Math 511: Linear Algebra

3.2 The Laplace Expansion

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3.2.1 Geometry of the Determinant

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The determinant of an $n\times n$ matrix is the scaling factor of the change in volume(area) in $n$-dimensional space taking orientation into account.

In two dimensions, we consider a matrix $A$, the unit square $S$ and the image of the unit square under the application of the linear transformation $A$. For example if

$$ A = \begin{bmatrix} -3\ &\ 2 \\ -1\ &\ 6 \end{bmatrix} $$

then the image of the unit square in $\mathbb{R}^2$ is defined by the vector where $\ihat$ lands, $\begin{bmatrix} -3\ \\ -1\ \end{bmatrix}$ and the vector where $\jhat$ lands $\begin{bmatrix} 2\ \\ 6\ \end{bmatrix}$.

The area of the parallelogram $A(S)$ is base time height.

First we will need to determine the interior angle $\theta$. Notice that this angle is the supplementary angle to the angle between our two vectors

$$ \begin{bmatrix} -3 \\ -1 \end{bmatrix}\text{ and }\begin{bmatrix} 2 \\ 6 \end{bmatrix} $$

We compute the angle between our two vectors using the dot product.

$$ \cos(\psi) = \dfrac{u\cdot v}{\|u\|\|v\|}. $$

Thus the interior angle of our parallelogram is given by

$$ \begin{align*} \theta &= \pi - \cos^{-1}\left(\dfrac{\begin{bmatrix} -3 \\ -1 \end{bmatrix}\cdot\begin{bmatrix} 2 \\ 6 \end{bmatrix}}{\left\|\begin{bmatrix} -3 \\ -1 \end{bmatrix}\right\|\left\|\begin{bmatrix} 2 \\ 6 \end{bmatrix}\right\|}\right) \\ \\ &= \pi - \cos^{-1}\left(\dfrac{-6-6}{\sqrt{10}\sqrt{40}}\right) = \cos^{-1}\left(\dfrac{-12}{\sqrt{400}}\right) \\ \\ &= \cos^{-1}\left(\dfrac{-12}{20}\right) = \cos^{-1}\left(-\dfrac{3}{5}\right) \end{align*} $$

It follows that the height $h$ of our parallelogram is

$$ h = \left\|\begin{bmatrix} -3 \\ -1 \end{bmatrix}\right\|\sin(\theta) = \sqrt{10}\sqrt{1-\dfrac{9}{25}} = \dfrac{4\sqrt{10}}{5} $$

The scaling factor of our matrix $A$ is the area of our parallelogram or

$$ \begin{align*} \text{Area}\left(A(S)\right) = -\text{base}\cdot\text{height} &= \left\|\begin{bmatrix} 2 \\ 6 \end{bmatrix}\right\|\,\left\|\begin{bmatrix} -3 \\ -1 \end{bmatrix}\right\|\sin(\theta) \\ \\ &= \sqrt{40}\cdot\dfrac{4\sqrt{10}}{5} = \dfrac{80}{5} = 16. \end{align*} $$

As our linear transformation $A$ changes the orientation of our basis vectors $\ihat$ and $\jhat$, the determinant of $A$ is negative;

$$ \text{det}(A) = -16. $$

Recall that the formula for the length of the cross product vector is also the area of the parallelogram image of the unit square.

$$ \|\mathbf{u}\times\mathbf{v}\| = \|u\|\|v\|\sin(\theta) = \text{area of parallelogram} $$

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3.2.2 Dot Products and Duality

You need to see the following two videos to understand an important application of determinants, the cross product of two vectors.

Dot Products, Duality

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3.2.3 Cross Products

Cross Products

We will revisit dot product, cross product and the geometry of these ideas in chapter 5. The main ideas here

• Determinant is defined geometrically like dot product and cross product.
• There are interesting connections between these geometrical ideas.
• If you fix all of the vectors except one, you obtain a linear transformation from $\mathbb{R}^n$ to the real (or complex) numbers.
• In the case of dot product, cross product or determinant, there is a dual vector involved.
• There is an easier way to compute these values using the dual vector and matrices than trigonometry.

