description/proof of that for finite-dimensional vectors space with inner product, components matrix of inner product w.r.t. basis is invertible
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of inner product on real or complex vectors space.
- The reader knows a definition of %ring name% matrices space.
- The reader knows a definition of determinant of square matrix over ring.
- The reader knows a definition of Gram-Schmidt orthonormalization of countable subset of vectors space with inner product.
- The reader admits the proposition that for any vectors space with any inner product, any orthonormal subset is linearly independent.
- The reader admits the proposition that for any finite-dimensional vectors space, any linearly independent subset with dimension number of elements is a basis.
- The reader admits the proposition that for any vectors space over any field and any square matrix over the field with dimension equal to or smaller than the dimension of the vectors space, the matrix is invertible if it maps a linearly-independent set of vectors to a linearly-independent set of vectors, and if the matrix is invertible, it maps any linearly-independent set of vectors to a linearly-independent set of vectors.
Target Context
- The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space with any inner product, the components matrix of the inner product with respect to any basis 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:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(d\): \(\in \mathbb{N} \setminus \{0\}\)
\(V\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\), with any inner product, \(\langle \bullet, \bullet \rangle\)
\(B\): \(\in \{\text{ the bases for } V\}\), \(= \{b_1, ..., b_d\}\)
\(M\): \(\in \{\text{ the } d \times d F \text{ matrices }\}\), such that \(M^j_l = \langle b_l, b_j \rangle\)
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Statements:
\(det M \neq 0\)
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2: Proof
Whole Strategy: Step 1: take the Gram-Schmidt orthonormalization of \(B\), \(B' = \{b'_1, ..., b'_d\}\), and see that \(M'\) with respect to \(B'\) is \(I\); Step 2: see that \(b_j = b'_l N^l_j\) where \(det N \neq 0\); Step 3: see that \(M = N^* M' N\); Step 4: see that \(det M = det (N^* M' N) \neq 0\).
Step 1:
Let us take the Gram-Schmidt orthonormalization of \(B\), \(B' = \{b'_1, ..., b'_d\}\).
\(B'\) is a basis for \(V\), by the proposition that for any vectors space with any inner product, any orthonormal subset is linearly independent and the proposition that for any finite-dimensional vectors space, any linearly independent subset with dimension number of elements is a basis.
Let \(M'\) be the \(d \times d\) \(F\) matrix such that \(M'^j_l = \langle b'_l, b'_j \rangle\).
\(M' = I\), because \(B'\) is orthonormal.
Step 2:
For each \(j \in \{1, ..., d\}\), \(b_j = b'_l N^l_j\) where \(N^l_j \in F\), because \(B'\) is a basis.
\(det N \neq 0\), by the proposition that for any vectors space over any field and any square matrix over the field with dimension equal to or smaller than the dimension of the vectors space, the matrix is invertible if it maps a linearly-independent set of vectors to a linearly-independent set of vectors, and if the matrix is invertible, it maps any linearly-independent set of vectors to a linearly-independent set of vectors.
Step 3:
\(M^j_l = \langle b_l, b_j \rangle = \langle b'_m N^m_l, b'_n N^n_j \rangle = \overline{N^n_j} \langle b'_m, b'_n \rangle N^m_l = \overline{N^n_j} M'^n_m N^m_l = (N^* M' N)^j_l\), so, \(M = N^* M' N\).
Step 4:
\(det M = det (N^* M' N) = det N^* det M' det N \neq 0\), because \(det N^* = \overline{det N} \neq 0\), \(det M' = det I = 1 \neq 0\), and \(det N \neq 0\).
