Editing Transpose (section)

=== Properties ===

Let {{math|'''A'''}} and {{math|'''B'''}} be matrices and {{mvar|c}} be a [[Scalar (mathematics)|scalar]]. 

* <math>\left(\mathbf{A}^\operatorname{T} \right)^\operatorname{T} = \mathbf{A}.</math>
*:The operation of taking the transpose is an [[Involution (mathematics)|involution]] (self-[[Inverse matrix|inverse]]).
*<math>\left(\mathbf{A} + \mathbf{B}\right)^\operatorname{T} = \mathbf{A}^\operatorname{T} + \mathbf{B}^\operatorname{T}.</math>
*:The transpose respects [[Matrix addition|addition]].
*<math>\left(c \mathbf{A}\right)^\operatorname{T} = c (\mathbf{A}^\operatorname{T}).</math>
*:The transpose of a scalar is the same scalar. Together with the preceding property, this implies that the transpose is a [[linear map]] from the [[Vector space|space]] of {{math|{{nowrap|''m'' × ''n''}}}} matrices to the space of the {{math|{{nowrap|''n'' × ''m''}}}} matrices.
*<math>\left(\mathbf{A B}\right)^\operatorname{T} = \mathbf{B}^\operatorname{T} \mathbf{A}^\operatorname{T}.</math>
*:The order of the factors reverses. By induction, this result extends to the general case of multiple matrices, so 
*::{{math|('''A'''<sub>1</sub>'''A'''<sub>2</sub>...'''A'''<sub>''k''−1</sub>'''A'''<sub>''k''</sub>)<sup>T</sup>&nbsp;{{=}}&nbsp;'''A'''<sub>''k''</sub><sup>T</sup>'''A'''<sub>''k''−1</sub><sup>T</sup>…'''A'''<sub>2</sub><sup>T</sup>'''A'''<sub>1</sub><sup>T</sup>}}.
*<math>\det \left(\mathbf{A}^\operatorname{T}\right) = \det(\mathbf{A}).</math>
*:The [[determinant]] of a square matrix is the same as the determinant of its transpose.
*The [[dot product]] of two column vectors {{math|'''a'''}} and {{math|'''b'''}} can be computed as the single entry of the matrix product<math display=block>\mathbf{a} \cdot \mathbf{b} = \mathbf{a}^{\operatorname{T}} \mathbf{b}.</math>
*If {{math|'''A'''}} has only real entries, then {{math|'''A'''<sup>T</sup>'''A'''}} is a [[positive-semidefinite matrix]].
*<math> \left(\mathbf{A}^\operatorname{T} \right)^{-1} = \left(\mathbf{A}^{-1} \right)^\operatorname{T}.</math>
*: The transpose of an invertible matrix is also invertible, and its inverse is the transpose of the inverse of the original matrix.<br>The notation {{math|'''A'''<sup>−T</sup>}} is sometimes used to represent either of these equivalent expressions.
*If {{math|'''A'''}} is a square matrix, then its [[Eigenvalue, eigenvector and eigenspace|eigenvalues]] are equal to the eigenvalues of its transpose, since they share the same [[characteristic polynomial]].
*<math> \left(\mathbf A\mathbf a\right) \cdot \mathbf b =\mathbf a \cdot \mathbf \left(A^T\mathbf b\right)</math> for two column vectors <math> \mathbf a, \mathbf b </math> and the standard [[dot product]].
*Over any field <math>k</math>, a square matrix <math>\mathbf{A}</math> is [[matrix similarity|similar]] to <math>\mathbf{A}^\operatorname{T}</math>.
*:This implies that <math>\mathbf{A}</math> and <math>\mathbf{A}^\operatorname{T}</math> have the same [[invariant factors]], which implies they share the same minimal polynomial, characteristic polynomial, and eigenvalues, among other properties.
*:A proof of this property uses the following two observations.
*:* Let <math>\mathbf{A}</math> and <math>\mathbf{B}</math> be <math>n\times n</math> matrices over some base field <math>k</math> and let <math>L</math> be a [[field extension]] of <math>k</math>. If <math>\mathbf{A}</math> and <math>\mathbf{B}</math> are similar as matrices over <math>L</math>, then they are similar over <math>k</math>. In particular this applies when <math>L</math> is the [[algebraic closure]] of <math>k</math>.
*:*If <math>\mathbf{A}</math> is a matrix over an algebraically closed field in [[Jordan normal form]] with respect to some basis, then <math>\mathbf{A}</math> is similar to <math>\mathbf{A}^\operatorname{T}</math>. This further reduces to proving the same fact when <math>\mathbf{A}</math> is a single Jordan block, which is a straightforward exercise.