This is a LOT to think about. You should probably take a break before continuing to the next cell.

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3.2.4 Vocabulary for Higher Dimensions

As always, first some vocabulary.

For $3\times 3$ matrices, the linear transformation maps the unit cube to a parallelepiped.

What terms do we use for $4\times 4$, $37\times 37$, $1280\times 1280$ or even larger matrices?

Mathematicians like the $\times$ symbol to describe mutli-dimensional sets so the unit square can be described as

$$ S = [0,1]\times[0,1] $$

This makes the unit cube

$$ S = [0,1]\times[0,1]\times[0,1] = {\large \times}_{k=1}^3 [0,1] $$

For larger values for $n$ one writes

$$ S = [0,1]\times[0,1]\times[0,1]\cdots\times[0,1] = {\large \times}_{k=1}^n [0,1] $$

• A higher dimensional polygon is referred to as a polytope.

$$ S\text{ is a polytope. } $$

• Similarly a higher dimensional parallelepiped is referred to as a parallelotope.

$$ S\text{ is a parallelotope. } $$

• A higher dimensional parallelotope ${\large \times}_{k=1}^n [0,1]$ is still referred to as a unit cube.

$$ S\text{ is a unit cube. } $$

• In higher dimensions the size of the interior of a polytope is its volume.

$$ \text{The volume of $S$ is one.} $$

These are generic terms. One uses the terms polytope and parallelotope even when referring to dimensions 2 and 3. One should also expect unit cube and volume to be used even when the dimension is two.

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3.2.5 Multiplication Symbols

Now that we see a connection between determinant, dot product and cross product, note also the use of the various multiplication symbols we use for scalars and vectors

Scalar Expression	Meaning	Vector Expression	Meaning
$3\times 2\qquad$	scalar multiplication	$\qquad\mathbf{u}\times\mathbf{v}\qquad$	cross product
$3\cdot 2\qquad$	scalar multiplication	$\qquad\mathbf{u}\cdot\mathbf{v}\qquad$	dot product
$3x\qquad$	scalar multiplication	$\qquad\mathbf{u}^T\mathbf{v}\qquad$	matrix multiplication

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3.2.6 Minor Matrices and Cofactors

Let us start with a $3\times 3$ matrix

$$ A = \begin{bmatrix}\ 4\ &\ \ 2\ & -3\ \\ \ 2\ & -1\ &\ \ 2\ \\ \ 2\ &\ \ 1\ & -4\ \end{bmatrix} $$

We can identify nine $2\times 2$ submatrices of $3\times 3$ matrix $A$. For example consider $a_{\color{#CC0099}{1}\color{darkblue}{1}} = 4$. If one removes row one and column one, from matrix $A$,

The result is a $2\times 2$ submatrix

$$ M_{11} = \begin{bmatrix} -1\ &\ \ 2\ \\ \ \ 1\ & -4 \end{bmatrix} $$

This submatrix is referred to as the minor for element $a_{11}$.

Notice that because the minor $M_{11}$ is a matrix, we use a capital letter. However there is a minor associated with every element of the original $n\times n$ matrix. As a result we use the denote the minor submatrix with the same subscripts that indicate the element of the original matrix.

In this way an $n\times n$ matrix will have $n^2$ minor submatrices, one for each element of the matrix. If we choose an element $a_{ij}$ of an $n\times n$ matrix, there is one submatrix and three scalars associated with it.

• The minor submatrix $M_{ij}$ associated with element $a_{ij}$
• The determinant of the minor submatrix $\left|M_{ij}\right|$
• The sgn associated with element $a_{ij}$
• The cofactor $A_{ij}$ associated with element $a_{ij}$.

