Axioms 1-3 and Properties I-III

Feb 19

Axioms 1-3 and Properties I-III

Axioms 1-3 and Properties I-III

Axioms

Axiom 1. Homogeneity. If $A^\prime$ denotes the matrix obtained from $A$ by multiplying one of its rows by a number $k,$ then $\det A^\prime=k\det A.$

This implies that for $k\neq 0,$ $A^\prime$ and $A$ are invertible simultaneously.

Axiom 2. If $A^\prime$ denotes the matrix obtained from $A$ by adding one of the rows of $A$ to another, then $\det A^\prime=\det A$ .

We remember that adding one of the rows of $A$ to another corresponds to adding one equation of the system $Ax=y$ to another and so this operation should not impact solvability of the system.

Axiom 3. $\det I=1.$

Adding this axiom on top of the previous two makes the determinant a unique function of a matrix.

Properties

Assuming that $A$ is of size $n\times n,$ it is convenient to partition it into rows:

$A=\left( \begin{array}{c} A_1 \\ ... \\ A_n \end{array} \right).$

Property I. If one of the rows of $A$ is zero, then $\det A=0.$

Indeed, if $A_i=0,$ then $A_i=0\times A_i$ and by Axiom 1 $\det A=0\times\det A=0.$

Property II. If we add to one of the rows of $A$ another row multiplied by some number $k,$ the determinant does not change.

Proof. If $k=0,$ there is nothing to prove. Suppose $k\neq 0$ and we want to add $kA_j$ to $A_i.$ This result can be achieved in three steps.

a) Multiply $A_j$ by $k.$ By Axiom 1, $\det A$ gets multiplied by $k.$

b) In the new matrix, add row $kA_j$ to $A_i.$ By Axiom 2, this does not change the determinant.

c) In the resulting matrix, divide row numbered $j$ by $k.$ The determinant gets divided by $k.$

The determinant of the very first matrix will be the same as that of the very last matrix, while the $i$ th row of the last matrix will be $A_i+kA_j.$

Property III. If rows of $A$ are linearly dependent, then $\det A=0.$

Proof. Suppose rows of $A$ are linearly dependent. Then one of the rows can be expressed as a linear combination of the others. Suppose, for instance, that $A_1=k_2A_2+...+k_nA_n.$ Multiply the $i$ th row by $-k_i$ and add the result to the first row, for $i=2,...,n.$ Thereby we make the first row equal to $0,$ while maintaining the determinant the same by Property II. Then by Property I $\det A=0.$