A third kind of “products” between two Euclidean vectors $\vec{a}$ and $\vec{b}$, besides the scalar product  $\vec{a}\!\cdot\!\vec{b}$ and the vector product $\vec{a}\!\times\!\vec{b}$, is the  dyad product$\vec{a}\,\vec{b}$,  which is usually denoted without any multiplication symbol.  The dyad products and the finite formal sums

 $\displaystyle\Phi:=\sum_{\mu}\vec{a}_{\mu}\vec{b}_{\mu}$ (1)

A dyad is not a vector, but an operator.  It on any vector $\vec{v}$ producing from it new vectors or new dyads according to the definitions

 $\displaystyle\Phi*\vec{v}:=\sum_{\mu}\vec{a}_{\mu}(\vec{b}_{\mu}*\vec{v}),% \quad\vec{v}*\Phi:=\sum_{\mu}(\vec{v}*\vec{a}_{\mu})\vec{b}_{\mu}.$ (2)

Here the asterisks empty, in which case the vector $\vec{v}$ must be replaced by a scalar $v$; the products $\Phi v$ and $v\Phi$ are dyads.

The dyad product obeys the distributive laws

 $\vec{a}(\vec{b}\!+\!\vec{c})=\vec{a}\,\vec{b}\!+\!\vec{a}\,\vec{c},\quad(\vec{% b}\!+\!\vec{c})\vec{a}=\vec{b}\,\vec{a}\!+\!\vec{c}\,\vec{a},$

which can be verified by multiplying an arbitrary vector $\vec{v}$ and both of these equations and then comparing the results.  Likewise, the scalar factor transfer rule is valid.  It follows that if we have  $\vec{a}=a_{1}\vec{e_{1}}+a_{2}\vec{e_{2}}+a_{3}\vec{e_{3}}$  and  $\vec{b}=b_{1}\vec{e_{1}}+b_{2}\vec{e_{2}}+b_{3}\vec{e_{3}}$  in the orthonormal basis$\{\vec{e_{1}},\,\vec{e_{2}},\,\vec{e_{3}}\}$ (for the brevity, we confine us to vectors of $\mathbb{R}^{3}$), their dyad product is the sum

 $\displaystyle\vec{a}\,\vec{b}=$ $\displaystyle a_{1}b_{1}\vec{e_{1}}\vec{e_{1}}+a_{1}b_{2}\vec{e_{1}}\vec{e_{2}% }+a_{1}b_{3}\vec{e_{1}}\vec{e_{3}}+$ $\displaystyle a_{2}b_{1}\vec{e_{2}}\vec{e_{1}}+a_{2}b_{2}\vec{e_{2}}\vec{e_{2}% }+a_{2}b_{3}\vec{e_{2}}\vec{e_{3}}+$ $\displaystyle a_{3}b_{1}\vec{e_{3}}\vec{e_{1}}+a_{3}b_{2}\vec{e_{3}}\vec{e_{2}% }+a_{3}b_{3}\vec{e_{3}}\vec{e_{3}},$

which shows that the dyad product has been formed similarly as the matrix product  of the vectors  $(a_{1},\,a_{2},\,a_{3})^{\mbox{\scriptsize{T}}}$  and $(b_{1},\,b_{2},\,b_{3})$.

 $\mbox{I}:=\vec{e_{1}}\vec{e_{1}}\!+\!\vec{e_{2}}\vec{e_{2}}\!+\!\vec{e_{3}}% \vec{e_{3}}=\nabla\vec{r},$

where $\vec{r}$ is the position vector,

 $\mbox{I}\!\cdot\!\vec{v}=\vec{v}\!\cdot\!\mbox{I}=\vec{v}$

and

 $\mbox{I}\!\times\!(\vec{u}\!\times\!\vec{v})=\vec{v}\,\vec{u}-\vec{u}\,\vec{v}$

for all vectors $\vec{u}$ and $\vec{v}$.

The product of two dyad products  $\vec{a}\,\vec{b}$  and  $\vec{c}\,\vec{d}$  is defined to be the dyad

 $\displaystyle(\vec{a}\,\vec{b})(\vec{c}\,\vec{d}):=(\vec{b}\!\cdot\!\vec{c})(% \vec{a}\,\vec{d})$ (3)

and the product of such dyads as (1) to be the formal sum of individual products (3).  The multiplication of dyads is associative and distributive over addition.  The unit dyad acts as unity in the ring of dyads:

 $\mbox{I}\Phi=\Phi\mbox{I}=\Phi\quad\forall\Phi$

## References

• 1 K. Väisälä: Vektorianalyysi.  Werner Söderström Osakeyhtiö, Helsinki (1961).