The sgn$\left(a_{ij}\right)$ depends only on the position of the element of the matrix. It is defined

$$ \text{sgn}\left(a_{ij}\right) := (-1)^{i+j} = \left\{\begin{array}{lcl}\ \ 1 & \ &\ \text{if } i+j\text{ is even} \\ -1 & \ &\ \text{if }i+j\text{ is odd} \end{array}\right. $$

This creates a pattern to for the sgn function in an $n\times n$ matrix

$$ \begin{bmatrix} \ +\ &\ -\ &\ +\ &\ -\ &\ \ &\ \cdots\ \\ \ -\ &\ +\ &\ -\ &\ +\ &\ \ &\ \cdots\ \\ \ +\ &\ -\ &\ +\ &\ -\ &\ \ &\ \cdots\ \\ \ -\ &\ +\ &\ -\ &\ +\ &\ \ &\ \cdots \ \\ \ \vdots\ &\ \vdots\ &\ \vdots\ &\ \vdots\ &\ \ddots\ &\ \vdots\ \\ \ \cdots\ &\ \cdots\ &\ \cdots\ &\ \cdots\ &\ \ &\ +\ \end{bmatrix} $$

Inexplicably, we denote the cofactor $A_{ij}$ associated with factor $a_{ij}$ using a capital letter,

$$ A_{ij} := \text{sgn}\left(a_{ij}\right)\text{det}\left(M_{ij}\right) = (-1)^{i+j}\left|M_{ij}\right|. $$

The cofactor matrix associated with an $n\times n$ matrix $A$ is

$$ \text{Cof}(A) := \begin{bmatrix} A_{ij} \end{bmatrix} = \begin{bmatrix} \ A_{11}\ &\ A_{12}\ &\ \cdots\ &\ A_{1n}\ \\ \ A_{21}\ &\ A_{22}\ &\ \cdots\ &\ A_{2n}\ \\ \ \vdots\ &\ \vdots\ &\ \ddots\ &\ \vdots\ \\ \ A_{n1}\ &\ A_{n2}\ &\ \cdots\ &\ A_{nn} \end{bmatrix} $$

The adjugate matrix associated with an $n\times n$ matrix $A$, the transpose of the cofactor matrix.

$$ \text{adj}(A) := \text{Cof}(A)^T = \begin{bmatrix} A_{ji} \end{bmatrix} = \begin{bmatrix} \ A_{11}\ &\ A_{21}\ &\ \cdots\ &\ A_{n1}\ \\ \ A_{12}\ &\ A_{22}\ &\ \cdots\ &\ A_{n2}\ \\ \ \vdots\ &\ \vdots\ &\ \ddots\ &\ \vdots\ \\ \ A_{1n}\ &\ A_{2n}\ &\ \cdots\ &\ A_{nn} \end{bmatrix} $$

The adjugate matrix or adjunct matrix was historically called the adjoint. There are still old publications that refer to the adjugate matrix as the classical adjoint. If you read of the adjoint of a matrix is an older publication insert the classical in your notes. The term adjoint is unfortunately used in a wide variety of situations include three in linear algebra. We will refrain from using the term adjoint in this course.

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3.2.7 The Laplace Expansion of the Determinant

The Laplace Exansion or cofactor expansion of a determinant computes the determinant of an $n\times n$ matrix as a linear combination of the determinants of submatrices.

To compute the determinant of an $n\times n$ matrix $A$ using the Laplace expansion select any row ${\color{#CC0099}i}$, where $1\le{\color{#CC0099}i}\le n$. The value of our determinant will be

$$ |A| = \displaystyle\sum_{j=1}^n a_{{\color{#CC0099}i}j}A_{{\color{#CC0099}i}j} = \displaystyle\sum_{j=1}^n (-1)^{{\color{#CC0099}i}+j}a_{{\color{#CC0099}i}j}\left|M _{{\color{#CC0099}i}j}\right| $$

We can also any column ${\color{#006666}j}$, where $1\le{\color{#006666}j}\le n$. In this case the value of the determinant of $A$ will be

$$ |A| = \displaystyle\sum_{i=1}^n a_{i{\color{#006666}j}}A_{i{\color{#006666}j}} = \displaystyle\sum_{i=1}^n (-1)^{i+{\color{#006666}j}}a_{i{\color{#006666}j}}\left|M_{i{\color{#006666}j}}\right| $$

Let us test this out by computing the determinant of our example matrix using two different expansions.

