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

Revision as of 23:44, 2 January 2019

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 name	Description in terms of eigenvectors	Description in terms of diagonalizability
$T$ 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)
$T$ 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 and the field of scalars is the complex numbers	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
$T$ is positive	There exists an orthonormal basis of $V$ consisting of eigenvectors of $T$ with positive real eigenvalues	$T$ is diagonalizable using an orthonormal basis and the diagonal entries are all positive real numbers

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 | <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> is an isometry ||
+| <math>T</math> is an isometry and the field of scalars is the complex numbers || 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>T</math> is positive ||
+| <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
 |}