector $vec v = (x, y)$, when its multiplied by a negative scalar $-|q|$, it not only scales the magnitude by a quantity equals to $|q|$, but also flips the vector $-|q|vec v = (-|q|x, -|q|y) = |q(x^2+y^2)|angleleft(pi+arctanfracyxright)$.



Many notable texts, claim the behavior to be valid. But, I find a few contradictions



  1. If by multiplying a vector $vec v$ scales a vector, what is the intuition of negative scaling?

  2. If the magnitude is absolute, and scalar has the only magnitude, how can scalar be negative?

  3. If multiplying by a scalar does not change the direction, why is multiplying a vector by a negative quantity and thus flipping it, a valid behavior?

The way, I am trying to understand is, multiplying by a negative scalar is actually two operations.



  1. Multiplying the vector by (-1) which flips a vector. This generates the negative inverse of a vector $because vec v + (-1)times vec v = 0$

  2. Scale the resultant vector by the appropriate scalar quantity.

So, scalar -q is actually $|q| times -1$ and thus multiplying with the vector $vec v = |v|angletheta$



$$Rightarrow -q times vec v = |q| times -1 vec v = |q| times |v|angleleft(pi + thetaright)$$



But, I cannot find a source to substantiate my reasoning and need the community support to help me with my understanding.







share|cite|improve this question













A scalar is a quantity which has the only magnitude $|m|$ in contrast to a vector which has both direction and magnitude $|m|angletheta$.
Intuitively, multiplying a vector by a scalar scales a vector by the respective magnitude. Moreover, various texts claim, that multiplying a vector by scalar only changes the magnitude but direction remains unaffected.



An unexpected result is noticed when we multiply a negative scalar by a vector. Notably, a vector $vec v = (x, y)$, when its multiplied by a negative scalar $-|q|$, it not only scales the magnitude by a quantity equals to $|q|$, but also flips the vector $-|q|vec v = (-|q|x, -|q|y) = |q(x^2+y^2)|angleleft(pi+arctanfracyxright)$.



Many notable texts, claim the behavior to be valid. But, I find a few contradictions



  1. If by multiplying a vector $vec v$ scales a vector, what is the intuition of negative scaling?

  2. If the magnitude is absolute, and scalar has the only magnitude, how can scalar be negative?

  3. If multiplying by a scalar does not change the direction, why is multiplying a vector by a negative quantity and thus flipping it, a valid behavior?

The way, I am trying to understand is, multiplying by a negative scalar is actually two operations.



  1. Multiplying the vector by (-1) which flips a vector. This generates the negative inverse of a vector $because vec v + (-1)times vec v = 0$

  2. Scale the resultant vector by the appropriate scalar quantity.

So, scalar -q is actually $|q| times -1$ and thus multiplying with the vector $vec v = |v|angletheta$



$$Rightarrow -q times vec v = |q| times -1 vec v = |q| times |v|angleleft(pi + thetaright)$$



But, I cannot find a source to substantiate my reasoning and need the community support to help me with my understanding.









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 27 at 4:32
























asked Jul 26 at 19:52









Abhijit

2,337719




2,337719











  • I think when they say that multiplying by a scalar does not change the direction, what they mean is the span of the vector remains unchanged.
    – gd1035
    Jul 26 at 20:01










  • Multiplying by a negative does "flip" and scale the vector, but the span of the vector remains unchanged.
    – gd1035
    Jul 26 at 20:02










  • Multiplying by a negative number changes the magnitude and direction, as you say. I wouldn't worry too much about the words the text used to describe this.
    – saulspatz
    Jul 26 at 20:02











  • When they say a scalar is "a quantity that has magnitude but not direction", they are speaking roughly. The term "magnitude" in this context is not meant to imply that a scalar cannot be negative. A scalar can be negative, and that is fine. When they say that multiplying by a scalar "does not change direction", they are also speaking roughly, because multiplying by a negative scalar does reverse the direction.
    – littleO
    Jul 26 at 20:04










  • For a vector $v$: Magnitude: $|v|$, Direction: $O=rv: rtext scalar$, Orientation: Fixing a vector $win O$, with $wneq0$, $v$ is positively oriented with respect to $w$ if there is $r>0$ such that $rv=w$, negatively oriented with respect to $w$ if there is $r<0$ such that $rv=w$. With these definitions, multiplying $v$ by $-2$, changes magnitude, and orientation with respect to a fixed $w$, and not direction.
    – user577471
    Jul 26 at 20:11
