First let us choose ${\color{#CC0099}i}$ to be the first row.

$$ \begin{align*} |A| &= \begin{vmatrix}\ 4\ &\ \ 2\ & -3\ \\ \ 2\ & -1\ &\ \ 2\ \\ \ 2\ &\ \ 1\ & -4\ \end{vmatrix} \\ \\ &= \displaystyle\sum_{j=1}^3 a_{{\color{#CC0099}1}j}A_{{\color{#CC0099}1}j} = \displaystyle\sum_{j=1}^3 (-1)^{{\color{#CC0099}1}+j}a_{{\color{#CC0099}1}j}\left|M_{{\color{#CC0099}1}j}\right| \\ \\ &= (-1)^{{\color{#CC0099}1}+1}(4)\begin{vmatrix} -1\ &\ \ 2\ \\ \ \ 1\ & -4\ \end{vmatrix} + (-1)^{{\color{#CC0099}1}+2}(2) \begin{vmatrix} \ \ 2\ &\ \ 2\ \\ \ \ 2\ & -4\ \end{vmatrix} + (-1)^{{\color{#CC0099}1}+3}(-3) \begin{vmatrix}\ \ 2\ & -1\ \\ \ \ 2\ &\ \ 1\ \end{vmatrix} \\ \\ &= (4)(4 - 2) - (2)(-8 - 4) + (-3)(2 + 2) \\ \\ &= (4)(2) - (2)(-12) - 3(4) = 8 + 24 - 12 = 20 \end{align*} $$

Second let us choose ${\color{#006666}j}$ to be the second column.

$$ \begin{align*} |A| &= \begin{vmatrix}\ 4\ &\ \ 2\ & -3\ \\ \ 2\ & -1\ &\ \ 2\ \\ \ 2\ &\ \ 1\ & -4\ \end{vmatrix} \\ \\ &= \displaystyle\sum_{i=1}^3 a_{i{\color{#006666}2}}A_{i{\color{#006666}2}} = \displaystyle\sum_{i=1}^3 (-1)^{i+{\color{#006666}2}}a_{i{\color{#006666}2}}\left|M_{i{\color{#006666}2}}\right| \\ \\ &= (-1)^{1+{\color{#006666}2}}(2)\begin{vmatrix}\ \ 2\ &\ \ 2\ \\ \ \ 2\ & -4\ \end{vmatrix} + (-1)^{2+{\color{#006666}2}}(-1) \begin{vmatrix}\ \ 4\ & -3\ \\ \ \ 2\ & -4\ \end{vmatrix} + (-1)^{3+{\color{#006666}2}}(1) \begin{vmatrix}\ \ 4\ & -3\ \\ \ \ 2\ &\ \ 2\ \end{vmatrix} \\ \\ &= (-2)(-8-4) - (-16 + 6) - (8 + 6) \\ \\ &= (-2)(-12) + 10 - 14 = 24 + 10 - 14 = 20 \end{align*} $$

It is important to understand that the Laplace Expansion of the determinant is not the definition of determinant; it is a property of the determinant. The Laplace or Cofactor expansion is introduced first in most textbooks, but it is only a computing strategy that results from the geometric definition.

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3.2.8 Properties of the Determinant with the Laplace Expansion

In this class use the Laplace expansion only for $2\times 2$ matrices, cross product, and after you reduce a row or column to one nonzero value.

Let us compute the determinant of our previous example using the properties of the determinant and then using the Laplace expansion only when it will save us a lot of computations.

$$ \begin{align*} |A| &= \begin{vmatrix}\ 4\ &\ \ 2\ & -3\ \\ \ 2\ & -1\ &\ \ 2\ \\ \ 2\ &\ \ 1\ & -4\ \end{vmatrix}\begin{array}{l} C_1 - 2C_2 \\ \ \\ \ \end{array} \\ \\ &= \begin{vmatrix}\ \ 0\ &\ \ 2\ & -3\ \\ \ 4\ & -1\ &\ \ 2\ \\ \ 0\ &\ \ 1\ & -4\ \end{vmatrix} \end{align*} $$

Remember that type III column operations don't change the value of the determinant.

$$ \begin{align*} |A| &= 0\left|M_{11}\right| - 4\begin{vmatrix}\ \ 2\ & -3\ \\ \ \ 1\ & -4 \ \end{vmatrix} + 0\left|M_{31}\right| \\ \\ &= -4(-8 + 3) = 20 \end{align*} $$

We do not compute the determinants of the minor submatrices $M_{11}$ and $M_{31}$ because their associated factors are zero.

