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. Below, we consider only complex operators, or the complexification of a real operator.

Operator kind	Description in terms of eigenvectors	Description in terms of diagonalizability	Geometric interpretation	Algebraic property	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? This result holds even if $V$ is merely a vector space with any field of scalars.	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		$TT^{*}=T^{*}T$
$T$ self-adjoint (aka 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		$T=T^{*}$
$T$ is anti-self-adjoint (aka skew-Hermitian or anti-Hermitian)	There exists an orthonormal basis of $V$ consisting of eigenvectors of $T$ with pure imaginary eigenvalues	$T$ is diagonalizable using an orthonormal basis and the diagonal entries are all pure imaginary		$T^{*}=-T$
$T$ is an isometry (aka unitary in a complex vector space, or orthogonal in a real vector space)	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		$TT^{*}=T^{*}T=TT^{-1}=I$	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	Polar decomposition says an arbitrary linear operator can be written as a positive operator followed by a rotation (isometry). In polar decomposition, the positive operator step chooses orthogonal directions in which to stretch or shrink, so that we have a tilted ellipse, and the isometry rotates that ellipse. So a positive operator is simply one that does not require the second step. In other words, for a positive operator you can find some orthogonal "coordinate axes" along which to scale.

Acknowledgments: Thanks to Philip B. for feedback on this page.