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I am not sure I'm asking the question correctly, but it's bugging me too much to let it go.

On this answer to a prior post by me I got stuck in what initially seemed to correspond to a dot product of vectors, and turned out to be a different operation.

My knowledge of linear algebra makes me understand a covariance matrix (as a prototypical case of symmetric semidefinite matrices, or $\bf A^\top A$ forms on my prior post) as a long list of returns for different stock ticker symbols on many days (as an example), each one representing a vector that after being scaled, and dotted with itself and the other stock company returns in the portfolio, is properly placed in a symmetric structure where the variances are in the diagonal, and covariances are off-diagonal.

Now, the operation in the answer is making reference to block matrices, with each element, $a_i$, corresponding to a row vector of norm $1$, such that

\begin{align} A^\top A & = \begin{bmatrix} \vdots & \vdots & \vdots & \cdots & \vdots \\ a_1^\top & a_2^\top & a_3^\top & \cdots & a_n^\top\\ \vdots & \vdots & \vdots & \cdots & \vdots\end{bmatrix} \begin{bmatrix} \cdots & a_1 & \cdots\\ \cdots & a_2 & \cdots \\ \cdots & a_3 & \cdots \\ & \vdots&\\ \cdots & a_n & \cdots \end{bmatrix}\\ &= a_1^\top a_1 + a_2^\top a_2 + a_3^\top a_3 + \cdots+a_n^\top a_n. \end{align}

I don't understand how this outer product (??) would work out to end up with the sum of outer products of vectors of identical indexes.

I know that an example (much like debugging a loop function) would clarify the overall structure of the resulting matrix and the operations involved. Alternatively, even the name of the operation possibly with a link or reference would be good.

Please note that the author explicitly clarifies in the comments, "I'm no longer saying they are orthogonal [the $a_i$ column vectors], only that the projections are orthogonal."

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    $\begingroup$ See en.wikipedia.org/wiki/Block_matrix#Block_matrix_multiplication. It may also help you to explicitly write down the size of the matrices involved (ie how many rows and columns does $a_i$ have? $a_i^T$? $a_i^Ta_i$?) $\endgroup$ – stewbasic Sep 9 '16 at 2:20
  • $\begingroup$ @stewbasic Thank you. Do you want to write something so I can approve it? I checked the link, and now it's clear. $\endgroup$ – Antoni Parellada Sep 9 '16 at 2:27
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    $\begingroup$ If $A$ has more rows than columns and the columns are linearly independent, then $A^\top A$ is invertible, and (1) $B = (A^\top A)^{-1} A^\top$ is a left inverse of $A$, in that $BA$ is an identity matrix (which has only as many rows or columns as $A$ has columns) and (2) $AB$ is a symmetric singular matrix that projects onto the column space of $A$, i.e. if $x$ is a column as big as those of $A$ then $ABx = \begin{cases} x & \text{if $x$ is in the column space of $A$}, \\ 0 & \text{if $x$ is orthogonal to the column space}. \end{cases}$ $\qquad$ $\endgroup$ – Michael Hardy Sep 9 '16 at 2:30

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