What is dyadic in linear algebra?

Question

Suppose a vector $\large\vec{a}$. It can be written as:

$$\large\sum_i \vec{e_i}\,\vec{e_i}\cdot \vec{a}=\vec{1}\cdot\vec{a}$$

where $\large\vec{e_i}$ are unit basis vectors.

I can't figue out what is $\large\vec{1}$ (called dyadic) and how to obtain it from $\large\vec{e_i}\vec{e_i}$. Reading Wikipedia gave no undestanding.

Would you help me?

Answer 1

Notation in the equation is misleading. We all agree that if $a$ and $b$ are column vectors, we can multiply them as $a^{\rm T} b$ and $ab^{\rm T}$, but certainly we can’t perform $ab$. On the other hand, I’m prone to believe $\vec e_i \cdot \vec a$ means $\vec e^{\rm T}_i \vec a$, the stantard inner product of vectors (which in this case, actually gives the $i$th entry of $\vec a$, $a_i$.) I would also add a parenthesis to the equation: $$\sum_i \vec e_i (\vec e_i \cdot \vec a) =\sum_i \vec e_i (\vec e^{\rm T}_i \vec a) =\sum_i (\vec e_i \vec e^{\rm T}_i) \vec a =\Big(\sum_i \vec e_i \vec e^{\rm T}_i\Big) \vec a = \text{Id } \vec a,$$ where Id is the identity matrix that has as many columns as $\vec a$ has entries. (By the way, $\vec 1$ usually denotes a vector with only ones and it shouln’t be there.)

Answer 2

The dot product $$\vec{1}\cdot \vec{a} = \sum_i \vec{e_i}\vec a$$

What you have instead is:

$$\sum_i \vec{e_i}(\vec{e_i} \cdot \vec a)$$

Which is like the sum of the projections of $\vec a$ onto the basis.