# A Basis for the space of covariant tensors ... ...Another Question ... Browder Theorem 12.9 ...

#### Peter

##### Well-known member
MHB Site Helper
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 12: Multilinear Algebra ... ...

I need some further help in order to fully understand the proof of Theorem 12.9 on pages 271-272 ... ...

The relevant text (Theorem 12.8 together with the preceding definition, Definition 12.8) reads as follows:

In the above proof by Browder we read the following:

" ... ... If $$\displaystyle c_{ j_1 \cdot \cdot \cdot j_r }$$ are scalars such that

$$\displaystyle \sum c_{ j_1 \cdot \cdot \cdot j_r } \tilde{u}^{j_1} \otimes \cdot \cdot \cdot \otimes \tilde{u}^{j_r} = 0$$

it follows that $$\displaystyle c_{ j_1 \cdot \cdot \cdot j_r } = 0$$ for all $$\displaystyle j_1 \cdot \cdot \cdot j_r$$. ... ... "

Can someone show explicitly why the above statement is true ... ... that is, demonstrate an explicit proof for the above statement ...

***NOTE***

The notation $$\displaystyle \tilde{u}^{j_i}$$ is explained in Browder Theorem 12.2 which is shown below ... ...

Help will be much appreciated ...

Peter

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I am providing the text of Chapter 12 up to and including Theorem 12.9 to give readers the context and notation of the Chapter ... ... and any necessary preliminary definitions and results ... as follows:

Hope that helps ... ...

Peter

Last edited:

#### steenis

##### Well-known member
MHB Math Helper
Let
$$v=\sum c_{ j_1 \cdot \cdot \cdot j_r } \tilde{u}^{j_1} \otimes \cdot \cdot \cdot \otimes \tilde{u}^{j_r} = 0$$

$$v \in V^{*}$$

Using 12.9, produce a $\underline{u} \in V \otimes \cdots \otimes V$ (r-times) such that

$$0=v(\underline{u}) = c_{ j_1 \cdot \cdot \cdot j_r }$$

Again, practice with low n and r