description/proof of that for commutative ring, inverse of transpose of matrix is transpose of inverse of matrix
Topics
About: matrices space
The table of contents of this article
Starting Context
Target Context
- The reader will have a description and a proof of the proposition that for any commutative ring, the inverse of the transpose of any invertible matrix is the transpose of the inverse of the matrix.
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:
\(R\): \(\in \{\text{ the commutative rings }\}\)
\(M\): \(\in \{\text{ the } m \times m \text{ invertible } R \text{ matrices } \}\)
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Statements:
\({M^t}^{-1} = {M^{-1}}^t\)
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2: Proof
Whole Strategy: Step 1: see that \({M^{-1}}^t M^t = I\) and see that \({M^t M^{-1}}^t = I\).
Step 1:
\({M^{-1}}^t M^t = (M M^{-1})^t\), by the proposition that for any commutative ring, the transpose of the product of any matrices is the product of the transposes of the constituents in the reverse order, \(= I^t = I\).
\({M^t M^{-1}}^t = (M^{-1} M)^t\), by the proposition that for any commutative ring, the transpose of the product of any matrices is the product of the transposes of the constituents in the reverse order, \(= I^t = I\).
That means that \({M^t}^{-1} = {M^{-1}}^t\).
