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The below expression has three summations (sigmas) and $L$ is a real-matrix and symmetric, $X$ is a real matrix with $n$ rows and $X_{p\mathbb{.}},X_{q\mathbb{.}}$ denote the $p$ and $q$ rows of matrix $X$.

$f(.)$ is a function acting on pairs of rows of $X$ and produces a scalar-real value. $c_p^t$ denotes the entry $t$ in a vector $c_p$ and $c_{q}^{s}$ denotes the entry $s$ in a vector $c_q$. $(L)^\dagger$ is the pseudo-inverse of $L$ and it is also a symmetric matrix in this problem.

I want the below expression to be simplified as much as possible!

ex: There may be redundancy in the summations and may be it can be expressed using two sigma's only? etc..

$\sum_{i,j=1}^{n} L_{ij} \sum_{p,q=1}^{n} \sum_{t,s=1}^{d}f(X_{p\mathbb{.}},X_{q\mathbb{.}})(L)^{\dagger }_{it}c_{p}^{t}(L)^{\dagger }_{js}c_{q}^{s}$

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What is $()^\dagger$? –  martini Nov 6 '12 at 13:20
    
$(L)^\dagger$ is the pseudo-inverse of $L$ and it is also a symmetric matrix in this problem. Also edited to include this. –  qlinck Nov 6 '12 at 13:26

1 Answer 1

up vote 2 down vote accepted

The best one can do here, I think is to see the matrix multiplications and rewrite them first, we have \[ \sum_t L^\dagger_{it}c^t_p = (L^\dagger \cdot c_p)_i,\quad \sum_s L^\dagger_{js}c^s_q = (L^\dagger \cdot c_q)_j \]

giving (after some reordering) $$ \sum_{p,q} f(X_{p\cdot}, X_{q\cdot}) \sum_{i,j} (L^\dagger \cdot c_p)_i L_{ij} (L^\dagger \cdot c_q)_j $$ Now the sum over $j$ is again a matrix product, we have $$ \sum_j L_{ij} (L^\dagger c_q)_j = (LL^\dagger c_q)_i $$ so we are left with $$ \sum_{p,q} f(X_{p\cdot}, X_{q\cdot}) \sum_{i} (L^\dagger \cdot c_p)_i (LL^\dagger \cdot c_q)_i. $$ The sum over $i$ is a scalar product, we write it as matrix product with the transposed vector, using the symmetry of $L^\dagger$: $$\sum_{i} (L^\dagger \cdot c_p)_i (LL^\dagger \cdot c_q)_i = c_p^\top L^\dagger L L^\dagger c_p$$ Now, by definition of the pseudoinverse, $L^\dagger L L^\dagger = L^\dagger$, giving $$ \sum_{p,q} f(X_{p\cdot}, X_{q\cdot}) \cdot c_p^\top L^\dagger c_q $$ As we don't know anything more about $f$ or the $c_\cdot$s, I don't think we can do more in general.

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@Martini-Thank you very much. –  qlinck Nov 6 '12 at 13:49

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