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I'm hoping to derive an equation for the dot product using Cramer's rule. Here I'm going to try in $\mathbb{R}^2$ and will generalize once I get this first issue cleared.

I'm hoping to arrive at an expression for the dot product by first asking how to describe some vector $\mathbf{r}$ in terms of two basis vectors $\mathbf{\hat{m}}, \mathbf{\hat{n}}$. If you can, then the coefficients of the linear combination ought to be equal to the projection of $\mathbf{r}$ onto one of its basis vectors.

Thus by solving the system $$\mathbf{r} = \alpha\mathbf{\hat{m}} + \beta\mathbf{\hat{n}}$$

in the unknowns $\alpha, \beta$, I want to see that $\alpha = \mathbf{r}\cdot \mathbf{\hat{m}}$ and $\beta = \mathbf{r} \cdot \mathbf{\hat{n}}$

Using Cramer's rule and the fact that $\det A = \det A^T$, we see that

$$\alpha = \frac{ \begin{array}{|cc|} r_x & r_y \\ n_x & n_y \end{array} } { \begin{array}{|cc|} m_x&m_y\\ n_x&n_y \end{array} } \qquad \beta = \frac{ \begin{array}{|cc|} m_x & m_y \\ r_x & r_y \end{array} } { \begin{array}{|cc|} m_x&m_y\\ n_x&n_y \end{array} } $$

Now it seems that $\alpha \ne \mathbf{r}\cdot\mathbf{\hat{m}} = r_xm_x + r_ym_y = \begin{array}{|cc|} r_x & r_y \\ -m_y & m_x \end{array} $ and similarly our expectations did not hold for $\beta$.

Can someone please explain what I am misunderstanding that leads to this unexpected conclusion

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  • $\begingroup$ (Slightly different way of putting Matt Samuel’s answer and the ensuing comments, below). Your idea that the coordinates relative to a basis are basically projections is a good one, but you’ve overlooked the fact that in general they’re not orthogonal projections (defining orthogonality via the standard Euclidean inner product, since you’re trying to recreate it). You have, however found a way to generate other inner products on this space, and with those inner products come alternative definitions of length and orthogonality. $\endgroup$ – amd May 9 '17 at 23:12
  • $\begingroup$ Wow pretty cool. I'm still in the dark about two things. Can you elaborate (1) what my alternative definitions of length and orthogonality are, (2) what this alternative inner product would be as an explicit formula, (3) under what algebraic conditions on $\mathbf{\hat{m}, \hat{n}}$ do we get that $\alpha=r\cdot m$ and $\beta = r\cdot n$? $\endgroup$ – theideasmith May 10 '17 at 0:16
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This only works if the two basis vectors are orthonormal. Putting that in your assumption presupposes the conclusion. However, there is another way to see it: this should define a new inner product, not necessarily the dot product, in which they are orthonormal.

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  • $\begingroup$ The reason I wanted to do this is to see the dot product as a projection from a purely algebraic standpoint. 2 questions (1) what would putting the two basis vectors as being orthonormal look like algebraically (as I haven't actually defined the dot product yet in this derivation), (2) can you explain to someone who hasn't actually studied linear what "a new inner product in which they are orthonormal" means $\endgroup$ – theideasmith May 9 '17 at 17:53
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    $\begingroup$ @theideasmith The dot product depends on a choice of basis. It's not natural in that sense, and there's no reason to expect it to just fall out of the equations. You will end up with an inner product, but it may not be the dot product. $\endgroup$ – Matt Samuel May 9 '17 at 17:55
  • $\begingroup$ @the en.wikipedia.org/wiki/Inner_product_space $\endgroup$ – Matt Samuel May 9 '17 at 17:56
  • $\begingroup$ Thanks for this. But I'm still at the bottom of the linear algebra ladder. Would you mind explaining how I can arrive at the dot product algebraically and what would the condition of orthonormal look like algebraically. $\endgroup$ – theideasmith May 9 '17 at 17:58
  • $\begingroup$ @the The dot product is generally defined by an explicit formula and is very simple. That's about as algebraic as it gets. Two vectors are orthonormal if they both have length $1$ and their dot product is $0$. $\endgroup$ – Matt Samuel May 9 '17 at 18:00

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