# Linear Operators and their representations

I am currently learning Quantum mechanics on a slightly advanced level. I am curious in knowing if there are Linear Operators (Linear Maps) in the Hilbert Space (finite dimensional ones) that don't have isomorphic with Matrices? In those case are there any other representations that we can choose?

• I am not sure what you're asking, but if you're asking if any linear transformation in a vector space can be given by a matrix in some coordinate system, the answer is yes. Jan 29 '14 at 5:20
• But I guess, I was talking about finite dimensional vector spaces. Momentum is an operator in continuously infinite dimensional Hilbert space. Jan 29 '14 at 5:44
• That's what I get for not reading.
– BMS
Jan 29 '14 at 5:51
• @user35952 No, momentum is not an operator in continuously infinite dimensional Hilbert space. The Hilbert space is $L^2(\mathbb R)$, which is a countably infinite dimensional Hilbert space. As it is separable, it admits a countable Hilbert basis and all bases have the same cardinality. Jan 29 '14 at 8:02
• @V.Moretti : I am sorry, I am at loss to understand you, can elucidate further about this separability Jan 29 '14 at 8:32

## 1 Answer

The space of linear operators on an $n$-dimensional vector space $V$ over a field $F$ is always isomorphic to the space of $n \times n$ matrices over $F$.

It is easy enough to see that any matrix is a linear map from $V$ to $V$ -- just left-multiply the column-vector representation of the input by the matrix. For the other direction, choose a basis $\{\hat{e}_i\}$ for $V$. Let the $k$-th column of a matrix $M$ be the column-vector representation of $\Omega(\hat{e}_k)$, where $\Omega$ is your operator. That is, $M_{ik} = \langle \hat{e}_i \vert \Omega \hat{e}_k\rangle$.

• Thanks. What about the existence of basis vectors for a space(might sound stupid, but just curious), are there any cases where they won't exist ? Jan 29 '14 at 5:33
• Also, I am interested in knowing is there some group to which linear operators in continuously infinite dimensional vector space are isomorphic to. Jan 29 '14 at 5:35
• @user35952 The existence of a basis for a vector space requires the axiom of choice! proofwiki.org/wiki/Vector_Space_has_Basis Jan 29 '14 at 5:40
• Note that your first sentence is true provided the field of the vector space is isomorphic to the field that the entries of the matrices are in. Jan 29 '14 at 7:13