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Re:Linear Algebra: What matrices are by philh on Saturday August 29, 2015 @11:07AM (#10140216)
Traditionally, multiplication takes two elements of a set and turns them into another element of that set. The dot product doesn't do that, it takes two vectors and turns them into a scalar.

The cross product only exists in three dimensions. And it's not associative (A×B×C gives a different answer depending which order you do it in), which is another thing multiplication usually satisfies.

There are two other not-quite-multiplication operators that I recall seeing. There's an analog of the cross product in two dimensions: (a,b,0)×(c,d,0) = (0,0,ad-bc), so it can be useful to have an operator (a,b)×(c,d) = ad-bc, again turning two vectors into a scalar.

And if the dot product is defined in terms of matrix multiplication by A·B = AᵀB, then you can also define an operator ABᵀ, turning two vectors into a matrix. These vectors don't even need to have the same length.

Re:Linear Algebra: What matrices are by tanderson92 on Saturday August 29, 2015 @12:51PM (#10140597)
> The cross product only exists in three dimensions

While true, the Wedge product is a useful concept that generalizes a cross product to arbitrary dimensions and is used in multivariable calculus for proving various integral theorems in high dimension. Here, generalized Stokes theorems apply despite the cross product not being defined. Admittedly, it isn't really a map on the vector space, but the fact that Stokes theorems still hold makes it pretty darn useful to me.

Re:Linear Algebra: What matrices are by jules on Saturday August 29, 2015 @07:44PM (#10141758)
FYI, in geometric algebra you can truly multiply two vectors and get another element of the geometric algebra out, no matter the dimensionality.
Re:Linear Algebra: What matrices are by tomadi on Saturday August 29, 2015 @09:43PM (#10141980)
But the product isn't a vector of the same shape as the inputs, so it isn't a vector multiplication. It is tensor multiplication.