Can a matrix be transformed like a vector?

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This discussion centers on the relationship between matrices and vector spaces, specifically exploring whether a transformation matrix B can be derived from matrix A in a manner analogous to matrix M transforming vectors in vector space V into vector space W. The conclusion is that such a transformation is possible if matrix M is an isomorphism; however, it is not guaranteed in general cases. A critical condition for this transformation is that matrix A must map the kernel of M into itself, allowing for the induction of a linear map between quotient spaces V/K and V/K.

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Suppose I have a vector space V and a matrix M such that multiplying every vector in V by M creates another vector space W. Now suppose I have another matrix A that I can also use to change vectors in V into other vectors. Does there exist a third matrix B such that - for any vector v1 in V - if Av1 = v2, Mv1 = w1 and Mv2 = w2 then Bw1 = w2 ? In other words, is there a way to transform matrix A into a matrix B analogous to the way M changes vectors in V into vectors in W, so that a kind of homomorphism is arrived at between the relationship between A and vectors in V and the relationship between B and the vectors in W?
 
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yes if M is an isomorphiosm, but no in general. e.g. if Mv1 = w1 = 0, then Bw1 must be zero, but there is no reason to expect Mv2 to be zero. I hope got this straight. but at least it looks as if you would need A to map the kernel of M into itself. Then probably you are ok. basically you are asking whether, given a subspace K of V, and a linear map A:V-->V, when does A induce a linear map of quotient spaces V/K --> V/K. and the necessary condition is that A map K into K.
 
Thanks mathwonk, that was very helpful. It made me
— look up the difference between homomorphism and isomorphism (it's been a while!), and
— see another way that matrix arithmetic is different from number arithmetic, in particular, if Av=w there does not necessarily exist a kind of multiplicative inverse matrix A' such that A'w=v.
 

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