1
$\begingroup$

A matrix can be regarded as a representation of a linear function from one vector space to another, with the condition that the basis vectors (and their order) are specified for both spaces.

Therefore a column matrix must necessarily be a representation of a vector of a space in relation to that space basis.

For example, let $\vec{u}=4\vec{e_1} + 1\vec{e_2}$. In matrix notation we have $\left[ \begin {matrix} 4 \\ 1 \\ \end {matrix} \right] = 4 \left[ \begin {matrix} 1 \\ 0 \\ \end {matrix} \right] + 1 \left[ \begin {matrix} 0 \\ 1 \\ \end {matrix} \right] $.

Let's assume the basis ($\vec{e_1}, \vec{e_2}$) looks like the vectors shown in the figure (The basis vectors and their linear combination).

My question is: what is the dot product of the basis vectors?

According to the matrix notation it should be zero: $ \left[ \begin {matrix} 1 \\ 0 \\ \end {matrix} \right] \cdot \left[ \begin {matrix} 0 \\ 1 \\ \end {matrix} \right] =0 $. But, according to the figure the dot product should not be zero as both vectors are clearly not orthogonal.

Obviously I am missing something. Any help would be appreciated.

$\endgroup$
  • $\begingroup$ Given two linearly independent vectors $\vec e_1,\vec e_2$ we can always define an inner product such that the two vectors are orthogonal ( and elements of an orthogonal basis). See:math.stackexchange.com/questions/1685621/… $\endgroup$ – Emilio Novati Jan 17 '18 at 22:16
  • $\begingroup$ I think you have it. In the standard basis, the basis vectors are orthogonal and $\bf {i\cdot j} = 0$ However, there is nothing that says that basis vectors must be orthogonal. In which case there is an inner product, in the new basis such that orthogonal vectors in the standard basis will have a $0$ inner product. However, it might not be the dot product your are familiar with. $\endgroup$ – Doug M Jan 17 '18 at 22:17
0
$\begingroup$

While the dot product in $\Bbb R^n$ is defined by the coordinates of the vectors with respect to the canonical basis, it is to be considered a function of the vectors, not of their coordinates. Thus when you change to a different basis, you do not change the value of their dot product. Instead, the dot product is expressed by a different formula. A concrete example may help.

In $\Bbb R^2$, let $\vec d_1 = (1,0), \vec d_2 = (0,1)$ be the canonical basis elements. And define $\vec e_1 = (1,1), \vec e_2 = (0,1)$, then $\vec e_1, \vec e_2$ are a non-orthogonal basis. Indeed $\vec e_1$ is not even normal: its length is $\sqrt 2$. Now let $\vec u = (u_1, u_2), \vec v = (v_1, v_2)$. We can express $\vec u, \vec v$ in terms of the $\vec e_1, \vec e_2$ basis: $$\vec u = u_1'\vec e_1 + u_2'\vec e_2\\\vec v = v_1'\vec e_1 + v_2'\vec e_2$$ If we solve the equations, it turns out that $u_1' = u_1$ and $u_2' = u_2 - u_1$, and similarly for $\vec v$. Now,

$$\begin{align}\vec u \cdot \vec v &= (u_1'\vec e_1 + u_2'\vec e_2) \cdot (v_1'\vec e_1 + v_2'\vec e_2)\\&=u_1'v_1'(\vec e_1 \cdot \vec e_1) + u_2'v_1'(\vec e_2 \cdot \vec e_1) + u_1'v_2'(\vec e_1 \cdot \vec e_2) + u_2'v_2'(\vec e_2 \cdot \vec e_2)\\&=u_1'v_1'(2) + u_2'v_1'(1) + u_1'v_2'(1) + u_2'v_2'(1)\end{align}$$

So in $\{\vec e_1, \vec e_2\}$ coordinates $$\begin{bmatrix}u_1' \\ u_2'\end{bmatrix}\cdot \begin{bmatrix}v_1' \\ v_2'\end{bmatrix} = \begin{bmatrix}u_1' & u_2'\end{bmatrix}\begin{bmatrix}2&1\\1&1\end{bmatrix}\begin{bmatrix}v_1' \\ v_2'\end{bmatrix}$$ while in the canonical $\{\vec d_1, \vec d_2\}$ coordinates $$\begin{bmatrix}u_1 \\ u_2\end{bmatrix}\cdot \begin{bmatrix}v_1 \\ v_2\end{bmatrix} = \begin{bmatrix}u_1 & u_2\end{bmatrix}\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}v_1 \\ v_2\end{bmatrix}$$

$\endgroup$
  • $\begingroup$ I see. My mistake stemmed from a naive approach to the dot product written in matrix form. Its only normal that the dot product does not change its value if one chooses to "see" the vector in a different way (that is, in another basis). If one were to work with a non-canonical basis, the change in coordinates must be reflected in the dot product as well. Thank you very much! $\endgroup$ – Iam Jan 18 '18 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.