Given 2 vectors $\mathbf{u}$ and $\mathbf{v}$ the following are equivalent: $\mathbf{u}\cdot\mathbf{v}$ $\mathbf{u}^T \mathbf{v}$ $u_i v_i$ $v_i u_i$ $\mathbf{v}\cdot\mathbf{u}$ $\mathbf{v}^T\mathbf{u}$Given a second order tensor (matrix) $\mathbf{A}$ and a vector $\mathbf{b}$, are the following also equivalent? $\mathbf{A}\cdot\mathbf{b}$ $\mathbf{A} \mathbf{b}$ $A_{ij} b_j$ $b_j A_{ij}$ $\mathbf{b}\cdot\mathbf{A}$ $\mathbf{b}^T\mathbf{A}^T$I know that the last one (and possibly the second last one) gives a row vector instead of a column vector. Is it possible to tell in index notation whether a vector is a row or column vector, or is that supposed to be clear based on context?
Vector and index notation equivalence
vector;tensor
Is it possible to tell in index notation whether a vector is a row or column vector, or is that supposed to be clear based on context?It seems the answer is actually lurking in your question itself here: if $u \cdot v \equiv u^T v$ then that doesn't leave much wiggle room. $u^T$ will have to be a row vector in order for the resulting product to be a scalar quantity - and as we all know, the dot product of two vectors is a scalar. The other scenario would give you a $n$ x $n$ system. Extending this logic to the case of $v^T u$ we see that both $v$ and $u$ are indeed column vectors - this is something I have observed is usually true regarding vector representation in literature. Now coming to the second part of your question, as Wolfgang has already mentioned, the dot notation between a matrix and a vector is not standard notation. However, if you must use it I think you will have to follow the guidelines of indicial notation in which a dot product essentially refers to a contraction, i.e. $A \cdot b \equiv A_{ij} b_j$. Which is once again a column vector (say $\mathbf{X}$). So there are some equivalencies in your second list, yes. For the last case, if you consider $(A b)^T = \mathbf{X}^T$, and apply the standard identity to the LHS you get $b^T A^T = \mathbf{X}^T$. So really, you are getting a row vector when you go out of your way to take a transpose of your entire system, which - if you think about it - is quite intuitive. Strictly speaking, your vectors are still column vectors, it is just that you have chosen to take their transpose. In summary:Non standard notation, but equals $A_{ij} b_j$Equivalent to (1) as per conventionEquivalent to (1)Equivalent to (3) as at this point, $A_{ij}$ and $b_j$ are merely scalars and their order is interchangeable. Also, note that $i$ is a dummy index here. I'm not sure what this yields but my best guess is $b_i A_{ij}$ which is NOT equivalent to the above. Transpose of solution to (1-4)Hope this helps!