So I was day dreaming about linear algebra today (in a class which had nothing to do with linear algebra), when I stumbled across an interesting relationship. I was thinking about how determinants are really the area spanned by column vectors, and I had the thought that one could measure linear independence (in $R^2$ in this case) on a scale from 0 to 1 by taking the ratio $\frac{|det(\vec{a},\vec{b})|}{|\vec{a}||\vec{b}|}$ where $det(\vec{a},\vec{b})$ is the determinant of a matrix whose column vectors are $\vec{a}$ and $\vec{b}$. Then I remembered the definition of the cross product in $R^2$, $\vec{a}\times\vec{b}= \begin{vmatrix}i&j&k\\a_1&a_2&0\\b_1&b_2&0 \end{vmatrix} = \begin{vmatrix}a_1&a_2\\b_1&b_2\end{vmatrix}k = det(\vec{a},\vec{b}) k$ . Taking the absolute value, we have $|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|sin(\theta) = |det(\vec{a},\vec{b})|$. Finally, rearranging we have $sin(\theta) = \frac{|det(\vec{a},\vec{b})|}{|\vec{a}||\vec{b}|}$. This can be thought of as comparing actual area spanned to maximum possible area spanned. With this definition, we can generalize this notion of measuring linear independence to higher dimensions (e.g. for $R^3$, $sin(\theta) = \frac{|det(\vec{a},\vec{b},\vec{c})|}{|\vec{a}||\vec{b}||\vec{c}|}$). Now clearly this isn't "really" $sin(\theta)$, but the notion should be the same, i.e. comparing actual spanned volume to maximum possible spanned volume. Finally we note that $|det(\vec{a},\vec{b},\vec{c})|$ may also be written as $|\vec{a}\wedge\vec{b}\wedge\vec{c}|$, allowing us to write $sin(\theta) = \frac{|\vec{a}\wedge\vec{b}\wedge\vec{c}|}{|\vec{a}||\vec{b}||\vec{c}|}$.

Which Leads Us To My Real Question:

Now, in $R^2$, we have the equation for cosine, $cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$, which looks an awful lot like the corresponding equation for sine, $sin(\theta) = \frac{|\vec{a}\wedge\vec{b}|}{|\vec{a}||\vec{b}|}$. The difference is, we have no problem generalizing the equation for sine to an arbitrary amount of dimensions. We simply need to tack on another wedge product, i.e. $sin(\theta) = \frac{|\vec{v_1}\wedge\vec{v_2}\cdots\wedge\vec{v_n}|}{|\vec{v_1}||\vec{v_2}|\cdots|\vec{v_n}|}$. My question is: Is there a similar way to generalize cosine using a simple combination of products in the numerator? Perhaps in geometric algebra? Maybe it would somehow extend the analogy of cosine being the ratio of actual projection to maximum possible projection?

Thanks in advance!

Edit: To anyone who is interested, I figured out since the denominator of the cosine generalization would be $|\vec{v}|^3$ for 3 vectors $\vec{v}$ that are all in the same direction. Therefore, the numerator would need to equal $|\vec{v}|^3$ to yield an answer of 1 in the case that all the vectors are parallel.


1 Answer 1


I think really the answer to your question is that, as we all know, the scalar product works perfectly well in any number of dimensions, so your definition $$ \cos{\theta} = \frac{a \cdot b}{|a||b|} $$ is valid in $n$ dimensions as well. This is saying that the notion of an "angle between rays" is well-defined in $n$ dimensions.

Your formula for the sine is really saying something about the "size" of volume forms (i.e. top-grade elements in the Grassmann algebra $\Lambda$) in terms of the size of their bewedged vectors. The dual of an $n$-form being a $0$-form, i.e. a scalar, this is easy to formulate sensibly. A more difficult question is how to formulate this on the other grades of the Grassmann algebra, to which I must alas protest mine own ignorance.

  • $\begingroup$ I think (perhaps due to my lack of emphasis) that you may have misunderstood my question. I mentioned that the regular definition of $sin(\theta)$ of two vectors, $sin(\theta) = \frac{|\vec{a}\wedge\vec{b}|}{|\vec{a}||\vec{b}|}$ can be thought of as a ratio of two areas: the area spanned by the two vectors (the determinant), and the maximum possible area allowable by the lengths of the vectors. Now, I proposed that you could somewhat extend the definition of $sin(theta)$ to THREE vectors, by thinking of it as a ratio of volumes instead. I was hoping to do the same for cosine. $\endgroup$
    – SSD
    Apr 8, 2015 at 23:15
  • $\begingroup$ You mean something along the lines of (using summation convention) $\frac{F_{ab}G^{ab}}{\sqrt{F^2} \sqrt{G^2}}$, where $F^2=F_{ab}F^{ab}$? $\endgroup$
    – Chappers
    Apr 8, 2015 at 23:22
  • $\begingroup$ Not quite. It must be a function of three vectors, and might be in the form (if we expect the trend to continue), of $\frac{actual projection}{maximum possible projection}$ where the maximum possible projection is simply the norms of the vectors multiplied together. What you're saying is actually an interesting question too though. $\endgroup$
    – SSD
    Apr 9, 2015 at 0:04
  • $\begingroup$ What I've written down there is basically a matrix norm. Well, as far as three vectors go, there's the scalar triple product in 3 dimensions, but after that you're going to start running into difficulties: the Grassmann algebra has a way of getting to a scalar by wedging $k$-forms and $n-k$-forms together (which you can use to build a scalar product using the Hodge dual), but beyond that I don't believe a canonical example can exist. To think about it another way, there's also the question of isotropic tensors on spaces of different dimension (which you need to be basis-independent). $\endgroup$
    – Chappers
    Apr 9, 2015 at 0:23

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