If I fix a basis, can I have different matrices representing the same linear transformation?

Question

let $T(x,y)=(x+y,x-y)$. Now if I fix my basis to be ${(1,-1),(0,1)}$, then we see that $T(1,-1)=(0,2)=0(1,-1)+2(0,1)$ and $T(0,1)=(1,-1)=(1,-1)+0(0,1)$. So we can say that the associated matrix is $$ \begin{pmatrix} 0 & 1 \\ 2 & 0 \\ \end{pmatrix} $$ but what stops me from expressing $ T(1,-1)=(0,2)=2(0,1)+0(1,-1)$ and $T(0,1)=(1,-1)=0(0,1)+(1,-1)$ leading to an associated matrix of $$ \begin{pmatrix} 2 & 0 \\ 0 & 1 \\ \end{pmatrix} $$

Answer 1

The matrix representing a linear transformation is unique for a given ordered basis. That is, changing the order of the basis elements may change the matrix representation, as you have shown.

A simpler thing to think about is that when the order of the basis elements changes, the vector representation for a given element of the vector space will change as well.

Answer 2

Yes of course the matrix representation, as for the vectors representation, depends upon the basis chosen. What is invariant is the rank and then the dimensions of $Im(T)$ and $Ker(T)$.