How can linear a operator have more than one matrix representation? Let $A$ be linear operator on a vector space $V$. That is $A : V \to V$. How can a linear operator have more than one matrix representation ? ( As suggested in the Book Neilson and Chuang that matrix representation depends on choice of input and output basis ). What I think is that if we specify how the operator maps a basis of the vector space , then it is a unique operator and has a unique matrix representation . Am I missing something ? 
 A: Once you fix a basis, the matrix representing the operator in that basis is unique. But if you change the basis, the matrix representation in the new basis will be different, although the operator is the same. One of the main points in linear algebra is how to choose a basis so that the matrix representation is as simple as possible, the simplest being a diagonal matrix.
A: Lets take for an example the derivation operator on real polynomials of degree at most 2.
$$V=\{p \in \mathbb{R}[t]: \deg(p)\leq 2\}$$
$$D: V\to V, p\mapsto p'$$
We can choose several basis for the space of polynomials of degree at most 2, 
one possibility is the monomial basis $(f,g,h)$ with $f(t)=1, u(t)=t, v(t)=t^2$; another
is the Bernstein basis $(q,r,s)$ with $q(t)=(1-t)^2, r(t)=2t(1-t), s(t)=t^2$.
For obtaining a matrix representation we apply the operator on the basis vectors of the domain and express the result in the basis of the codomain. In my example I will chose
the same basis in the domain and codomain.
Monomial Basis
The derivatives of the basis vectors are
$$f'(t)=0,\quad g'(t)=1, \quad h'(t)=2t$$ 
therefore we have
$$D(f) = \mathbf{0},\quad D(g)=f,\quad D(h)=2\cdot g.$$
As a matrix:
$$\left(\begin{array}{ccc}0&1&0\\0&0&2\\0&0&0\end{array}\right)$$
note that the first column is $f'$ represented in $(f,g,h)$, the 
second column is $g'$ expressed in $(f,g,h)$ and the last column is $h'$ expressed in 
$(f,g,h)$.
Bernstein Basis
In the Bernstein Basis the derivations are a bit more involved:
$$q'(t) = 2t-2, \quad r'(t) = -4t+2,\quad s'(t) = 2t$$
you can verify that
$$ q' = -2q-r,\quad r'=2q-2s,\quad s'=r+2s$$
and therefore the matrix representation is
$$\left(\begin{array}{ccc}-2&2&0\\-1&0&1\\0&-2&2\end{array}\right)$$
Both matrices represent the same linear operator, derivation, but are expressed for
different basis of domain and codomain spaces.
You can verify that the two matrices share properties, because these properties are not only properties of the matrix, but of the underlying operator.
For example the two matrices have same rank, determinant and trace.
