description/proof of that determinant of square matrix over field is nonzero iff set of columns or rows is linearly independent
Topics
About: matrices space
The table of contents of this article
Starting Context
- The reader knows a definition of field.
- The reader knows a definition of determinant of square matrix over ring.
- The reader knows a definition of linearly independent subset of module.
- The reader knows Cramer's rule for any system of linear equations.
Target Context
- The reader will have a description and a proof of the proposition that the determinant of any square matrix over any field is nonzero if and only if the set of the columns or the rows is linearly independent.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\text{ the fields }\}\)
\(M\): \(\in \{\text{ the } n \times n F \text{ matrices }\}\)
//
Statements:
(
\(det M \neq 0\)
\(\iff\)
\(\{\text{ the columns of } M\} \in \{\text{ the linearly independent subsets of the } n \text{ -dimensional } F \text{ columns vectors space }\}\)
)
\(\land\)
(
\(det M \neq 0\)
\(\iff\)
\(\{\text{ the rows of } M\} \in \{\text{ the linearly independent subsets of the } n \text{ -dimensional } F \text{ rows vectors space }\}\)
)
//
2: Proof
Whole Strategy: apply Cramer's rule for any system of linear equations; Step 1: take \(M (c^1, ..., c^n)^t = 0\) and conclude for the columns; Step 2: take \(M^t (c^1, ..., c^n)^t = 0\) and conclude for the rows.
Step 1:
Let the \(j\)-th column of \(M\) be denoted as \(M_j\).
Let us take \(M (c^1, ..., c^n)^t = 0\).
It is nothing but \(c^1 M_1 + ... + c^n M_n = 0\).
So, the columns' being linearly independent is nothing but \(M (c^1, ..., c^n)^t = 0\)'s having only the 0 solution.
By Cramer's rule for any system of linear equations, that is nothing but \(det M \neq 0\): if \(det M = 0\), the rank would be smaller than \(n\), and at least 1 of \(c^j\) s could be taken arbitrarily.
So, \(det M \neq 0\) if and only if the set of the columns is linearly independent.
Step 2:
By Step 1, \(det M^t \neq 0\) if and only if the set of the columns is linearly independent.
But the set of the columns of \(M^t\) is linearly independent if and only if the set of the rows of \(M\) is linearly independent, obviously.
On the other hand, \(det M^t = det M\).
So, \(det M \neq 0\) if and only if the set of the rows of \(M\) is linearly independent.
