Is it true: $\mathbf{a.B}=\mathbf{B^{T}.a}$

Question

Important notion

I know that this question may seem too simple for you mathematicians, but I know nowhere than here to ask it.

Question

Is it true: $$\mathbf{a.B}=\mathbf{B^{T}.a}$$

Assumptions

Assuming $\mathbf{a}$ as a Cartesian vector and $\mathbf{B}$ as a second-order tensor.

My route

It states that:$$a_ib_{ij}=b_{ji}a_i$$Expanding LHS yields$$a_1b_{1j}+a_2b_{2j}+a_3b_{3j}$$and doing so for RHS$$a_1b_{j1}+a_2b_{j2}+a_3b_{j3}$$which for a generally non-symmetric $\mathbf{B}$ is not true. So, which one is false, my understanding or the theorem?

Answer 1

Notice that you need to transpose $B$, i.e. $(B^t)_{ij}=B_{ji}$.