description/proof of that if square ring matrix has inverse, inverse is unique
Topics
About: matrices space
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 an element has an inverse, the inverse is unique.
Target Context
- The reader will have a description and a proof of the proposition that if any square ring matrix has an inverse, the inverse is the unique inverse.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M_n (R)\): \(= \text{ the ring of the } n \times n R \text{ matrices }\)
\(M\): \(\in M_n (R)\)
//
Statements:
\(\exists M' \in M_n (R) (M' M = M M' = I) \land \exists M'' \in M_n (R) (M'' M = M M'' = I)\)
\(\implies\)
\(M' = M''\)
//
2: Proof
Whole Strategy: Step 1: apply the proposition that for any ring, if an element has an inverse, the inverse is unique.
Step 1:
As \(M_n (R)\) is a ring, by Note for the definition of ring of \(n \times n\) ring matrices, the proposition holds, by the proposition that for any ring, if an element has an inverse, the inverse is unique.
