5 ms·
There is a brute force solution, you get equations in eight variables (the matrix entries) which you can solve. Not elegant, but it works. Also, WLOG (without
by impendia 15y ago
There is a brute force solution, you get equations in eight variables (the matrix entries) which you can solve. Not elegant, but it works.
Also, WLOG (without loss of generality) the top left entry of C and D is either 0 or 1, because you can multiply either matrix by any number other than 0 without changing whether the equation is true.
- gruseom 15y agoRight, the brute force way is better than sheer guesswork, but it's still too low-level. Kind of like solving the problem in assembly language. The intention of the text is clearly to get one thinking about these operations at a higher level, "at the level of the problem" as we would say in software. Your WLOG is elegant, though, in that it gives a simpler starting point that's guaranteed to work if anything does. I'll have to remember that trick.
- impendia 15y agoI feel like there is unlikely to be a solution which doesn't involve (1) brute force, or (2) higher-level theory (i.e. eigenvectors and eigenvalues). Prove me wrong, and I will be tripping over myself to upvote you ;)
- gruseom 15y agoCertainly it's artificial to restrict oneself to primitive concepts, unless of course you don't know about eigenvalues yet or forgot it all. :) And it is an exercise from the earliest chapter, so we're supposed to stay in simple-land. Did you look at my solution (http://news.ycombinator.com/item?id=3129023 http://news.ycombinator.com/item?id=3129023)? There's no higher-level theory in it. Is it brute force? Not sure. At least it doesn't require solving any element-wise equations.
- impendia 15y agoIt's elegant, but as far as I can see, your argument only holds for the single matrix C which you describe.