# The subtle difference between matrices and linear operators

1. May 15, 2009

### 1230wc

For example, if I were to prove that all symmetric matrices are diagonalizable, may I say "view symmetric matrix A as the matrix of a linear operator T wrt an orthonormal basis. So, T is self-adjoint, which is diagonalizable by the Spectral thm. Hence, A is also so."

Is it a little awkward to specify a basis in the proof? Are linear operators and matrices technically two different classes of objects that may be linked by some "matrix representation function" wrt a basis? Thanks!

2. May 15, 2009

### HallsofIvy

Staff Emeritus
Let L be a linear transformation from vector space U to vector space V. If $\{u_1, u_2, ..., u_n}$ is a basis for U and ${v_1, v_2, ..., v_m} is a basis for V. Apply L to each [itex]u_i$ in turn and write it as a linear combination of the V basis. The coefficients give the ith column of the matrix representation of L.
(The bases do not have to be orthonormal. You just have to have some inner product defined on the space to talk about self-adjoint.)

But to go the other way, you don't have to say "view matrix A as a linear transformation". An n by m matrix is a linear transformation from vector space $R^n$ to $R^m$.