>>11096015>diagonalizationThat's just reorienting your basis so that your matrix becomes simple and recognizeable. All your equations become as simple as c_i x_i = b_i (no sum convention). Very simple to solve.
>find eigenvectorsOnce you know evalue c, that's just solving Ax = cx, or equivalently (A-cI)x=0, which reduces to finding a nontrivial vector in the kernel. For that, just do rref, multiplying by invertible elementary matrices on the left doesn't change the solution space.
>Change of basis is freaky with the calculation because it won't stick in my head but the abstract concept of it is clear to meIf the abstract concept is clear to you, the actual calculation should be clear to you too. Think about it this way, change of basis matrix P tells you what your new vectors are in terms of the old vectors in the bases. Pe_i is the i'th new vector in terms of old basis. If matrix A tells you how to take in old basis vectors and outputs old basis vectors, AP works like this (read from the right to left) vector in terms of new basis -(P)-> in terms of old basis --A--> output in terms of old basis. So all you need in the end is to convert the old basis to new basis , which is exactly what multiplying by P^-1 on the left means. So the linear map represented by A in the new basis is P^-1 A P. Easy really, when you think of matrices as representations of linear maps in terms of some basis.
>The larger issue is these types of exercises won't melt into my knowledge base because I can't connect them with the rest of the systemThey will once you understand them.