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3.2.8 Exercises

Compute each determinant by inspection. This means use the properties of the determinant to compute the value of the determinant in your head.

1. $\begin{vmatrix} -3\ & -4\ \\ \ \ 2\ & -2\ \end{vmatrix}$

View Solution

$$ \begin{vmatrix} -3\ & -4\ \\ \ \ 2\ & -2\ \end{vmatrix} = -3(-2) - (-4)(2) = 6 + 8 = 14 $$

2. $\begin{vmatrix}\ \ 2\ &\ \ 0\ &\ \ 2\ \\ \ \ 3\ &\ \ 0\ &\ \ 2\ \\ -3\ &\ \ 1\ & -2\ \end{vmatrix}$

View Solution

$$ \begin{vmatrix}\ \ 2\ &\ \ 0\ &\ \ 2\ \\ \ \ 3\ &\ \ 0\ &\ \ 2\ \\ -3\ &\ \ 1\ & -2\ \end{vmatrix} = -0\left|M_{21}\right| + 0\left|M_{22}\right| - 1\begin{vmatrix}\ \ 2\ &\ \ 2\ \\ \ \ 3\ &\ \ 2\ \end{vmatrix} = -\left( 2(2) - 3(2) \right) = -\left( 4 - 6 \right) = 2 $$

3. $\begin{vmatrix} -3\ &\ \ 0\ &\ \ 0\ \\ -2\ &\ \ 3\ &\ \ 0\ \\ -3\ &\ \ 1\ &\ \ 3\ \end{vmatrix}$

View Solution

$$ \begin{vmatrix} -3\ &\ \ 0\ &\ \ 0\ \\ -2\ &\ \ 3\ &\ \ 0\ \\ -3\ &\ \ 1\ &\ \ 3\ \end{vmatrix} = -3(3)(3) = -27 $$

4. $\begin{vmatrix}\ \ 2\ & -2\ &\ \ 2\ \\ -2\ &\ \ 3\ & -2\ \\ \ \ 1\ &\ \ 4\ &\ \ 1\ \end{vmatrix}$

View Solution

Columns one and three are equal so $$ \begin{vmatrix}\ \ 2\ & -2\ &\ \ 2\ \\ -2\ &\ \ 3\ & -2\ \\ \ \ 1\ &\ \ 4\ &\ \ 1\ \end{vmatrix} = 0 $$

5. $\begin{vmatrix} -1\ &\ \ 4\ & -1\ \\ -4\ &\ \ 0\ &\ \ 4\ \\ \ \ 4\ &\ \ 0\ & -1\ \end{vmatrix}$

View Solution

$$ \begin{vmatrix} -1\ &\ \ 4\ & -1\ \\ -4\ &\ \ 0\ &\ \ 4\ \\ \ \ 4\ &\ \ 0\ & -1\ \end{vmatrix} = -4\begin{vmatrix} -4\ &\ \ 4\ \\ \ \ 4\ & -1\ \end{vmatrix} + 0\left|M_{22}\right| - 0\left|M_{32}\right| = -4\left(\,(-4)(-1) - 4(4)\,\right) = -4\left( 4 - 16 \right) = 48 $$

6. $\begin{vmatrix} -3\ &\ \ 0\ & -3\ &\ \ 1\ \\ \ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 0\ \\ \ \ 2\ &\ \ 0\ & -3\ & -4\ \\ \ \ 1\ &\ \ 2\ & -1\ &\ \ 4\ \end{vmatrix}$

