punnoose Posted February 23, 2011 Share Posted February 23, 2011 i am a newbie to linear algebra My question is Let T of dimension mxn, be a linear map from A -> B. if n<m and rank of T is n, which basically means all of the vector space A is mapped,and null space is empty. Then why there is no inverse for T?? My understanding is that for A_inv to exist, all the vector space B should be one to one mapped back to A. This is not possible because vector space B is of high dimension(m) and thus have more elements than A, which is of low dimension(n). Is this view right?? Link to comment Share on other sites More sharing options...
ajb Posted February 23, 2011 Share Posted February 23, 2011 For a matrix to be invertible (have a left and right inverse) it needs to be square. Thinking about how to construct the identity matrix of a given dimension should convince you of this. However, not all square matrices are invertible, so just matching the dimensions will not in general be enough. Link to comment Share on other sites More sharing options...
Bignose Posted February 23, 2011 Share Posted February 23, 2011 There is a concept of a pseudo-inverse. However, it does not have a unique solution. More info here: http://en.wikipedia.org/wiki/Moore%E2%80%93Penrose_pseudoinverse Link to comment Share on other sites More sharing options...
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