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Let $B = \left(\vec{e}_1, \vec{e}_2, \vec{e}_3\right)$ a direct orthonormal basis of $V^3$

and let $\vec{a}$ and $\vec{b}$ be two nonzero vectors of $V^3$ such that

$\vec{a}=a_{1}\vec{e}_1+a_{2}\vec{e}_2+a_{3}\vec{e}_2$

$\vec{b}=b_{1}\vec{e}_1+b_{2}\vec{e}_2+b_{3}\vec{e}_2$

Give the matrix associated with the following linear transformations

$T(\vec{u})=(\vec{a}\cdot\vec{u})\vec{b}$

$T(\vec{u})=\vec{a}\times\vec{u}$

Can anyone please give me an idea how to find these matrix?

I know the answers are

$$ \begin{bmatrix} a_{1}b_{1} & a_{2}b_{1} & a_{3}b_{1} \\ a_{1}b_{2} & a_{2}b_{2} & a_{3}b_{2} \\ a_{1}b_{3} & a_{2}b_{3} & a_{3}b_{3} \\ \end{bmatrix} $$

$$ \begin{bmatrix} 0 & -a_{3} & a_{2} \\ a_{3} & 0 & -a_{1} \\ -a_{2} & a_{1} & 0 \\ \end{bmatrix} $$

but I have no idea what I need to do to get them

Thanks in advance

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My suggestion would be to evaluate the transformation in $e_1$, $e_2$ and $e_3$. Also you must note that the matrix associated to a transformation in a given basis $\{e_1,e_2,e_3\}$ is $[T(e_1)T(e_2)T(e_3)]$ where each $T(e)$ is a column vector.

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  • $\begingroup$ I considered B like $(\vec{i}, \vec{j}, \vec{k})$, but I found \begin{bmatrix} a_{1}b_{1} & 0 & 0 \\ 0 & a_{2}b_{2} & 0 \\ 0 & 0 & a_{3}b_{3} \\ \end{bmatrix} I'm about two hours stuck in this problem... $\endgroup$ – user78723 May 30 '13 at 20:17
  • $\begingroup$ What's your result for $T(e_1)$? I feel your getting dot (inner) product mixed up with scalar product. $\endgroup$ – Julio Cáceres May 30 '13 at 20:21
  • $\begingroup$ Well I did this $(a_{1}u_{1} + a_{2}u_{2} + a_{3}u_{3})\vec{b}$ $(a_{1}u_{1} + a_{2}u_{2} + a_{3}u_{3})b_{1} + (a_{1}u_{1} + a_{2}u_{2} + a_{3}u_{3})b_{2} + (a_{1}u_{1} + a_{2}u_{2} + a_{3}u_{3})b_{3}$ My $T(\vec{e}_{1})$ will be something like that... \begin{bmatrix} a_{1}b_{1} \\ 0 \\ 0 \\ \end{bmatrix} $\endgroup$ – user78723 May 30 '13 at 20:40
  • $\begingroup$ The second product is not a dot product but a scalar product, so if $a \cdot u = \lambda$, then $\lambda(b_1,b_2,b_3) = (\lambda b_1,\lambda b_2 , \lambda b_3)$ $\endgroup$ – Julio Cáceres May 30 '13 at 20:45
  • $\begingroup$ But how $(\lambda b_1,\lambda b_2 , \lambda b_3)$ Became \begin{bmatrix} \lambda b_1\\ \lambda b_3 \\ \lambda b_3 \\ \end{bmatrix} And not \begin{bmatrix} \lambda b_1\\ 0 \\ 0 \\ \end{bmatrix} If my basis is $(\vec{i}, \vec{j}, \vec{k})$ ? $\endgroup$ – user78723 May 30 '13 at 21:11

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