Inversion of a matrix of the given form 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?
 A: 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:
$$ \eqalign{(X^{-1})_{ii} &= - \frac{n-2}{(n-1) w_i^2}\cr
            (X^{-1})_{ij} &= \frac{1}{(n-1) w_i w_j} \ \text{otherwise}\cr} $$
A: 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.
