• 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?

  • 1
    $\begingroup$ Hint: what is $a \times ( x+y)$? Use the properties! $\endgroup$ – Sean Roberson Aug 11 '18 at 3:54
  • $\begingroup$ @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? $\endgroup$ – Kourosh Aug 11 '18 at 4:02
  • 1
    $\begingroup$ First, I recommend changing your scalar from "a" to $\alpha$ since you also define "a" to be a fixed vector. You can see that $T(\alpha x) = a\times \alpha x$ and that $\alpha T(x) = \alpha (a\times x)$. If you can show that $ a\times \alpha x = \alpha (a\times 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. $\endgroup$ – JessicaK Aug 11 '18 at 4:15
  • $\begingroup$ @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? $\endgroup$ – Kourosh Aug 11 '18 at 5:58

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 $x, y \in \mathbb{R}^3$. Now:

\begin{align*} T(kx + y) &= a \times (kx + y) \\ &= a \times (kx) + a \times y \\ &= k(a \times x) + a \times y \\ &= kT(x) + T(y) \end{align*}

Done! So this map is linear!

  • $\begingroup$ 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? $\endgroup$ – Kourosh Aug 11 '18 at 5:53
  • $\begingroup$ Notice I also showed homogeneity. "...I"ll show both conditions at once." $\endgroup$ – Sean Roberson Aug 11 '18 at 6:21
  • $\begingroup$ My bad, I missed that, but you are the real MVP, I appreciate your help :) $\endgroup$ – Kourosh Aug 11 '18 at 6:35
  • $\begingroup$ This proof is kind-of tautologic. The algebraic properties that you use are, in effect, equivalent to the linearity of the cross product. $\endgroup$ – Federico Poloni Aug 11 '18 at 8:45

Even if you consider $$T(p,q) = p\times q,$$ 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.


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