View Solution

$$ \begin{vmatrix} -3\ &\ \ 0\ & -3\ &\ \ 1\ \\ \ \ 1\ &\ \ 0\ &\ \ 0\ &\ \ 0\ \\ \ \ 2\ &\ \ 0\ & -3\ & -4\ \\ \ \ 1\ &\ \ 2\ & -1\ &\ \ 4\ \end{vmatrix} = (-1)(2)\begin{vmatrix} -3\ &\ \ 1\ \\ -3\ & -4\ \end{vmatrix} = -2\left( (-3)(-4) - (-3)(1) \right) = -2\left( 12 + 3 \right) = -30 $$

7. $\begin{vmatrix} -3\ & -2\ &\ \ 0\ & -1\ \\ \ \ 1\ &\ \ 1\ &\ \ 0\ & -2\ \\ \ \ 1\ & -2\ &\ \ 0\ &\ \ 4\ \\ \ \ 3\ &\ \ 3\ &\ \ 0\ & -4\ \end{vmatrix}$

View Solution

This matrix has a column of zeros so $$ \begin{vmatrix} -3\ & -2\ &\ \ 0\ & -1\ \\ \ \ 1\ &\ \ 1\ &\ \ 0\ & -2\ \\ \ \ 1\ & -2\ &\ \ 0\ &\ \ 4\ \\ \ \ 3\ &\ \ 3\ &\ \ 0\ & -4\ \end{vmatrix} = 0 $$

8. $\begin{vmatrix} -3\ &\ \ 3\ & -1\ & -1\ \\ -3\ & -4\ &\ \ 1\ & -4\ \\ \ \ 0\ &\ \ 0\ &\ \ 0\ &\ \ 0\ \\ \ \ 3\ & -1\ & -2\ & -3\ \end{vmatrix}$

View Solution

This matrix has a row of zeros so $$ \begin{vmatrix} -3\ &\ \ 3\ & -1\ & -1\ \\ -3\ & -4\ &\ \ 1\ & -4\ \\ \ \ 0\ &\ \ 0\ &\ \ 0\ &\ \ 0\ \\ \ \ 3\ & -1\ & -2\ & -3\ \end{vmatrix} = 0 $$

9. $\begin{vmatrix} -3\ &\ \ 4\ &\ \ 1\ \\ \ \ 3\ & -4\ & -1\ \\ \ \ 0\ & -1\ & -1\ \end{vmatrix}$

View Solution

Row two is the negative of row 1 so $$ \begin{vmatrix} -3\ &\ \ 4\ &\ \ 1\ \\ \ \ 3\ & -4\ & -1\ \\ \ \ 0\ & -1\ & -1\ \end{vmatrix} = 0 $$

10. $\begin{vmatrix} -1\ &\ \ 0\ &\ \ 0\ &\ \ 0\ \\ \ \ 0\ &\ \ 3\ & -4\ &\ \ 0\ \\ \ \ 0\ &\ \ 2\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 0\ & -2\ \end{vmatrix}$

View Solution

$$ \begin{vmatrix} -1\ &\ \ 0\ &\ \ 0\ &\ \ 0\ \\ \ \ 0\ &\ \ 3\ & -4\ &\ \ 0\ \\ \ \ 0\ &\ \ 2\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ &\ \ 0\ & -2\ \end{vmatrix} = -1\begin{vmatrix}\ \ 3\ & -4\ &\ \ 0\ \\ \ \ 2\ &\ \ 1\ &\ \ 0\ \\ \ \ 0\ &\ \ 0\ & -2\ \end{vmatrix} = -1(-2)\begin{vmatrix}\ \ 3\ & -4\ \\ \ \ 2\ &\ \ 1\ \end{vmatrix} = 2\left( 3(1) - 2(-4) \right) = 22 $$

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3.2.9 Complex-valued Matrices and Determinants