  • I think when they say that multiplying by a scalar does not change the direction, what they mean is the span of the vector remains unchanged.
    – gd1035
    Jul 26 at 20:01










  • Multiplying by a negative does "flip" and scale the vector, but the span of the vector remains unchanged.
    – gd1035
    Jul 26 at 20:02










  • Multiplying by a negative number changes the magnitude and direction, as you say. I wouldn't worry too much about the words the text used to describe this.
    – saulspatz
    Jul 26 at 20:02











  • When they say a scalar is "a quantity that has magnitude but not direction", they are speaking roughly. The term "magnitude" in this context is not meant to imply that a scalar cannot be negative. A scalar can be negative, and that is fine. When they say that multiplying by a scalar "does not change direction", they are also speaking roughly, because multiplying by a negative scalar does reverse the direction.
    – littleO
    Jul 26 at 20:04










  • For a vector $v$: Magnitude: $|v|$, Direction: $O=rv: rtext scalar$, Orientation: Fixing a vector $win O$, with $wneq0$, $v$ is positively oriented with respect to $w$ if there is $r>0$ such that $rv=w$, negatively oriented with respect to $w$ if there is $r<0$ such that $rv=w$. With these definitions, multiplying $v$ by $-2$, changes magnitude, and orientation with respect to a fixed $w$, and not direction.
    – user577471
    Jul 26 at 20:11















I think when they say that multiplying by a scalar does not change the direction, what they mean is the span of the vector remains unchanged.
– gd1035
Jul 26 at 20:01




I think when they say that multiplying by a scalar does not change the direction, what they mean is the span of the vector remains unchanged.
– gd1035
Jul 26 at 20:01












Multiplying by a negative does "flip" and scale the vector, but the span of the vector remains unchanged.
– gd1035
Jul 26 at 20:02




Multiplying by a negative does "flip" and scale the vector, but the span of the vector remains unchanged.
– gd1035
Jul 26 at 20:02












Multiplying by a negative number changes the magnitude and direction, as you say. I wouldn't worry too much about the words the text used to describe this.
– saulspatz
Jul 26 at 20:02





Multiplying by a negative number changes the magnitude and direction, as you say. I wouldn't worry too much about the words the text used to describe this.
– saulspatz
Jul 26 at 20:02













When they say a scalar is "a quantity that has magnitude but not direction", they are speaking roughly. The term "magnitude" in this context is not meant to imply that a scalar cannot be negative. A scalar can be negative, and that is fine. When they say that multiplying by a scalar "does not change direction", they are also speaking roughly, because multiplying by a negative scalar does reverse the direction.
– littleO
Jul 26 at 20:04




When they say a scalar is "a quantity that has magnitude but not direction", they are speaking roughly. The term "magnitude" in this context is not meant to imply that a scalar cannot be negative. A scalar can be negative, and that is fine. When they say that multiplying by a scalar "does not change direction", they are also speaking roughly, because multiplying by a negative scalar does reverse the direction.
– littleO
Jul 26 at 20:04












For a vector $v$: Magnitude: $|v|$, Direction: $O=rv: rtext scalar$, Orientation: Fixing a vector $win O$, with $wneq0$, $v$ is positively oriented with respect to $w$ if there is $r>0$ such that $rv=w$, negatively oriented with respect to $w$ if there is $r<0$ such that $rv=w$. With these definitions, multiplying $v$ by $-2$, changes magnitude, and orientation with respect to a fixed $w$, and not direction.
– user577471
Jul 26 at 20:11




For a vector $v$: Magnitude: $|v|$, Direction: $O=rv: rtext scalar$, Orientation: Fixing a vector $win O$, with $wneq0$, $v$ is positively oriented with respect to $w$ if there is $r>0$ such that $rv=w$, negatively oriented with respect to $w$ if there is $r<0$ such that $rv=w$. With these definitions, multiplying $v$ by $-2$, changes magnitude, and orientation with respect to a fixed $w$, and not direction.
– user577471
Jul 26 at 20:11










2 Answers
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We have that for $lambdain mathbbR$



  • for $lambda>0$ we have that the operation $lambdavec v$ scale $vec v$ of a factor $|lambda|$ with the same orientation


  • for $lambda<0$ we have that the operation $lambdavec v$ scale $vec v$ of a factor $|lambda|$ with the opposite orientation


enter image description here



Note that in both cases the direction doesn't change since both vectors belong to the same line.






share|cite|improve this answer




























    up vote
    1
    down vote













    In a real vector space where the scalars are real numbers, a scalar has a magnitude and a sign.



