Determinant

Define Determinant through axioms

We find a function $\det \{A\in {\mathbb{R}}^{n\times n}\}\to \mathbb{R}$ (Note that $A$ must be a square matrix) with following properties

1. $\det (A)=\det (A^{\top })$
2. $\det (I)=1$
3. $\det (A)=0$ if the columns are linear dependent
4. $\det ([a_1|\dots |a_{k-1}|a_k|\lambda a+\mu b|\dots |a_{k-1}])=\lambda \det ([\dots |a_{k-1}|a|\dots ])+\mu \det ([\dots |a_{k-1}|b|\dots ])$

Simple Example

$\det \left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]=\det \left[\begin{array}{cc}\left(\begin{array}{c}0\\ 1\end{array}\right)+\left(\begin{array}{c}1\\ 0\end{array}\right)& \left(\begin{array}{c}1\\ 1\end{array}\right)\end{array}\right]=\det \left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]+\det \left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]=0$ $0=\det \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\det \left[\begin{array}{cc}0& 0\\ 1& 1\end{array}\right]+\det \left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]+\det \left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$ $0=1+0+0+\det \left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$ $\det \left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]=-1$

general example for n=2

$\det \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ $=\det \left[\begin{array}{cc}a& 0\\ 0& d\end{array}\right]+\det \left[\begin{array}{cc}a& b\\ 0& 0\end{array}\right]+\det \left[\begin{array}{cc}0& 0\\ c& d\end{array}\right]+\det \left[\begin{array}{cc}0& b\\ c& 0\end{array}\right]$ $=\det \left[\begin{array}{cc}a\left(\begin{array}{c}1\\ 0\end{array}\right)& d\left(\begin{array}{c}0\\ 1\end{array}\right)\end{array}\right]+\dots$ $=ad+0+0+(-bc)$ $=ad-bc$

For $n=3$ $\det \left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]=-\det \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=-1$ the matrix $\left[\begin{array}{cc}{a}_{22}& a_{23}\\ a_{32}& a_{33}\end{array}\right]$ is flipped and thus must multiply $-1$, same principle to example We can see that if we exchange two rows, the determinant must multiply $-1$

Another Perspective

computing the determinant of a $2\times 2$ matrix as the area of the image of the unit square after a linear transformation

General case

Definition 7.2.1

Given a permutation $\pi :\{1\dots n\}\to \{1\dots n\}$ of n elements

$$sgn(\pi )=\{\begin{array}{l}1\text{ if }|\{(i,j)|i<j\text{ and }\pi (i)>\pi (j)\}|\text{ is even}\\ -1\text{ if }|\{(i,j)|i<j\text{ and }\pi (i)>\pi (j)\}|\text{ is odd}\end{array}$$

Example

$\det \left[\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right]$ $\pi (1)=3>\pi (2)=1$ $\pi (1)=3>\pi (3)=2$ $\pi (2)=1<\pi (3)=2$ $sgn(\pi )=1$ $\det \left[\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right]$ is positive (actually $\det \left[\begin{array}{ccc}0& 0& 1\\ 1& 0& 0\\ 0& 1& 0\end{array}\right]=1$)

Definition: determinant

$$\det (A)=\sum _{\sigma \in {\prod }_n}sgn(\sigma )\prod _{i=1}^n{A}_{i,\sigma (i)}$$

Properties

• $\det (A^{\top })=\det (A)$
• $\det (I)=1$
• The columns are linear dependent $⟹\det (A)=0$
• linear
• Let $\pi$ be a permutation, and $P(\pi )$ is the permutation matrix
  • $\det (P(\pi ))=sgn(\pi )$
• For a triangular matrix $T$
  • $\det (T)={\prod }_{i=1}^n{T}_{{}_i}$
• For a orthogonal matrix $Q$
  • $\det (Q)\in \{-1,1\}$
• $\det (AB)=\det (A)\det (B)$
• $\det (A^{-1})=\frac{1}{\det (A)}$
• $A$ invertible $⟺$ $\det (A)\ne 0$