PX153 - K5 - trace and determinants

only for square matrices
some properties of a matrix can be characterized by a single number

trace

T r A = a_{11} + a_{22} + \dots + a_{n n} = \sum_{i = 1}^{\infty} a_{i i}

T r (A + B) = T r A + T r B

T r (A B) = \sum_{j, i} a_{i j} b_{j i} = \sum_{j, i} b_{j i} a_{i j} = T r (B A)

eg:

T r [\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}] = 1 - 1 = 0

determinant

let $A$ be a $n \times n$ matrix. the determinant of $A$ , written as $| A |$ or $det A$ , is a number defined as:

det [\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}] = a_{11} a_{22} - a_{12} a_{21}

det [\begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix}] = a_{11} det [\begin{matrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{matrix}] - a_{12} det [\begin{matrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{matrix}] + det [\begin{matrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{matrix}]

eg:

det [\begin{matrix} 1 & 0 & 2 \\ - 1 & 1 & 3 \\ 0 & 2 & 1 \end{matrix}] = 1 \times (- 5) - 0 + 2 \times (- 2) = - 9

cofactor

more generally, for a $n \times n$ matrix, $A = (a_{i j})_{n \times n}$ , the co-factor, $c_{i j}$ , for $a_{i j}$ is defined to be $(- 1)^{i + j} \times det A_{(n - 1) \times (n - 1)}$
the determinant can be written as:

det A = a_{11} c_{11} + a_{12} c_{12} + \dots + a_{1 n} c_{1 n}

in 3D:

\begin{aligned} det A & = a_{11} c_{11} + a_{12} c_{12} + a_{13} c_{13} \\ = a_{11} (- 1)^{1 + 1} det [\begin{array}{c} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}] + a_{12} (- 1)^{1 + 2} det [\begin{array}{c} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array}] + a_{13} det [\begin{array}{c} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}] \end{aligned}

if $A$ is a $n \times n$ matrix, then $det A$ contains of the order of $n!$ products of $n$ matrix elements that are in different rows and columns

connection between determinant and cross product

PX153 - A6 - advanced vector operations#scalar triple product

consider 3 vectors:
$\vec{e} = (e_{1}, e_{2}, e_{3})$
$\vec{f} = (f_{1}, f_{2}, f_{3})$
$\vec{g} = (g_{1}, g_{2}, g_{3})$
the cross product:

\vec{f} \times \vec{g} = (f_{2} g_{3} - g_{2} f_{3}, g_{1} f_{3} - g_{3} f_{1}, f_{1} g_{2} - g_{1} f_{2})

\begin{aligned} \vec{e} \cdot (\vec{f} \times \vec{g}) & = (e_{1} (f_{2} g_{3} - g_{2} f_{3}), e_{2} (g_{1} f_{3} - g_{3} f_{1}), e_{3} (f_{1} g_{2} - g_{1} f_{2})) \\ = e_{1} c_{11} + e_{2} c_{12} + e_{3} c_{13} \\ = det [\begin{array}{c} e_{1} & e_{2} & e_{3} \\ f_{1} & f_{2} & f_{3} \\ g_{1} & g_{2} & g_{3} \end{array}] \end{aligned}

this gives the volume of the parallelepiped, the base area, $| \vec{f} \times \vec{g} |$ , times height, $| \vec{e} | \cos θ$ :

\vec{e} \cdot (\vec{f} \cdot \vec{g}) = | \vec{f} \times \vec{g} | | \vec{e} | \cos θ