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

Revision as of 05:32, 17 January 2020

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	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?	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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 | <math>T</math> is an isometry || There exists an orthonormal basis of <math>V</math> consisting of eigenvectors of <math>T</math> whose eigenvalues all have absolute value 1 || <math>T</math> 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
 |-
-| <math>T</math> is positive (positive semidefinite) || There exists an orthonormal basis of <math>V</math> consisting of eigenvectors of <math>T</math> with nonnegative real eigenvalues || <math>T</math> is diagonalizable using an orthonormal basis and the diagonal entries are all nonnegative real numbers ||
+| <math>T</math> is positive (positive semidefinite) || There exists an orthonormal basis of <math>V</math> consisting of eigenvectors of <math>T</math> with nonnegative real eigenvalues || <math>T</math> 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.
 |}