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??

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.

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