# Does dot product always commute?

I thought the dot product was commutative but there must be something about it that I don't understand. Perhaps the dot product is commutative only for vectors and not for tensors generally?

In Kusse and Westwig, p70, it says that the order of terms matters because, in general,

$$\hat{e}_j \hat{e}_k\cdot\hat{e}_l$$ does not equal $$\hat{e}_l\cdot\hat{e}_j\hat{e}_k$$

(the e's are all basis vector e's but I did not know how to show that)

[Kusse and Westwig, Mathematical Physics 2e (Wiley 2006)]

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I just realized something, after posting my question. The reason the dot product is not generally commutative probably has to do with whether the vectors or tensors are orthogonal.
Is that it? I think what confused me was that the book uses subscripts j, k, and l. That should not have confused me but it probably did and I assumed orthogonality.

I do not know about tensors, but the dot product of vectors is commutative. However, the dot product of orthogonal vectors = 0

kuruman
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I thought the dot product was commutative but there must be something about it that I don't understand. Perhaps the dot product is commutative only for vectors and not for tensors generally?

In Kusse and Westwig, p70, it says that the order of terms matters because, in general,

$$\hat{e}_j \hat{e}_k\cdot\hat{e}_l$$ does not equal $$\hat{e}_l\cdot\hat{e}_j\hat{e}_k$$

(the e's are all basis vector e's but I did not know how to show that)

[Kusse and Westwig, Mathematical Physics 2e (Wiley 2006)]
If your basis vectors are orthogonal, then the two expressions that you show are both zero, therefore equal. However, I am not sure what you mean by "dot product". If you mean "inner product" then it is strictly a scalar or zero rank tensor. The two expressions that you show are vectors (first rank tensors) pointing in different directions, one along "j" and the other along "k" so they are not generally equal.