User:IssaRice/Linear algebra/Classification of operators

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 $\beta =(v_1,\dots ,v_n)$ 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 (positive semidefinite)	There exists an orthonormal basis of $V$ consisting of eigenvectors of $T$ with nonnegative real eigenvalues	$T$ is diagonalizable using an orthonormal basis and the diagonal entries are all nonnegative real numbers