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I have a matrix with data, every dataset is a column vector in my matrix. I want to know the dot product of the transpose of each column vector with the original column vector. If I transpose the matrix and then do matrix multiplication with itself, what I want is on the diagonal. Is there a way to perform this calculation and only get the vector consisting of those diagonal elements? The diagonal elements are the sums of the squares of the column vectors.

Original $m \times n$ matrix $M$ with $m$ sets at $n$ points:

$M =\left[ \begin{array} {cc}M_{1,1}&...&M_{1,n}\\.&.&.\\.&.&.\\.&.&.\\M_{m,1}&...&M_{m,n} \end{array} \right] = \left[ \begin{array} {cc}.&.&.\\.&.&.\\c1&c...&cn\\.&.&.\\.&.&. \end{array} \right] $

Resulting $1 \times m$ matrix with resulting dot products of $c_n^T \cdot c_n$, e.g. the sum of its squares:

$R = \left[ \begin{array}{cc}c_1^T \cdot c_1, ..., c_n^T \cdot c_n \end{array} \right]$

Which are the diagonal elements of:

$M^T\cdot M = S = \left[ \begin{array} {cc}R_1&.&.\\.&...&.\\.&.&R_n \end{array} \right] $

How do I calculate just the diagonals, and get that vector, instead of the whole matrix $S$.

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    $\begingroup$ What do you need this for? In Matlab you could say R = sum(M.^2); $\endgroup$
    – littleO
    Commented Oct 3, 2013 at 9:16
  • $\begingroup$ Indeed I am using this in a matlab script, but when I was writing it down I thought, 'how do I describe this mathematically?' and I didnt know so thats why I asked this question. $\endgroup$
    – Leo
    Commented Oct 3, 2013 at 12:35

1 Answer 1

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Do you need the answer in terms of computational complexity? I can only come up with a non-closed form like that (which is pretty useless, but answers your question to some extent): $$ R = \sum_{i=1}^n (0 \ldots 1_i \ldots 0) \cdot M^T \cdot M \cdot (0 \ldots 1_i \ldots 0)^T \cdot (0 \ldots 1_i \ldots 0). $$

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  • $\begingroup$ Im looking for a linear algbra expression to reach my end goal, preferably but not necessarily without calculating the whole matrix $S$. $\endgroup$
    – Leo
    Commented Oct 2, 2013 at 14:59

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