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User:IssaRice/Linear algebra/Classification of operators

< User:IssaRice

Let V be a finite-dimensional inner product space, and let T : V \to V be a linear transformation. Then in the table below, the statements within the same row are equivalent.

Operator kind Description in terms of eigenvectors Description in terms of diagonalizability Notes Examples
T is diagonalizable There exists a basis of V consisting of eigenvectors of T T is diagonalizable (there exists a basis \beta of V with respect to which [T]_\beta^\beta is a diagonal matrix) This basis is not unique because we can reorder the vectors and also scale eigenvectors by a non-zero number to obtain an eigenvector. But there are at most \dim V distinct eigenvalues so the diagonal matrix should be unique up to order? If T is the identity map, then every non-zero vector v \in V is an eigenvector of T with eigenvalue 1 because Tv = 1v. Thus every basis \beta = (v_1,\ldots,v_n) diagonalizes T. The matrix of T with respect to \beta is the identity matrix.
T is normal There exists an orthonormal basis of V consisting of eigenvectors of T T is diagonalizable using an orthonormal basis
T self-adjoint (T is Hermitian) There exists an orthonormal basis of V consisting of eigenvectors of T with real eigenvalues T is diagonalizable using an orthonormal basis and the diagonal entries are all real
T is an isometry There exists an orthonormal basis of V consisting of eigenvectors of T whose eigenvalues all have absolute value 1 T is diagonalizable using an orthonormal basis and the diagonal entries all have absolute values 1 This only works when the field of scalars is the complex numbers
T is positive There exists an orthonormal basis of V consisting of eigenvectors of T with positive real eigenvalues T is diagonalizable using an orthonormal basis and the diagonal entries are all positive real numbers