description/proof of that for ring of \(n \times n\) field matrices, if element multiplied by another element from left or right is \(1\), another element is inverse
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of ring of \(n \times n\) ring matrices.
- The reader admits the proposition that for any ring, if any element has an inverse and the element multiplied by any another element from left or right is \(1\), the another element is the inverse.
- The reader admits the proposition that over any commutative ring, the determinant of the product of any square matrices is the product of the determinants of the matrices.
- The reader admits the proposition that over any field, any square matrix has the inverse if and only if its determinant is nonzero, and the inverse is this.
Target Context
- The reader will have a description and a proof of the proposition that for any ring of \(n \times n\) field matrices, if any element multiplied by any another element from left or right is \(1\), the another element is the inverse of the element.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(M_n (F)\): \(= \text{ the ring of the } n \times n F \text{ matrices }\)
\(M\): \(\in M_n (F)\)
//
Statements:
(
\(\exists M' \in M_n (F) (M' M = I)\)
\(\implies\)
\(M' = M^{- 1}\)
)
\(\land\)
(
\(\exists M' \in M_n (F) (M M' = I)\)
\(\implies\)
\(M' = M^{- 1}\)
)
//
2: Note
Compare with the proposition that for any ring, if any element has an inverse and the element multiplied by any another element from left or right is \(1\), the another element is the inverse: for this proposition, \(M\) does not need to be presupposed to be invertible, which is due to \(F\) being a field.
3: Proof
Whole Strategy: see that \(M\) is invertible and apply the proposition that for any ring, if any element has an inverse and the element multiplied by any another element from left or right is \(1\), the another element is the inverse; Step 1: suppose that \(M' M = I\); Step 2: take \(det (M' M) = det I\) and see that \(M\) is invertible; Step 3: suppose that \(M M' = I\); Step 4: take \(det (M M') = det I\) and see that \(M\) is invertible.
Step 1:
Let us suppose that \(M' M = I\).
Step 2:
\(det (M' M) = det I = 1\), but the left hand side is \(det M' det M\), by the proposition that over any commutative ring, the determinant of the product of any square matrices is the product of the determinants of the matrices
So, \(det M \neq 0\).
\(M\) has an inverse, by the proposition that over any field, any square matrix has the inverse if and only if its determinant is nonzero, and the inverse is this.
So, by the proposition that for any ring, if any element has an inverse and the element multiplied by any another element from left or right is \(1\), the another element is the inverse, \(M' = M^{- 1}\).
Step 3:
Let us suppose that \(M M' = I\).
Step 4:
\(det (M M') = det I = 1\), but the left hand side is \(det M det M'\), by the proposition that over any commutative ring, the determinant of the product of any square matrices is the product of the determinants of the matrices
So, \(det M \neq 0\).
\(M\) has an inverse, by the proposition that over any field, any square matrix has the inverse if and only if its determinant is nonzero, and the inverse is this.
So, by the proposition that for any ring, if any element has an inverse and the element multiplied by any another element from left or right is \(1\), the another element is the inverse, \(M' = M^{- 1}\).
