9
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 ith 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 ith 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.

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