# Determinants

Aug-Nov 2020

## Recap

• Vector space $V$ over a scalar field $F$
• $F$: real field $\mathbb{R}$ or complex field $\mathbb{C}$ in this course
• $m\times n$ matrix A represents a linear map $T:F^n\to F^m$
• null$(A)=$ null $T=\{v\in V:Tv=0\}$, colspace$(A)$ = range $T=\{Tv:v\in V\}$
• dim null $T$ + dim range $T$ = dim $V$
• Linear equation: $Ax=b$
• Solved using elementary row operations
• Solution (if it exists): $u+$ null$(A)$
• Linear map $T$ induces a one-to-one map $V/\text{null }T\to \text{range }T$
• Four fundamental subspaces of a matrix
• Column space, row space, null space, left null space

## Determinants

$A:n\times n$ matrix written as $A=[v_1;\ldots;v_n]$, where $v_j$ is the $j$-th row of $A$.

$v_j$: row vector of length $n$

det: $F^{n,n}\to F$ is a function from square matrices to the field $F$ satisfying the following conditions or defining properties:

1. Identity: det$(I)=1$

2. Row scaling: det$([v_1;\ldots;cv_j;\ldots,v_n])=c$ det$([v_1;\ldots;v_j;\ldots;v_n])$

3. Row linearity: det$([v_1;\ldots;v_j+v'_j;\ldots;v_n])=$ det$([v_1;\ldots;v_j;\ldots;v_n])+$ det$([v_1;\ldots;v'_j;\ldots;v_n])$

4. Equal rows: det$([v_1;\ldots;v_n])=0$, if any two rows are equal

• why such an intricate definition?

• is there such a function?

## Geometric connection

$A=\begin{bmatrix} a&b\\ c&d \end{bmatrix}$, det$(A)=ad-bc$

• Area of parallelogram formed by $(a,b),(c,d)$ $=$ $|$det$(A)|$
• Defining properties satisfied by area function

$A=\begin{bmatrix} a&b&c\\ d&e&f\\ g&h&i \end{bmatrix}$, det$(A)=aei+bfg+cdh-gec-hfa-idb$

• Volume of cuboid formed by $(a,b,c),(d,e,f),(g,h,i)$ $=$ $|$det$(A)|$
• Defining properties satisfied by volume function

$A:n\times n$ matrix, det$(A)=?$

• Volume of $n$-dimensional parallelepiped formed by rows
• Defining properties satisfied

## Further properties of determinants

Zero row: det$([v_1;\ldots;v_n])=0$, if any of the rows is equal to all zeros

Proof

• Use row scaling property with $c=-1$ on all-zero row

Row operation: det$([v_1+cv_j;\ldots;v_j;\ldots;v_n])=\text{det}([v_1;\ldots;v_j;\ldots;v_n])$

Proof

• Use row addition, scaling and equal row properties

Dependent rows: det$([v_1;\ldots;v_n])=0$, if the rows or columns are linearly dependent

Proof

• Replace $v_j$ by $v_j+a_1v_1+\cdots+a_{j-1}v_{j-1}=0$
• Take determinant and use defining properties

Row swap: If two rows are interchanged, determinant gets multiplied by $-1$

Proof

• Consider det$([v_1+v_2;v_1+v_2;\ldots])$

## Determinants and elementary row operations

Elementary row operators: Let $E$ be an elementary row operator $\text{det}(E)=\begin{cases} c&\text{ if row scaling by }c\\ -1&\text{ if row swap}\\ 1&\text{ if row }i = \text{row }i + c(\text{row }j) \end{cases}$

Product of elementary row operators and a matrix $\text{det}(EA)=\text{det}(E)\text{det}(A)$

$\text{det }\left(\left(\prod_i E_i\right) A\right) = \left(\prod_i \text{det}(E_i)\right)\text{det}(A)$

• Proof of last result by induction on $i$

## Compute determinants using properties

Diagonal matrix: det $\begin{bmatrix}d_1&0&\cdots&0\\ 0&d_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\cdots&d_n\end{bmatrix}=d_1d_2\cdots d_n$

Non-invertible matrix: det $=0$

Invertible matrix: Row reduce to identity

There exist elementary row operators $E_i$ such that $\left(\prod_i E_i\right) A = I$

• Take determinants and use elementary row operator property

det$(A)=(-1)^{n_s}\dfrac{1}{\left(\prod_j c_j\right)}$

## Determinant of product of two matrices

$A,B$: two square matrices

det$(AB)=$ det$(A)$ det$(B)$

Proof

• $A$ or $B$ non-invertible

• $AB$ is also non-invertible

• det$(AB)=0=$ det$(A)$ det$(B)$

• $A$ and $B$ invertible

• $\left(\prod_i E_i\right) A = I$, $\left(\prod_j F_j\right) B = I$

• $\left(\prod_j F_j\right)\left(\prod_i E_i\right) AB = I$

Corollary: If $A$ is invertible, det$(A^{-1})=\dfrac{1}{\text{det}(A)}$

## Elementary column operations and transpose

Elementary column operators: $E^T$, where $E$ is an elementary row operator

det$(E^T)=$ det$(E)$

det$(A^T)=\text{det}(A)$

Proof

• $A$: non-invertible

• $A^T$ is non-invertible

• det$(A^T)=0=\text{det}(A)$

• $A$: invertible

• $E_1\cdots E_L A = I$ implies $I=A^T E^T_L\cdots E^T_1$

• Take determinants

## Determinant function definition

det: $F^{n,n}\to F$ is a function from square matrices to the field $F$ satisfying the following conditions or defining properties:

1. Identity: det$(I)=1$

2. Row scaling: det$([v_1;\ldots;cv_j;\ldots,v_n])=c$ det$([v_1;\ldots;v_j;\ldots;v_n])$

3. Row linearity: det$([v_1;\ldots;v_j+v'_j;\ldots;v_n])=$ det$([v_1;\ldots;v_j;\ldots;v_n])+$ det$([v_1;\ldots;v'_j;\ldots;v_n])$

4. Equal rows: det$([v_1;\ldots;v_n])=0$, if any two rows are equal

• Unique function that satisfies all of the above properties

• Co-factor expansion along any row or column

• Permutation formula