If $A$ is a matrix and $B$ is a tensor (for example 3 by 3/rank 2, and with the same components) and $v$ is a 3 by 1 vector,

  • Is there any difference between $A.v$ and $B.v$(in terms of the formula to compute them), where $.$ is the dot product. Are they both a normal "matrix multiplication" which result a 3 by 1 vector?

Thank you for your help.

  • $\begingroup$ Do you understand the difference between a "matrix" and a "tensor"? It is the same as the difference between a "triple of numbers" and a vector. A tensor can be represented by a matrix in a given coordinate system. If we change the coordinate system the matrix may change but the tensor is the same, just represented by a different matrix. $\endgroup$
    – user247327
    Jun 25, 2021 at 19:26
  • $\begingroup$ any of your matrices are used as $Av$ to give you a linear transformation but also $v^{\top}Aw$ to have a bilinear map into the scalar field $\endgroup$
    – janmarqz
    Jun 25, 2021 at 20:41

1 Answer 1


Depending on uses of indexations.

If $A$ is used to represent a linear transformation then ones could use $$w^k=A_s{}^kv^s,$$ to get $n$ quantities (since $1\le k\le n$) for the components of $w$, from those of $v$.

Or for a bilinear map $B$, where two vectors $u,v$ are paired via $$B_{st}u^sv^t,$$ to assign a number.

The repetition of one index above and one below is to indicate summation (The Einstein's Sum Convention).

However other indexation's conventions could be employed.


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