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

Latest revision as of 13:45, 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	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 $β$ 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		$T T^{} = T^{} T$	A normal operator has the additional property that it can be written as $T = S + A$ , where $S$ is a self-adjoint operator and $A$ is an anti-self-adjoint operator, and where $S$ and $A$ are simultaneously diagonalizable using a single orthonormal basis
$T$ 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		$T^{*} = - T$		90-degree rotation of the plane?
$T$ 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		$T T^{} = T^{} T = T T^{- 1} = I$	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.

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-Let <math>V</math> be a finite-dimensional inner product space, and let <math>T : V \to V</math> be a linear transformation. Then in the table below, the statements within the same row are equivalent.
+Let <math>V</math> be a finite-dimensional inner product space, and let <math>T : V \to V</math> 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.
 {| class="sortable wikitable"
 |-
-! Operator name !! Description in terms of eigenvectors !! Description in terms of diagonalizability !! Notes
+! Operator kind !! Description in terms of eigenvectors !! Description in terms of diagonalizability !! Geometric interpretation !! Algebraic property !! Notes !! Examples
 |-
-| <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)
+| <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.
 |-
-| <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
+| <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 || || <math>TT^* = T^*T</math> || A normal operator has the additional property that it can be written as <math>T = S + A</math>, where <math>S</math> is a self-adjoint operator and <math>A</math> is an anti-self-adjoint operator, and where <math>S</math> and <math>A</math> are simultaneously diagonalizable using a single orthonormal basis ||
 |-
-| <math>T</math> self-adjoint (<math>T</math> is Hermitian) || There exists an orthonormal basis of <math>V</math> consisting of eigenvectors of <math>T</math> with real eigenvalues || <math>T</math> is diagonalizable using an orthonormal basis and the diagonal entries are all real
+| <math>T</math> self-adjoint (aka Hermitian) || There exists an orthonormal basis of <math>V</math> consisting of eigenvectors of <math>T</math> with real eigenvalues || <math>T</math> is diagonalizable using an orthonormal basis and the diagonal entries are all real || || <math>T = T^*</math> ||
 |-
-| <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 anti-self-adjoint (aka skew-Hermitian or anti-Hermitian) || There exists an orthonormal basis of <math>V</math> consisting of eigenvectors of <math>T</math> with pure imaginary eigenvalues || <math>T</math> is diagonalizable using an orthonormal basis and the diagonal entries are all pure imaginary || || <math>T^* = -T</math> || || 90-degree rotation of the plane?
 |-
-| <math>T</math> is positive || There exists an orthonormal basis of <math>V</math> consisting of eigenvectors of <math>T</math> with positive real eigenvalues || <math>T</math> is diagonalizable using an orthonormal basis and the diagonal entries are all positive real numbers
+| <math>T</math> is an isometry (aka unitary in a complex vector space, or orthogonal in a real vector space) || 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 || || <math>TT^* = T^*T = TT^{-1} = I</math> || 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 || 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.