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I feel very similar. Do you have any recommendations for resources to learn about why matrix multiplication is defined that way?
by t3nary 10y ago
I feel very similar. Do you have any recommendations for resources to learn about why matrix multiplication is defined that way?
- darrickw 10y agoCheck out this series of videos "The Essence of Linear Algebra"[1] for a really powerful visual and intuitive explanation. It starts with vectors and builds to matrix multiplication and further to several other topics. [1]: https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2x...
- bandrami 10y ago"All the mathematics you missed but need to know for graduate school"[1] helped me a lot (and, in fact, I had and did). Once I had finished that, Cullen's "Matrices and linear transformations"[2] was really helpful too. But I wouldn't do Cullen if you're still, as I was, floundering with the concepts of why you're doing this in the first place. It's great once you have those concepts down. [1]: https://www.amazon.com/All-Mathematics-You-Missed-Graduate/dp/0521797071 https://www.amazon.com/All-Mathematics-You-Missed-Graduate/d... [2]: http://store.doverpublications.com/0486663280.html http://store.doverpublications.com/0486663280.html
- cousin_it 10y agoLet's say you have a system with N possible states, evolving in discrete steps. At each step, the system has some probability of switching from any state to any other state. That gives you an NxN matrix of switching probabilities. For example, if the system always stays in the same state as it started, the switching probabilities are an identity matrix (1 on the diagonal and 0 everywhere else). Now let's see what happens after two steps. If the system started out in state i, what's the probability that after two steps it will end up in state k? Well, it's the sum over all possible paths. In other words, the sum of probabilities of i->j->k for all possible j. In other words, the sum of p_{ij} times p_{jk} for j from 1 to N. But that's exactly the definition of multiplying a matrix by itself. Now it should be easy to understand that whenever you have matrices that represent transformations of some object, composing transformations will correspond to multiplying matrices.
- ww2 10y agoIf you compose two linear transformations into one step, and apply distributive property, you get the matrix multiplication rule. http://math.stackexchange.com/questions/31725/intuition-behind-matrix-multiplication http://math.stackexchange.com/questions/31725/intuition-behi...
- jperras 10y agoAxler's book, "Linear Algebra Done Right", is widely considered to be one of the best texts on linear algebra that focuses less on the mechanics and more on the structure of linear operators on vector spaces. http://linear.axler.net/ http://linear.axler.net/
- CalChris 10y agoYes, and the abridged version (no proofs, examples, and exercises) is downloadable for free. He avoids determinants altogether (well, until the 10th chapter). http://linear.axler.net/LinearAbridged.html http://linear.axler.net/LinearAbridged.html