1024: For Tensor Product of $k$ Finite-Dimensional Vectors Spaces over Field, Transition of Standard Bases w.r.t. Bases for Vectors Spaces Is This|T.B.P.

k

finite-dimensional vectors spaces over field, transition of standard bases w.r.t. bases for vectors spaces is this

Topics

About: vectors space

The reader will have a description and a proof of the proposition that for the tensor product of any $k$ finite-dimensional vectors spaces over any field, the transition of the standard bases with respect to any bases for the vectors spaces is this.

F

\in {the fields}

{V_{1}, . . ., V_{k}}

\subseteq {the finite-dimensional F vectors spaces}

V_{1} \otimes . . . \otimes V_{k}

= the tensor product

{B_{1}, . . ., B_{k}}

B_{j} \in {the bases for V_{j}} = {{b_{j}}_{l} | 1 \leq l \leq d i m V_{j}}

{B_{1}^{'}, . . ., B_{k}^{'}}

B_{j}^{'} \in {the bases for V_{j}} = {{b_{j}^{'}}_{l} | 1 \leq l \leq d i m V_{j}}

B

= {[(({b_{1}}_{l_{1}}, . . ., {b_{k}}_{l_{k}}))] | {b_{j}}^{l_{j}} \in B_{j}}

\in {the bases for V_{1} \otimes . . . \otimes V_{k}}

B^{'}

= {[(({b_{1}^{'}}_{l_{1}}, . . ., {b_{k}^{'}}_{l_{k}}))] | {b_{j}^{'}}^{l_{j}} \in B_{j}^{'}}

\in {the bases for V_{1} \otimes . . . \otimes V_{k}}

//

Statements:

{b_{j}^{'}}_{l} = {b_{j}}_{m} {M_{j}}_{l}^{m}

⟹

[(({b_{1}^{'}}_{l_{1}}, . . ., {b_{k}^{'}}_{l_{k}}))] = [(({b_{1}}_{m_{1}}, . . ., {b_{k}}_{m_{k}}))] {M_{1}}_{l_{1}}^{m_{1}} . . . {M_{k}}_{l_{k}}^{m_{k}}

Whole Strategy: Step 1: see that

B

and

B^{'}

are some bases for

V_{1} \otimes . . . \otimes V_{k}

; Step 2: conclude the proposition.

Step 1:

B

and

B^{'}

are indeed some bases for

V_{1} \otimes . . . \otimes V_{k}

k

[(({b_{1}^{'}}_{l_{1}}, . . ., {b_{k}^{'}}_{l_{k}}))] = [(({b_{1}}_{m_{1}} {M_{1}}_{l_{1}}^{m_{1}}, . . ., {b_{k}}_{m_{k}} {M_{k}}_{l_{k}}^{m_{k}}))]

.

We note the fact that in general,

[((v_{1}, . . ., r v_{j} + r^{'} v_{j}^{'}, . . ., v_{k}))] = r [((v_{1}, . . ., v_{j}, . . ., v_{k}))] + r^{'} [((v_{1}, . . ., v_{j}^{'}, . . ., v_{k}))]

k

[(({b_{1}}_{m_{1}} {M_{1}}_{l_{1}}^{m_{1}}, . . ., {b_{k}}_{m_{k}} {M_{k}}_{l_{k}}^{m_{k}}))] = [(({b_{1}}_{m_{1}}, . . ., {b_{k}}_{m_{k}}))] {M_{1}}_{l_{1}}^{m_{1}} . . . {M_{k}}_{l_{k}}^{m_{k}}

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