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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?

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  • $\begingroup$ I'm not that sure that the fifth is equivalent. And yes, the last two give as result a row vector $\endgroup$ – nicoguaro♦ Jun 30 '15 at 14:26
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    $\begingroup$ We don't typically use dot notation for matrix-vector products. It is strictly used for dot products between vectors. $\endgroup$ – Wolfgang Bangerth Jun 30 '15 at 14:57
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    $\begingroup$ Typically, in the matrix/vector notation, vectors are always column vectors, and you use $v^{T}$ to turn $v$ into a row vector. In tensor notation, superscripted and subscripted indices are often used to distinguish between covariant and contravariant tensors. You'll also want to watch out for the handling of complex vectors- in most cases you'll want a Hermitian transpose and transposes and takes the conjugate. This is sometimes indicated with a superscript H rather than T. $\endgroup$ – Brian Borchers Jun 30 '15 at 15:35
  • $\begingroup$ @WolfgangBangerth One of my books on fluids uses $\nabla \mathbf{u} \cdot\mathbf{n}$ (where $\mathbf{u}$ is the fluid velocity and $\mathbf{n}$ is the unit normal on the boundary), it is an older book so maybe that's not standard notation anymore, but in any case would it be correct to say that is equivalent to $\mathbf{n}\cdot\nabla\mathbf{u}$? Because clearly the vector matrix product $\mathbf{n}\nabla\mathbf{u}$ doesn't exist. What would be the current standard way of writing this? $\endgroup$ – Lukas Bystricky Jun 30 '15 at 16:23
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    $\begingroup$ We would write it as $\nabla \mathbf u \; \mathbf n$. That will create a column vector. If you write it as $\mathbf n^T \nabla \mathbf u^T$, you get a row vector (the transpose of the first one), whereas $\mathbf n^T \nabla \mathbf u$ is a different row vector. You need to choose column/row vectors consistely if you want to add them together. $\endgroup$ – Wolfgang Bangerth Jun 30 '15 at 16:32
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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:

  1. Non standard notation, but equals $A_{ij} b_j$
  2. Equivalent to (1) as per convention
  3. Equivalent to (1)
  4. 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.
  5. I'm not sure what this yields but my best guess is $b_i A_{ij}$ which is NOT equivalent to the above.
  6. Transpose of solution to (1-4)

Hope this helps!

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No.

  1. Non-standard notation as noted by @WolfgangBangerth
  2. Column vector
  3. Same as 2
  4. Same as 3
  5. Non-standard notation
  6. Row vector
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