# Proof that the tensor product is associative .... Browder Section 12.7 and 12.8 ... ...

#### 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 am uncertain regarding a formal and rigorous proof that the tensor product is associative ... so I will give my attempt at a proof and then I hope that someone will kindly critique the proof for me ...

I will use the definitions and notation of Sections 12.7 and 12.8 ...

Sections 12.7 and 12.8 read as follows:

We have to show that $$\displaystyle ( \alpha \otimes \beta ) \otimes \gamma = \alpha \otimes ( \beta \otimes \gamma)$$

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Let $$\displaystyle \alpha$$ be such that $$\displaystyle \alpha : V^r \to \mathbb{R}$$

Let $$\displaystyle \beta$$ be such that $$\displaystyle \beta : V^s \to \mathbb{R}$$

Let $$\displaystyle \gamma$$ be such that $$\displaystyle \gamma : V^t \to \mathbb{R}$$

Then we have ...

$$\displaystyle ( \alpha \otimes \beta ) \otimes \gamma (v_1 \ ... \ ... \ v_{ r + s + t } )$$

$$\displaystyle = ( \alpha \otimes \beta ) (v_1 \ ... \ ... \ v_{ r + s } ) \gamma (v_{ r + s + 1} \ ... \ ... \ v_{ r + s + t } )$$

$$\displaystyle = \alpha (v_1 \ ... \ ... \ v_r ) \beta (v_{ r + 1 } \ ... \ ... \ v_{ r + s } ) \gamma (v_{ r + s + 1 } \ ... \ ... \ v_{ r + s + t } )$$ ... ... ... (1)

... and we also have ...

$$\displaystyle \alpha \otimes ( \beta \otimes \gamma) (v_1 \ ... \ ... \ v_{ r + s + t } )$$

$$\displaystyle = \alpha (v_1 \ ... \ ... \ v_r ) ( \beta \otimes \gamma) (v_{ r + 1 } \ ... \ ... \ v_{ r + s + t } )$$

$$\displaystyle = \alpha (v_1 \ ... \ ... \ v_r ) \beta (v_{ r + 1 } \ ... \ ... \ v_{ r + s } ) \gamma (v_{ r + s + 1 } \ ... \ ... \ v_{ r + s + t } )$$ ... ... ... (2)

Now (1), (2) $$\displaystyle \Longrightarrow$$ $$\displaystyle ( \alpha \otimes \beta ) \otimes \gamma = \alpha \otimes ( \beta \otimes \gamma)$$

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NOTE: ... ... in the above proof I calculated/processed the tensor product sign in the parenthesis second ... not first as I believe the notation is supposed to mean ...

I did this because I felt that when expanding a tensor product we needed to keep the form

tensor $$\displaystyle \otimes$$ tensor (ordered list of variables) = tensor (ordered list) tensor (ordered list)

and processing the tensor product within the parentheses first did not appear to allow this form to be preserved ...

Can someone please critique the above proof ...

Peter

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#### GJA

##### Well-known member
MHB Math Scholar
Looks great to me. Nicely done!

#### steenis

##### Well-known member
MHB Math Helper
By definition 12.8, you can only multiply two tensors. If you want to multiply 3 tensors, $\alpha \otimes \beta \otimes \gamma$, the trick is to compute $\alpha \otimes ( \beta \otimes \gamma )$ because you can compute the product of two tensors $\beta \otimes \gamma = \delta$ and you can compute the product of two tensors $\alpha \otimes \delta$

The evaluation of this is, as we read in the four lines above definition 12.8:
$$[\alpha \otimes ( \beta \otimes \gamma )] (\underline{u}, \underline{v}, \underline{w}) =$$
$$\alpha( \underline{u} ) [[ (\beta \otimes \gamma )] (\underline{v}, \underline{w})] =$$
$$\alpha( \underline{u} ) [ \beta( \underline{v} ) \gamma( \underline{w} )] =$$
$$\alpha( \underline{u} ) (\beta( \underline{v} ) \gamma( \underline{w} )$$

You can say that you evaluate $(\beta \otimes \gamma)$ first, before evaluating the whole

