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Assume $A$ is a square matrix defined as follow: $$A=\sum_{i} u_{i}u_{i}^T$$ where for each $i$, $u_i$ is a non-negative column vector.

Do the matrices of these forms have any special name?

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  • $\begingroup$ So $A$ is $U\cdot U^T$ where $U$ is a nonnegative square matrix. $\endgroup$ – Louis Nov 15 '13 at 0:09
  • $\begingroup$ No it's not. Note that in the definition I have $u_{i}u_{i}^T$ not $u_{i}u_{j}^T$. $\endgroup$ – user54626 Nov 15 '13 at 0:12
  • $\begingroup$ In general, if $a_i$ are the column vectors of $A$ and $b_i$ are the column vectors of $B$, then $A\cdot B^T = \sum a_i \cdot b_i^T$ $\endgroup$ – Louis Nov 15 '13 at 0:17
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    $\begingroup$ If you drop the condition that $u$ is non-negative, the name is "symmetric positive-definite". The extra condition is certainly important. Maybe the answer is "symmetric positive-definite non-negative"? Do you have an example that disproves this guess? $\endgroup$ – apt1002 Nov 15 '13 at 0:22
  • $\begingroup$ Not entirely relevant, but a matrix of the form $u v^T$ is called a dyad. $\endgroup$ – copper.hat Nov 15 '13 at 0:31
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Yes, these are the completely positive matrices.

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Let's assumes there are $n$ such $u_i$'s and there are exactly $k$ linearly independent vectors among them, then matrix $A$ would be a rank-k matrix.
Each of the $u_iu_i^t$ is a rank-1 matrix and they would sum up to a rank-k matrix if exactly k out of them are linearly independent.
As $x^t(u_iu_i^t)x=(x^tu_i)(x^tu_i)^t \gt0$ , each of the $u_iu_i^t$ is a symmetric positive definite matrix, thus A is also a symmetric positive definite matrix.

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There is a problem with the term 'non-negative column vector'. The notion of non-negative is defined for square matrices.

We can require $u_i$'s such that $M=[u_{ij}]$ is positive (non-negative) that is such that $M=UU^*$ for some square matrix $U$.

There is also a different notion of positivity (again for square matrices) which requires the entries to be non-negative.

One vaguely related idea is the following. If $M$ is a symmetric matrix that is orthogonally diagonalised by

$P=[u_1\space{} u_2\space{}\dots u_n]$

and if $\lambda_i$'s are eigenvectors of $M$ corresponding to the unit eigenvectors $u_i$'s, then

$M=\lambda_1 u_1u_1^T+\dots +\lambda_nu_nu_n^T$

is called the spectral decomposition of $M$.

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  • $\begingroup$ I'm assuming the components of the column vectors are nonnegative. Perhaps the OP should clarify. $\endgroup$ – Stefan Smith Nov 15 '13 at 1:00
  • $\begingroup$ I don't know what do you mean by "$A$, as it is written, denotes a single positive number". I mentioned that each $u_i$ is a column vector so $A$ is a matrix. $\endgroup$ – user54626 Nov 15 '13 at 3:22
  • $\begingroup$ Fixed that. Will look it up. $\endgroup$ – AnyAD Nov 15 '13 at 4:20

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