description/proof of that positive-definite Hermitian matrix is invertible
Topics
About: matrices space
The table of contents of this article
Starting Context
- The reader knows a definition of positive-definite Hermitian matrix.
- 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.
- The reader admits the proposition that any positive-definite Hermitian matrix can be transformed to the identity by a unitary matrix multiplied by a positive diagonal matrix from right.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that any positive-definite Hermitian matrix is invertible.
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:
\(M\): \(\in \{\text{ the positive-definite Hermitian matrices }\}\)
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Statements:
\(\exists M^{-1}\)
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2: Proof
Whole Strategy: Step 1: see that being invertible equals having nonzero determinant; Step 2: use the proposition that any positive-definite Hermitian matrix can be transformed to the identity by a unitary matrix multiplied by a positive diagonal matrix from right.
Step 1:
Being invertible equals having nonzero determinant, 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, we are going to see that \(det M \neq 0\).
Step 2:
By the proposition that any positive-definite Hermitian matrix can be transformed to the identity by a unitary matrix multiplied by a positive diagonal matrix from right, \((U D)^* M (U D) = I\) where \(U\) is a unitary matrix and \(D\) is a positive diagonal matrix.
So, \(det ((U D)^* M (U D)) = det I = 1 \neq 0\), but \(det ((U D)^* M (U D)) = det ((U D)^*) det M det (U D)\), 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, \(det M \neq 0\).