#### Peter

##### Well-known member
MHB Site Helper
By definition 12.8, you can only multiply two tensors. If you want to multiply 3 tensors, $\alpha \otimes \beta \otimes \gamma$, the trick is to compute $\alpha \otimes ( \beta \otimes \gamma )$ because you can compute the product of two tensors $\beta \otimes \gamma = \delta$ and you can compute the product of two tensors $\alpha \otimes \delta$

The evaluation of this is, as we read in the four lines above definition 12.8:
$$[\alpha \otimes ( \beta \otimes \gamma )] (\underline{u}, \underline{v}, \underline{w}) =$$
$$\alpha( \underline{u} ) [[ (\beta \otimes \gamma )] (\underline{v}, \underline{w})] =$$
$$\alpha( \underline{u} ) [ \beta( \underline{v} ) \gamma( \underline{w} )] =$$
$$\alpha( \underline{u} ) (\beta( \underline{v} ) \gamma( \underline{w} )$$

You can say that you evaluate $(\beta \otimes \gamma)$ first, before evaluating the whole

Thanks Steenis ... your post really got me thinking ...

Yes ... important point that ... ... " ... By definition 12.8, you can only multiply two tensors. ... "

But just a couple of questions ...

Question 1

In your evaluation above you write ...

" ... $$\alpha( \underline{u} ) [ \beta( \underline{v} ) \gamma( \underline{w} )] =$$
$$\alpha( \underline{u} ) (\beta( \underline{v} ) \gamma( \underline{w} )$$ ... "

Why is this true ...?

Is it because $$\alpha( \underline{u} ), \beta( \underline{v} ) \text{ and } \gamma( \underline{w} )$$ are real numbers when evaluated and real numbers are associative ... ... ?

Question 2

You write ...:

" ... You can say that you evaluate $(\beta \otimes \gamma)$ first, before evaluating the whole ... "

Can you justify this statement ...

It looks to me as if $(\beta \otimes \gamma)$ is evaluated second which violates the meaning of the parentheses (which mean that the operation in the parentheses is evaluated first ... ... ) ... ...

Can you please comment ...

Thanks again for your help ...

Peter

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#### steenis

##### Well-known member
MHB Math Helper
Q1: that is correct

Q2: I understand what you mean regarding the parenthess and I tried to give an answer in that direction. But I was not happy with my own answer too.

Strictly speaking, if you leave the parentheses away then there is something that is not defined yet: $\alpha \otimes \beta \otimes \gamma$. It is defined after you have proven the associativity

The parentheses indicate that you have to do more steps to evaluate the expression $\beta \otimes \gamma$ within $\alpha \otimes ( \beta \otimes \gamma )$.
So you evaluate $\alpha$, giving $\alpha (\underline{u})$ and then you evaluate $\beta \otimes \gamma$, giving $\beta (\underline{v}) \gamma (\underline{w})$ in two steps

#### Peter

##### Well-known member
MHB Site Helper
Q1: that is correct

Q2: I understand what you mean regarding the parenthess and I tried to give an answer in that direction. But I was not happy with my own answer too.

Strictly speaking, if you leave the parentheses away then there is something that is not defined yet: $\alpha \otimes \beta \otimes \gamma$. It is defined after you have proven the associativity

The parentheses indicate that you have to do more steps to evaluate the expression $\beta \otimes \gamma$ within $\alpha \otimes ( \beta \otimes \gamma )$.
So you evaluate $\alpha$, giving $\alpha (\underline{u})$ and then you evaluate $\beta \otimes \gamma$, giving $\beta (\underline{v}) \gamma (\underline{w})$ in two steps

Thanks for such a clear and honest post, Hugo ...

Still reflecting on your new answer to the issue ...

Does anyone else have an answer ...?

Peter

#### steenis

##### Well-known member
MHB Math Helper
I understand your hesitation.

Right now, I am on holidays in Saigon, Vietnam. I have time to follow MHB, but I have trouble focusing. What do you say, it is 33-35 degrees C here, and I do not like that. In a week I hope to be home again.

#### Peter

##### Well-known member
MHB Site Helper
I understand your hesitation.

Right now, I am on holidays in Saigon, Vietnam. I have time to follow MHB, but I have trouble focusing. What do you say, it is 33-35 degrees C here, and I do not like that. In a week I hope to be home again.

Hi Hugo ...

Thanks again for your help ...

Hope that despite the heat, you enjoy your holiday ...

It gets hot down here sometimes ... even though I live in the southern-most part of Australia ...

Peter