User:IssaRice/Linear algebra/Classification of operators: Difference between revisions

Revision as of 13:28, 28 December 2021

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	Notes	Examples
$T$ is diagonalizable	There exists a basis of $V$ consisting of eigenvectors of $T$	$T$ is diagonalizable (there exists a basis $β$ of $V$ with respect to which $[T]_{β}^{β}$ 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 $T v = 1 v$ . Thus every basis $β = (v_{1}, \dots, v_{n})$ diagonalizes $T$ . The matrix of $T$ with respect to $β$ 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	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.

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 ! Operator kind !! Description in terms of eigenvectors !! Description in terms of diagonalizability !! Geometric interpretation !! Notes !! Examples
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-| <math>T</math> is diagonalizable || There exists a basis of <math>V</math> consisting of eigenvectors of <math>T</math> || <math>T</math> is diagonalizable (there exists a basis <math>\beta</math> of <math>V</math> with respect to which <math>[T]_\beta^\beta</math> 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 <math>\dim V</math> distinct eigenvalues so the diagonal matrix should be unique up to order? This result holds even if <math>V</math> is merely a vector space. || If <math>T</math> is the identity map, then every non-zero vector <math>v \in V</math> is an eigenvector of <math>T</math> with eigenvalue <math>1</math> because <math>Tv = 1v</math>. Thus every basis <math>\beta = (v_1,\ldots,v_n)</math> diagonalizes <math>T</math>. The matrix of <math>T</math> with respect to <math>\beta</math> is the identity matrix.
+| <math>T</math> is diagonalizable || There exists a basis of <math>V</math> consisting of eigenvectors of <math>T</math> || <math>T</math> is diagonalizable (there exists a basis <math>\beta</math> of <math>V</math> with respect to which <math>[T]_\beta^\beta</math> 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 <math>\dim V</math> distinct eigenvalues so the diagonal matrix should be unique up to order? This result holds even if <math>V</math> is merely a vector space with any field of scalars. || If <math>T</math> is the identity map, then every non-zero vector <math>v \in V</math> is an eigenvector of <math>T</math> with eigenvalue <math>1</math> because <math>Tv = 1v</math>. Thus every basis <math>\beta = (v_1,\ldots,v_n)</math> diagonalizes <math>T</math>. The matrix of <math>T</math> with respect to <math>\beta</math> is the identity matrix.
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 | <math>T</math> is normal || There exists an orthonormal basis of <math>V</math> consisting of eigenvectors of <math>T</math> || <math>T</math> is diagonalizable using an orthonormal basis ||