# Dot Product

1. Dec 11, 2008

### Apteronotus

If you look up dot product in Wikipedia, under 'properties' it states the following:

"The dot product is not associative, however with the help of the matrix-multiplication one can derive:
$$\left(\vec{a} \cdot \vec{b}\right) \vec{c} = \left(\vec{c}\vec{b}^{T}\right)\vec{a}$$"​

I simply dont see how this can be true for any vector $$\vec{c}$$. Is it?

2. Dec 11, 2008

### tiny-tim

Hi Apteronotus!

cTb is a scalar, c.b, but cbT is a matrix.

(cbT)a = ∑∑(cbT)ijajei
= ∑∑cibjajei
= (∑bjaj)∑ciei = (b.a)c

3. Dec 11, 2008

### HallsofIvy

Staff Emeritus
cTb, the dot product, is often called the "inner product" and is a scalar while bcT is called the "outer product" and is a matrix.

4. Dec 22, 2008

### Apteronotus

Thank you both for your replies. But I still think there may be a problem.
Consider three vectors $$\vec{a}, \vec{b}$$ and $$\vec{c}$$, where

$$\vec{a}=\left[a_{1}, a_{1}, a_{3}\right]$$
$$\vec{b}=\left[b_{1}, b_{1}, b_{3}\right]$$

and
$$\vec{c}=\left[c_{1}, c_{1}, c_{3}\right]$$

then
$$\left(\vec{a}\cdot\vec{b}\right)\vec{c}=\left(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}\right)\left[c_{1}, c_{1}, c_{3}\right]$$

and

$$\left(\vec{c}\vec{b}^{T}\right)\vec{a}=\left(c_{1}b_{1}+c_{2}b_{2}+c_{3}b_{3}\right)\left[a_{1}, a_{2}, a_{3}\right]$$

Now these vectors are not necessarily equal. Consider the first element of each.
It is clear that in general,

$$\left(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}\right)c_{1}\neq\left(c_{1}b_{1}+c_{2}b_{2}+c_{3}b_{3}\right)a_{1}$$

5. Dec 22, 2008

### tiny-tim

But tcbT is not (c1b2+c2b2+c3b3) …

it's a matrix.

6. Dec 23, 2008

Specifically, it is the matrix
$$cb^T = \begin{bmatrix} b_1 c_1 & b_2 c_1 & b_3 c_1 \\ b_1 c_2 & b_2 c_2 & b_3 c_2 \\ b_1 c_3 & b_2 c_3 & b_3 c_3 \end{bmatrix}.$$

Last edited: Dec 23, 2008
7. Dec 25, 2008

### Apteronotus

Yes! Thank you both very much.
My error was in taking the vectors as row vectors.
Thanks again,

8. Dec 27, 2008

### HallsofIvy

Staff Emeritus
This is more abstract and more advanced than the "inner product" but if you are using the "outer product", you may want to think about this.

The "dual" of a finite dimensional vector space, V, (the space of linear functionals from V to the base field) is isomorphic to v with a "natural" isomorphism: given a basis ${e_1, e_2, \cdot\cdot\cdot, e_n}$, map each basis vector $e_i$ to the functional, $f_{e_i}(v)$ that maps $e_i$ to 1, all other $e_j$ to 0. Then extend it to the entire space by "linearity": if $v= a_1e_1+ a_2e_2+ \cdot\cdot\cdot a_ie_i+ \cdot\cdot\cdot+ a_ne_n$, $f(v)= a_1 f(e_1)+ a_2f(e_2)+\cdot\cdot\cdot+ a_if(e_i)+ \cdot\cdot\cdot+ a_nf(e_n)$$= a_1(0)+ a_2(0)+ \cdot\cdot\cdot+ a_i(1)+ \cdot\cdot\cdot+ a_n(0)= a_i$.

Since that is an isomorphism, given any vector u, that isomorphism maps it to f_u(x). Given any two vectors, u, and v, the functional f_u(v) takes v to the real number $u\cdot v$, their dot product as defined in that particular basis. On the other hand "$v f_v(x)$ can be interpreted as a linear transformation that maps each vector , w, into the vector $(f_v(w))u$ an numeric multiple of u. If we agree to write vectors as column matrices, say
$$v= \left[\begin{array}{c}a_1 \\ a_2\\ \cdot \\ \cdot \\ \cdot \\ a_n\end{array}\right]$$
and functionals in the dual space as row matrices, say
$$f_u= \left[\begin{array}{ccccc}b_1 & b_2 & \cdot\cdot\cdot & b_n\end{array}\right]$$
Then the operation of the functional, $f_u$ on v is the matrix product
$$\left[\begin{array}{ccccc}b_1 & b_2 & \cdot\cdot\cdot & b_n\end{array}\right]\left[\begin{array}{c}a_1 \\ a_2\\ \cdot \\ \cdot \\ \cdot \\ a_n\end{array}\right]$$
while the linear transformation corresponding to $v f_u$ is give by the matrix product

$$\left[\begin{array}{c}a_1 \\ a_2\\ \cdot \\ \cdot \\ \cdot \\ a_n\end{array}\right]\left[\begin{array}{ccccc}b_1 & b_2 & \cdot\cdot\cdot & b_n\end{array}\right]$$

Last edited: Dec 27, 2008