definition of non-degenerate \((0, 2)\)-tensor of vectors space
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of \((p, q)\)-tensors space of vectors space.
- The reader knows a definition of interior multiplication of tensor by vector.
Target Context
- The reader will have a definition of non-degenerate \((0, 2)\)-tensor of vectors space.
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:
\( F\): \(\in \{\text{ the fields }\}\)
\( V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\( T^0_2 (V)\):
\( *t\): \(\in T^0_2 (V)\)
//
Conditions:
\(\forall v \in V (\forall w \in V (t (v, w) = 0) \implies v = 0)\)
//
The condition, \(\forall v \in V \text{ such that } v \neq 0 (\exists w \in V (t (v, w) \neq 0))\), is an equivalent condition (called "the 1st alternative condition"), as will be seen in Note.
When \(V\) is finite-dimensional, the condition, \(\widehat{t}: V \to V^*, v \mapsto i_v (t) \in \{\text{ the bijections }\}\)
, is an equivalent condition (called "the 2nd alternative condition"), as will be seen in Note: in fact, \(\widehat{t}\) is a 'vectors spaces - linear morphisms' isomorphism.
When \(V\) is finite-dimensional, the condition, the representative matrix of \(\widehat{t}\) with respect to any basis for \(V\) and the dual basis for \(V^*\) is invertible, is an equivalent condition (called "the 3rd alternative condition"), as will be seen in Note.
2: Note
Usually, \(t \in \Lambda_2 (V)\) or \(t \in \Sigma_2 (V)\) is supposed, but the definition itself does not need it.
Let us see that the 1st alternative condition is an equivalent condition.
Let us suppose the condition of this definition.
Let \(v \in V\) be any such that \(v \neq 0\).
If there was no \(w \in V\) such that \(t (v, w) \neq 0\), for each \(w \in V\), \(t (v, w) = 0\), which implies \(v = 0\), a contradiction.
So, there is a \(w\).
Let us suppose that 1st alternative condition.
Let \(v \in V\) be any.
Let us suppose that for each \(w \in V\), \(t (v, w) = 0\).
If \(v \neq 0\), there would be a \(w \in V\) such that \(t (v, w) \neq 0\), a contradiction against that \(t (v, w) = 0\) for each \(w \in V\).
So, \(v = 0\).
Let us suppose that \(V\) is finite-dimensional hereafter.
Let us see that the 2nd alternative condition is an equivalent condition.
\(\widehat{t}\) is valid and \(F\)-linear, by the proposition that for any \((0, 2)\)-tensor of any vectors space, a linear map from the vectors space into the covectors space is induced, and when the vectors space is finite-dimensional, the induced map's representative matrix with respect to any basis and the dual basis is this.
Let us suppose this definition.
Let us see that \(\widehat{t}\) is injective.
Let \(v, v' \in V\) be any such that \(v \neq v'\).
Let us suppose that \(\widehat{t} (v) = \widehat{t} (v')\).
For each \(w \in V\), \(\widehat{t} (v) (w) = \widehat{t} (v') (w)\), so, \(t (v, w) = t (v', w)\), so, \(t (v, w) - t (v', w) = 0\), but the left hand side was \(t (v - v', w)\), which would imply that \(v - v' = 0\), a contradiction.
So, \(\widehat{t} (v) \neq \widehat{t} (v')\).
So, \(\widehat{t}\) is injective.
So, \(\widehat{t}\) is an injective linear map into \(V^*\).
But as \(V\) is finite-dimensional, \(V^*\) has the dimension of \(V\): refer to the definition of canonical 'vectors spaces - linear morphisms' isomorphism between finite-dimensional vectors space and its covectors space with respect to original space basis.
So, \(\widehat{t}\) is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that any linear injection between any same-finite-dimensional vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.
Especially, \(\widehat{t}\) is a bijection.
Let us suppose that 2nd alternative condition.
Let \(v \in V\) be any such that \(v \neq 0\).
\(\widehat{t} (v) \neq 0\), because \(\widehat{t}\) is linear and injective: \(\widehat{t} (0) = 0\).
That means that there is a \(w \in V\) such that \(\widehat{t} (v) (w) \neq 0\), but the left hand side is \(t (v, w)\).
So, the 1st alternative condition is satisfied.
So, this definition is satisfied.
Let us see that the 3rd alternative condition is an equivalent condition.
Let \(B = \{b_1, ..., b_d\} \subseteq V\) be any basis for \(V\).
Let \(B^* = \{b^1, ..., b^d\} \subseteq V\) be the dual basis of \(B\) for \(V^*\).
Let \(M\) be the representative matrix of \(\widehat{t}\) with respect to \(B\) and \(B^*\) by the definition of representative matrix of linear map between finite-dimensional vectors spaces with respect to bases with the canonical 'vectors spaces - linear morphisms' isomorphism from \(V\) onto the components vectors space with respect to \(B\), \(f_1: V \to F^d\), and the canonical 'vectors spaces - linear morphisms' isomorphism from \(V^*\) onto the components vectors space with respect to \(B^*\), \(f_2: V^* \to F^d\).
\(M\) is the canonical representative matrix of \(f_2 \circ \widehat{t} \circ {f_1}^{-1}\).
Let us suppose this definition.
The 2nd alternative condition is satisfied.
So, \(\widehat{t}\) is a 'vectors spaces - linear morphisms' isomorphism.
So, \(f_2 \circ \widehat{t} \circ {f_1}^{-1}\) is a 'vectors spaces - linear morphisms' isomorphism.
As a representative of it, \(M\) is invertible.
Let us suppose that the 3rd alternative condition is satisfied.
Then, \(f_2 \circ \widehat{t} \circ {f_1}^{-1}\) is bijective.
So, \(\widehat{t} = {f_2}^{-1} \circ f_2 \circ \widehat{t} \circ {f_1}^{-1} \circ f_1\) is bijective.
So, the 2nd alternative condition is satisfied.
So, this definition is satisfied.
