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Assume that $w$ is a probability-vector with positive components and $X$ denotes a matrix of the form $ww^T$, but with zeros on its diagonal. So, $x_{ij} = (1 - \delta_{ij}))w_iw_j$.

Is this matrix invertible in the general case? If so, how to find its inversion?

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    This is a good question. I think the answer involves looking at eigenvectors, but I cannot solve it. – 2017-01-13
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    it does not seem to be invertible in the general case. First the 0 matrix has the required form. More generally if $w$ has a 0 coefficient you get a 0 column in your matrix. Consequently we have to consider only $w$ vectors with non-zero entries – 2017-01-13
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    Clearly not all such matrices are invertible but some are. Your Question would be improved by giving examples of each case. The topic can be compared with rank one updates to invertible matrices. E.g. what about when $w$ has a zero entry? – 2017-01-13
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    @hardmath you were right. I have missed important point. The vector $w$ define a probability distribution and has positive components. – 2017-01-13
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    @fonfonx, thank you, you're right. I've edited my post – 2017-01-13

2 Answers 2

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As fonfonx remarked, you must assume all $w_i \ne 0$. That turns out to be sufficient. In fact, $X^{-1}$ has a closed form:

$$ \eqalign{(X^{-1})_{ii} &= - \frac{n-2}{(n-1) w_i^2}\cr (X^{-1})_{ij} &= \frac{1}{(n-1) w_i w_j} \ \text{otherwise}\cr} $$

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    Wow, thank you! Is it a known fact or you've used some intuition? – 2017-01-13
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Let $v$ be an $n{\times}1$ column vector, and let $A$ be the matrix obtained from the matrix $vv^T$ by setting all diagonal entries to $0$.

Here's an answer to your first question ("When is $A$ invertible?").

If $n = 1$ then, regardless of the choice of $v$, $A$ is the $1{\times}1$ zero matrix, hence $A$ is not invertible.

Next, assume $n > 1$.

Claim: $A$ is invertible if and only if all components of $v$ are nonzero.

Justification . . .

Note that every generalized diagonal of $A$ has product equal to either $p^2$ or $-p^2$, where ${\displaystyle p = {\prod_{i=1}^n v_i}}$.

Then for each fixed $n$, there is an integer $m$, independent of $v$, such that $\text{det}(A) = mp^2$.

Claim $m \ne 0$.

Let $v$ be the $n{\times}1$ column vector with all entries equal to $1$.

For this choice of $v$, we have $p^2 = 1$, hence $\text{det}(A) = m$.

Then $A + I$ is the $n{\times}n$ matrix with all entries equal to $1$, hence

\begin{align*} &(A + I)^2 = n(A + I)\\ \implies\; &A^2 + 2A + I = nA + nI\\ \implies\; &A^2 - (n-2)A = (n-1)I\\ \implies\; &A(A - (n-2)I)) = (n-1)I\\ \implies\; &\text{det}(A) \ne 0\\ \implies\; &m \ne 0\\ \end{align*}

Returning to the general case with $v$ unknown, then since $\text{det}(A) = mp^2$ and $m \ne 0$, it follows that $A$ is invertible if and only if all components of $v$ are nonzero, as claimed.