description/proof of that quadratic form by \(2\)-symmetric-tensor determines tensor
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of quadratic form by \(2\)-symmetric-tensor.
Target Context
- The reader will have a description and a proof of the proposition that the quadratic form by any \(2\)-symmetric-tensor determines the tensor.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(\{V, W\}\): \(\subseteq \{\text{ the } F \text{ vectors spaces }\}\)
\(\Sigma_2 (V: W)\): \(= \text{ the symmetric-tensors space with respect to } F \text{ and } 2 \text{ same vectors spaces and vectors space over } F\)
\(t\): \(\in \Sigma_2 (V: W)\)
\(t'\): \(\in \Sigma_2 (V: W)\)
\(\widetilde{t}\): \(: V \to W, v \mapsto t (v, v)\)
\(\widetilde{t'}\): \(: V \to W, v \mapsto t' (v, v)\)
//
Statements:
\(\widetilde{t} = \widetilde{t'}\)
\(\implies\)
\(t = t'\)
//
2: Note
This proposition holds because \(t\) and \(t'\) are presupposed to be symmetric.
If \(t\) and \(t'\) are not presupposed to be symmetric, \(\widetilde{t} = \widetilde{t'}\) will not imply \(t = t'\).
For example, let \(F = \mathbb{R}\), \(V = \mathbb{R}^2\), \(W = \mathbb{R}\), with the standard basis of \(V\), the components of \(t\) is \(\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\), and the components of \(t'\) is \(\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}\), then \(\widetilde{t} (v) = v^t \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} v = (v^1, v^2) (v^1 + v^2, v^2)^t = v^1 (v^1 + v^2) + {v^2}^2 = {v^1}^2 + v^1 v^2 + {v^2}^2\) and \(\widetilde{t'} (v) = v^t \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} v = (v^1, v^2) (v^1, v^1 + v^2)^t = {v^1}^2 + v^2 (v^1 + v^2) = {v^1}^2 + v^1 v^2 + {v^2}^2\), so, \(\widetilde{t} = \widetilde{t'}\) but \(t \neq t'\).
3: Proof
Whole Strategy: Step 1: for each \(v_1, v_2 \in V\), take \(t (v_1 + v_2) = t' (v_1 + v_2)\) and use the fact that \(t\) and \(t'\) are symmetric.
Step 1:
Let \(v_1, v_2 \in V\) be any.
\(\widetilde{t} (v_1 + v_2) = \widetilde{t'} (v_1 + v_2)\).
\(\widetilde{t} (v_1 + v_2) = t (v_1 + v_2, v_1 + v_2) = t (v_1 + v_2, v_1) + t (v_1 + v_2, v_2) = t (v_1, v_1) + t (v_2, v_1) + t (v_1, v_2) + t (v_2, v_2) = t (v_1, v_1) + t (v_1, v_2) + t (v_1, v_2) + t (v_2, v_2)\), because \(t\) is symmetric, \(= t (v_1, v_1) + 2 t (v_1, v_2) + t (v_2, v_2)\).
\(\widetilde{t'} (v_1 + v_2) = t' (v_1, v_1) + 2 t' (v_1, v_2) + t' (v_2, v_2)\), likewise.
But as \(t (v_1, v_1) = t' (v_1, v_1)\) and \(t (v_2, v_2) = t' (v_2, v_2)\), \(\widetilde{t} (v_1 + v_2) = \widetilde{t'} (v_1 + v_2)\) implies that \(t (v_1, v_2) = t' (v_1, v_2)\).
That means that for any arguments set, \(t\) and \(t'\) give the same result, which means that \(t = t'\).