We will eventually be required to consider matrices and vectors of complex numbers. Recall that a complex number $a + bi$ has real part $\mathfrak{Re}(a+bi) = a$ and imaginary part $\mathfrak{Im}(a + bi) = b$. The conjugate of complex number $a + bi$ is given by

$$ \text{conj}(a + bi) = (a + bi)^* = a - bi $$

Often the complex conjugate of a number is denoted with an overbar for symbol, however if we use the overbar to denote a vector, we will need to use some else. Another common symbol is the asterisk or splat. The Hermitian operator or dagger $\dagger$ denotes the complex conjugate transpose of a complex-valued matrix. If

$$ A = \begin{bmatrix}\ 1-2i &\ 2+3i&\ 3+i\ \\ \ 4-i &\ 5+2i&\ 6-3i\ \\ \ 7+2i &\ 8-i &\ 9+2i\ \end{bmatrix} $$

then the conjugate matrix of matrix $A$ is given by

$$ A^* = \text{conj}(A) = \begin{bmatrix}\ 1+2i &\ 2-3i&\ 3-i\ \\ \ 4+i &\ 5-2i&\ 6+3i\ \\ \ 7-2i &\ 8+i &\ 9-2i\ \end{bmatrix} $$

and the Hermitian of matrix $A$ is given by

$$ A^{H} = A^{\dagger} = \left(A^*\right)^T = \left(A^T\right)^* = \begin{bmatrix} \ 1+2i &\ 4+i &\ 7-2i\ \\ \ 2-3i &\ 5-2i &\ 8+i\ \\ \ 3-i &\ 6+3i &\ 9-2i\ \end{bmatrix} $$

Since the transpose of a $1\times 1$ matrix or scalar is itself, we can denote complex conjugation using the Hermitian operator

$$ (a + bi)^H = (a + bi)^* = a - bi $$

Recall that if $z$ and $w$ are complex numbers then

1. $ (z + w)^* = z^* + w^* = w^* + z^* = (w + z)^*$


2. $ (zw)^* = z^*w^* = w^*z^* = (wz)^*$


3. $ \left(\dfrac{z}{w}\right)^* = \dfrac{z^*}{w^*}$


4. $ \left(z^*\right)^* = z$

These same properties work for complex-valued matrices. If $A$ and $B$ are complex-valued matrices and $\alpha$ is a complex number (scalar), then when the size of the matrices permit matrix operations

1. $ \left(A+B\right)^H = A^H + B^H = B^H + A^H = (B + A)^H$


2. $ \left(AB\right)^H = B^HA^H$


3. $ \left(\alpha A\right)^H = \alpha^H\,A^H = \alpha^*\,A^H$


4. $ \left(A^H\right)^H = A$

If a complex number is equal to its own conjugate then it is a real number.

$$ \begin{align*} a + bi &= (a + bi)^* = a - bi \\ \\ bi &= -bi \\ \\ b &= -b \\ \\ b &= 0 \end{align*} $$

If $n\times n$ matrix $A$ is equal to its own Hermitian, $A = A^H$, then all of the diagonal entries are real numbers and we call matrix $A$ an Hermitian matrix.

$$ B = \begin{bmatrix}\ 1 &\ 2-i &\ 3+2i\ \\ \ 2+i &\ 4 &\ 5-3i\ \\ \ 3-2i &\ 5+3i &\ 6 \end{bmatrix} $$

is an Hermitian matrix. Notice that this matrix is not symmetric.

If a matrix is both real and symmetric, then it is Hermitian

Consider a $2\times 2$ matrix

$$ A = \begin{bmatrix}\ a\ &\ b\ \\ \ c\ &\ d\ \end{bmatrix} $$

and the conjugate matrix

$$ A^* = \begin{bmatrix}\ a^*\ &\ b^*\ \\ \ c^*\ &\ d^*\ \end{bmatrix} $$

The determinant

$$ \text{det}\left(A^*\right) = \begin{vmatrix}\ a^*\ &\ b^*\ \\ \ c^*\ &\ d^*\ \end{vmatrix} = a^*d^* - b^*c^* = (ad)^* - (bc)^* = (ad - bc)^* = |A|^* $$

Since all determinates are sums of products of elements of a matrix we have

$\text{det}\left(A^*\right) = \text{det}(A)^*$

We can also write this

$\left|\,A^*\,\right| = |A|^*$

Finally, using the properties we have for the determinant

$\left|A^{\dagger}\right| = \left|A^H\right| = \left|\,\left(A^T\right)^*\,\right| = \left|\,A^T\,\right|^* = |A|^*$

The determinant of the Hermitian of matrix $A$ is the complex conjugate of the determinant of matrix $A$.

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