4 ms·
> A vector is pretty much by definition also a matrix, and there is a standard way to multiply matrices. A standard way to multiply a MxN with a NxK matrix, bu
by ReleaseCandidat 5y ago
> A vector is pretty much by definition also a matrix, and there is a standard way to multiply matrices.
A standard way to multiply a MxN with a NxK matrix, but none for a 1xN with a 1xN or a Nx1 with a Nx1 matrix - the two possible ways to describe a vector.
You have to transpose exactly one of the two vectors.
And then you have two possible results, 1xN multiplied with Nx1 yields a scalar (that's actually the 'usual' dot-product/scalar product/whatever you call it) and Nx1 multiplied with 1xN, where the result is a NxN matrix.
- thaumasiotes 5y agoDoesn't matter. We already didn't have a unique inverse, but it's perfectly possible to find a left pseudoinverse and a right pseudoinverse, bearing in mind that they're not unique. Though thinking about it more, it seems like the outer-product-inverse of a vector (a) must be unique if it exists; and (b) is highly unlikely to exist. > 1xN multiplied with Nx1 yields a scalar (that's actually the 'usual' dot-product/scalar product/whatever you call it) I'm aware of this, but there are two ways we might conceive of an "inverse": - Since a vector is a matrix, the inverse of a vector might be defined by matrix multiplication, where A is the inverse of B if AB is "the" identity matrix. This is only strictly defined for square matrices, but the pseudoinverse concept extends it to nonsquare matrices. - Or, we could go for a more basic sense of "multiplicative inverse", where the concept is that if AB = C, then B = A⁻¹C. This is what I was thinking of when saying that the concept of an inverse doesn't make sense when multiplication is the inner product - if I give you a vector v, and its inner product with some other vector u, there is no way of recovering what u was.