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# complex conjugate

# 1 Definition

# 1.1 Scalar Complex Conjugate

Let $z$ be a complex number with real part $a$ and imaginary part $b$,

$z=a+bi$ |

Then the *complex conjugate* of $z$ is

$\bar{z}=a-bi$ |

Complex conjugation represents a reflection about the real axis on the Argand diagram representing a complex number.

$\int_{{-\infty}}^{{\infty}}\Psi^{{*}}\Psi dx=1$ |

where $\Psi^{*}$ is the complex conjugate of a wave function.

# 1.2 Matrix Complex Conjugate

Let $A=(a_{{ij}})$ be a $n\times m$ matrix with complex
entries. Then the *complex conjugate* of $A$ is the matrix
$\overline{A}=(\overline{a_{{ij}}})$. In particular, if
$v=(v^{1},\ldots,v^{n})$ is a complex row/column vector, then
$\overline{v}=(\overline{v^{1}},\ldots,\overline{v^{n}})$.

Hence, the matrix complex conjugate is what we would expect: the same matrix with all of its scalar components conjugated.

# 2 Properties of the Complex Conjugate

# 2.1 Scalar Properties

If $u,v$ are complex numbers, then

1. $\overline{uv}=(\overline{u})(\overline{v})$

2. $\overline{u+v}=\overline{u}+\overline{v}$

3. $\big(\overline{u}\big)^{{-1}}=\overline{u^{{-1}}}$

4. $\overline{(\overline{u})}=u$

5. If $v\neq 0$, then $\overline{(\frac{u}{v})}={\overline{u}}/{\overline{v}}$

6. Let $u=a+bi$. Then $\overline{u}u=u\overline{u}=a^{2}+b^{2}\geq 0$ (the complex modulus).

7. If $z$ is written in polar form as $z=re^{{i\phi}}$, then $\overline{z}=re^{{-i\phi}}$.

# 2.2 Matrix and Vector Properties

Let $A$ be a matrix with complex entries, and let $v$ be a complex row/column vector.

Then

1. $\overline{A^{T}}=\big(\overline{A}\big)^{T}$

2. $\overline{Av}=\overline{A}\overline{v}$, and $\overline{vA}=\overline{v}\overline{A}$. (Here we assume that $A$ and $v$ are compatible size.)

Now assume further that $A$ is a complex square matrix, then

1. $\operatorname{trace}\overline{A}=\overline{(\operatorname{trace}\ A)}$

2. $\det\overline{A}=\overline{(\det A)}$

3. $\big(\overline{A}\big)^{{-1}}=\overline{A^{{-1}}}$

## Mathematics Subject Classification

12D99*no label found*30-00

*no label found*32-00

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## Comments

## Do matrix-vector properties apply for a general linear opera...

Hello,

Does the $\overline{Av} = \overline{A}\overline{v}$ result rely upon $A$ being finite dimensional, or will it hold for a general linear operator?

Thanks,

Rhys

## Re: Do matrix-vector properties apply for a general linear o...

Hi Rhys,

In my opinion the answer is yes, but it is necessary to define bases in the spaces between which the linear operator maps. What does this mean?

Let A,B be spaces of dimension n, m resp., and let L be a linear operator such that

L: A \mapsto B.

So for every element x in A we have y = Lx, where y is some element in B.

Next let's choose bases e_j (j=1,...,n), f_i (i=1,...,m) in A,B resp. Thus, if x=(x_1,...,x_n), y=(y_1,...,y_m), we may writing out the linear system

y_i = L_{ij} x_j.

A,B are defined over the field F where the entries L_{ij} belong.

Then we say that the matrix ||L_{ij}|| is associated to the operator L. Obviously, if we change any basis, then another matrix will represent the operator L.

perucho

## Re: Do matrix-vector properties apply for a general linear o...

What do you mean by \overline{Av}, \overline{A}, and \overline{v}? Are these representations of the vector Av, the matrix A, and the vector v with respect to some basis for your space? What do you mean by "$A$ being finite dimensional?" Do you mean that the space under considered (on which A acts) is finite dimensional?

- Keenan