    Thus when you multiply a vector by a scalar you scale the magnitude of the vector and change or not change the direction of your vector based on the sign of the scalar.



    ( The vector stays on the same line, which sometimes is interpreted as not changing the direction )



    The concept is clear but the terminology is sometimes confusing.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
      2






      active

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      active

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      up vote
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      We have that for $lambdain mathbbR$



      • for $lambda>0$ we have that the operation $lambdavec v$ scale $vec v$ of a factor $|lambda|$ with the same orientation


      • for $lambda<0$ we have that the operation $lambdavec v$ scale $vec v$ of a factor $|lambda|$ with the opposite orientation


      enter image description here



      Note that in both cases the direction doesn't change since both vectors belong to the same line.






      share|cite|improve this answer

























        up vote
        1
        down vote













        We have that for $lambdain mathbbR$



        • for $lambda>0$ we have that the operation $lambdavec v$ scale $vec v$ of a factor $|lambda|$ with the same orientation


        • for $lambda<0$ we have that the operation $lambdavec v$ scale $vec v$ of a factor $|lambda|$ with the opposite orientation


        enter image description here



        Note that in both cases the direction doesn't change since both vectors belong to the same line.






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          We have that for $lambdain mathbbR$



          • for $lambda>0$ we have that the operation $lambdavec v$ scale $vec v$ of a factor $|lambda|$ with the same orientation


          • for $lambda<0$ we have that the operation $lambdavec v$ scale $vec v$ of a factor $|lambda|$ with the opposite orientation


          enter image description here



          Note that in both cases the direction doesn't change since both vectors belong to the same line.






          share|cite|improve this answer













          We have that for $lambdain mathbbR$



          • for $lambda>0$ we have that the operation $lambdavec v$ scale $vec v$ of a factor $|lambda|$ with the same orientation


          • for $lambda<0$ we have that the operation $lambdavec v$ scale $vec v$ of a factor $|lambda|$ with the opposite orientation


          enter image description here



          Note that in both cases the direction doesn't change since both vectors belong to the same line.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 26 at 20:11









          gimusi

          65k73583




          65k73583




















              up vote
              1
              down vote













              In a real vector space where the scalars are real numbers, a scalar has a magnitude and a sign.



              Thus when you multiply a vector by a scalar you scale the magnitude of the vector and change or not change the direction of your vector based on the sign of the scalar.



              ( The vector stays on the same line, which sometimes is interpreted as not changing the direction )



              The concept is clear but the terminology is sometimes confusing.






              share|cite|improve this answer

























                up vote
                1
                down vote













                In a real vector space where the scalars are real numbers, a scalar has a magnitude and a sign.



                Thus when you multiply a vector by a scalar you scale the magnitude of the vector and change or not change the direction of your vector based on the sign of the scalar.



                ( The vector stays on the same line, which sometimes is interpreted as not changing the direction )



                The concept is clear but the terminology is sometimes confusing.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  In a real vector space where the scalars are real numbers, a scalar has a magnitude and a sign.



                  Thus when you multiply a vector by a scalar you scale the magnitude of the vector and change or not change the direction of your vector based on the sign of the scalar.



                  ( The vector stays on the same line, which sometimes is interpreted as not changing the direction )



                  The concept is clear but the terminology is sometimes confusing.






                  share|cite|improve this answer













                  In a real vector space where the scalars are real numbers, a scalar has a magnitude and a sign.



                  Thus when you multiply a vector by a scalar you scale the magnitude of the vector and change or not change the direction of your vector based on the sign of the scalar.



                  ( The vector stays on the same line, which sometimes is interpreted as not changing the direction )



                  The concept is clear but the terminology is sometimes confusing.







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Jul 26 at 20:12









                  Mohammad Riazi-Kermani

                  27.3k41851




                  27.3k41851






















                       

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