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It's funny because the whole discussion only exists because there is no division operator in "matrix math". For scalars we can simply write a / b (instead of a^
by rsp1984 10y ago
It's funny because the whole discussion only exists because there is no division operator in "matrix math". For scalars we can simply write a / b (instead of a^-1 * b) and everyone knows how to calculate it. But for some reason there's no such thing when dealing with matrices.
Well it turns out there is something like division for matrices and it's called factorization (followed by back-substitution) and there are actually many ways to do it, not just one (each having different numerical properties and complexity).
So whenever I see A^-1 * b in an equation I view it as something like b / A and I know there's the choice of "division methods" to use for this.
- cpburns2009 10y agoI think you meant "b / a" in "a / b (instead of a^-1 * b)".
- jacobolus 10y agoBy b / A he is implying a horizontal fraction bar with b above and A below. Of course, since matrix multiplication is not commutative, it’s probably better to write something like /A * b (or Matlab uses the cute notation A \ b) instead of b / A, but the intention is fairly clear from context. (In general, most of our mathematical notation conventions are designed for the real number system, and start falling over when we have “number-like objects” which behave a bit differently.)
- GFK_of_xmaspast 10y agoIt's a little more complicated than: """It's funny because the whole discussion only exists because there is no division operator in "matrix math".""" Matrices in general (insert caveats on size here) form a ring, but not a division ring (https://en.wikipedia.org/wiki/Division_ring https://en.wikipedia.org/wiki/Division_ring). This means that not all (square) matrices will have a multiplicative inverse. This is why we often restrict to things like GL(n), the group of non-singular square matrices of size n. Furthermore when you say "scalars", they have multiplicative inverses because they're elements of a field (and sometimes the inverse might be a pain like if you're in a number field or something). This isn't even getting into the Moore-Penrose pseudoinverse (https://en.wikipedia.org/wiki/Moore–Penrose_pseudoinverse https://en.wikipedia.org/wiki/Moore–Penrose_pseudoinverse) or linear operators or so on, but I'll let someone else shout about algebra.