Mathematics-Online course: Linear Algebra-Matrices-Matrix Operations-Transpose of and adjoint Matrix

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	Mathematics-Online course: Linear Algebra - Matrices - Matrix Operations
	Transpose of and adjoint Matrix

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Interchanging rows and columns of matrix

we obtain the so called transposed matrix $B=A^{\operatorname t}$ , that is,

$\displaystyle b_{i,j} = a_{j,i}\, .$

If we, in addition, replace the complex entries of $A\in \mathbb{C}^{m\times n}$ by their complex conjugates, then we obtain the so called adjoint matrix $C=A^\ast = \bar A^{\operatorname t}$ , that is,

$\displaystyle c_{i,j} = \bar a_{j,i}\, .$

A matrix satisfying $A=A^{\operatorname t}$ is called symmetric. A matrix satisfying $A=A^\ast$ is called self-adjoint or Hermitian. For real matrices the notions 'hermitian' and 'symmetric' have the same meaning.

The following rules hold true:

$(A B)^{\operatorname t}= B^{\operatorname t}A^{\operatorname t}$ and $(A B)^\ast = B^\ast A^\ast$ ,
$(A^{\operatorname t})^{-1} = (A^{-1})^{\operatorname t}$ and $(A^\ast)^{-1} = (A^{-1})^\ast$ and

For the entries of $C^{\operatorname t}=(AB)^{\operatorname t}$ we have

$\displaystyle c^{\operatorname t}_{ji} = c_{ij} = \sum\limits_ka_{ik}b_{kj} = ... ...ame t}_{jk} = \sum\limits_k b^{\operatorname t}_{jk}a^{\operatorname t}_{kj}$

and, thus, $C^{\operatorname t}= B^{\operatorname t}A^{\operatorname t}$ . Since $\overline{a+b}=\bar a + \bar b$ and $\overline{ab}=\bar a \bar b$ , the respective assertion for adjoint matrices also holds true.

From

$\displaystyle E=E^{\operatorname t}= \left(AA^{-1}\right)^{\operatorname t}= \left(A^{-1}\right)^{\operatorname t} A^{\operatorname t}$

it follows that $\left(A^{-1}\right)^{\operatorname t}= \left(A^{\operatorname t}\right)^{-1}$ . We find similar results for adjoint matrices.

(Authors: Burkhardt/Höllig/Hörner)

automatically generated 4/21/2005