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My lecturer has used some notation that I've never seen before: it is a (matrix) product symbol with a left-to-right arrow over the top. Does anybody know what this means?

Thanks in advance.

Edit: It looks like this:

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  • $\begingroup$ Perhaps you could upload it as an image... to see what it actually looks like $\endgroup$ – Tolaso Aug 10 '15 at 11:56
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    $\begingroup$ I think the simplest answer is to ask your lecturer what he/she means by it. $\endgroup$ – wltrup Aug 10 '15 at 12:21
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    $\begingroup$ @Tolaso not statistics. It was really just linear algebra. $\endgroup$ – Ziggy Aug 10 '15 at 12:23
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    $\begingroup$ For whatever it's worth: you can format the symbol as \prod^\curvearrowright to get $$ \prod^\curvearrowright $$ not sure how to add the indices in there though $\endgroup$ – Omnomnomnom Aug 10 '15 at 12:41
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    $\begingroup$ @Omnomnomnom \overset{\curvearrowright}{\prod^{k}} produces $$ \overset{\curvearrowright}{\prod^{k}}$$ $\endgroup$ – wltrup Aug 10 '15 at 13:11
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Matrix multiplication is not commutative, so (maybe) that symbol means that you are considering the product adding the next matrix to the right: $$ X_1 X_2 \cdots X_k. $$ Anyway, I am just guessing, I've never seen it before!

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    $\begingroup$ You mean "Matrix multiplication is not commutative, ..." :) $\endgroup$ – wltrup Aug 10 '15 at 13:00
  • $\begingroup$ Sure! Thanks ;) $\endgroup$ – Paglia Aug 10 '15 at 13:05
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The arrow notation removes ambiguity when certain procedures are represented by successive matrix multiplication.

For example, we can find $LU$ decomposition of the matrix $A$ by running Gaussian elimination on $A$. The reduced matrix is the upper triangular matrix $U$, where the reduction is represented by left multiplication of transformation matrices $E_i$ for $i \in\{1,\ldots, n-1\}$,

$$U = \left(\overset{\curvearrowleft}{\prod_{i=1}^{n-1}} E_i\right) A = E_{n-1}\cdots E_1 A.$$

Further, it can be shown that

$$L = \overset{\curvearrowright}{\prod_{i=1}^{n-1}} E_i^{-1} = E_1^{-1}\cdots E_{n-1}^{-1}$$

is a lower triangular matrix. Whence we obtain $A = LU$.

Not great for exposition, but fine as short hand.

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The picture suggests that these are perhaps just two lines, which give $$\curvearrowright \prod_{i=1}^k X_i=Y$$ if you write it in one line. Then it would mean that the equation $ \prod_{i=1}^k X_i=X_1X_2\cdots X_k=Y$ follows from another relation in the text before (i.e., $\cal{X} \curvearrowright \cal{Y})$.

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