Let $V$ be a vector space over $F$ and $a\in V$ is a fixed vector. Let's define a function $f:V \to V$ with $f(v) = a \times v$. Check if $f$ is a linear operator. If it is, determine it's transformation matrix in respect to standard basis.

So, it is indeed a linear operator because:

$(\forall x, y \in V) \space \space f(x+y)=a \times(x+y)=a\times x + a\times y = f(x) + f(y)$

$(\forall x \in V)(\forall \alpha \in F) \space \space f(\alpha x) = a \times (\alpha x) = \alpha (a\times x) = \alpha f(x)$

Now I need to find it's transformation matrix. Since we don't know the dimension of $V$ we have:

$a= \begin{bmatrix} a_{1}\\ a_{2}\\ \vdots\\ a_{n} \end{bmatrix} $ and $e_{1} = \begin{bmatrix} 1\\ 0\\ \vdots\\ 0 \end{bmatrix} $. How do I find $a\times e_{1}$ and the transformation matrix?

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    $\begingroup$ What's $a\times v$? There is no vector multiplication on a generic vector space. And a generic vector space has no “standard basis”. $\endgroup$ – egreg Nov 9 '17 at 21:55
  • $\begingroup$ Well I got the problem written like this, I assumed it's a vector product since it doesn't say otherwise and $\times$ in vector spaces usually means vector product. That is what confuses me. $\endgroup$ – Nebeski Nov 9 '17 at 22:01

Vector product can only be defined over three-dimensional spaces, so probably your space is $F^3$; if $\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\}$ is the standard basis, then the vector product $\mathbf{a}\times \mathbf{v}$ is the “formal determinant” $$ \mathbf{a}\times \mathbf{v}= \det\begin{bmatrix} a_1 & v_1 & \mathbf{e}_1\\ a_2 & v_2 & \mathbf{e}_2 \\ a_3 & v_3 & \mathbf{e}_3 \end{bmatrix}= (a_2v_3-a_3v_2)\mathbf{e}_1-(a_1v_3-a_3v_1)\mathbf{e}_2+(a_1v_2-a_2v_1)\mathbf{e}_3 $$ to be expanded with Laplace’s rule with respect to the third column. Then $$ \mathbf{a}\times \mathbf{e}_1=a_3\mathbf{e}_2-a_2\mathbf{e}_3 \qquad \mathbf{a}\times \mathbf{e}_2=-a_3\mathbf{e}_1+a_1\mathbf{e}_3 \qquad \mathbf{a}\times \mathbf{e}_3=a_2\mathbf{e}_1-a_1\mathbf{e}_2 $$ so the matrix is $$ \begin{bmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{bmatrix} $$

  • $\begingroup$ I guess that is the only logical solution, I don't know why it says only $F$. $\endgroup$ – Nebeski Nov 9 '17 at 23:52

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