How to show matrix of probabilities is non-singular Let $\pi_1, \pi_2, \cdots, \pi_{k+1}$ be probabilities where $\pi_i\neq 0$ for all $i$ and $\pi_1+\cdots+\pi_{k+1}=1$. Consider the following matrix. Show it is a non-sigular matrix.
\begin{equation}
A_{k-1}=
\begin{bmatrix}
\pi_1(1-\pi_1) & -\pi_1\pi_2 & -\pi_1\pi_3& \cdots & -\pi_1\pi_{k-1}\\
-\pi_1\pi_2 & \pi_2(1-\pi_2) & -\pi_2\pi_3 & \cdots & -\pi_2\pi_{k-1}\\
\vdots & & &  \vdots&\\
-\pi_{k-1}\pi_1 & -\pi_{k-1}\pi_2 & -\pi_{k-1}\pi_3 & \cdots & \pi_{k-1}(1-\pi_{k-1})
\end{bmatrix}
\end{equation}
My try: I have calculated $|A_2| = \pi_1\pi_2(1-\pi_1-\pi_2)$ for $k=3$, $|A_3| = \pi_1\pi_2\pi_3(1-\pi_1-\pi_2-\pi_3)$ for $k=4$, and $|A_4| = \pi_1\pi_2\pi_3\pi_4(1-\pi_1-\pi_2-\pi_3-\pi_4)$ for $k=5$. So the proof should be running an induction. I do not know how to connect $|A_{k-1}| = \pi_1\pi_2\cdots\pi_{k-1}(1-\pi_1-\cdots-\pi_{k-1})$ to proof determinant of $|A_{k-1}|$.
 A: Here is a simple proof using the matrix-determinant lemma:
$$\displaystyle \det \left(\mathbf {A} +\mathbf {uv} ^{\textsf {T}}\right)=\left(1+\mathbf {v} ^{\textsf {T}}\mathbf {A} ^{-1}\mathbf {u} \right)\,\det \left(\mathbf {A} \right)$$
with $A=diag(\pi_1, \pi_2,\cdots,\pi_{n-1})$, $u=\begin{pmatrix}\pi_1\\ \pi_2\\ \cdots \\\pi_{n-1}\end{pmatrix}$,  $v=-u$,
giving:
$$\displaystyle\left(1-\begin{pmatrix}\pi_1&\pi_2& \cdots &\pi_{n-1}\end{pmatrix} diag(\dfrac{1}{\pi_1}, \dfrac{1}{\pi_2},\cdots,\dfrac{1}{\pi_{n-1}})\begin{pmatrix}\pi_1\\ \pi_2\\ \cdots \\\pi_{n-1}\end{pmatrix}\right)\det(A)$$
$$=(1-(\pi_1+\pi_2+\cdots+\pi_{n-1}))\prod_{k=1}^{n-1} \pi_k$$ as desired.
A: By factoring out  the common coefficient on each row we can rewrite the determinant as $$\det(A_k) = \pi_1\pi_2\ldots\pi_{k-1} \cdot\det\begin{bmatrix}
1-\pi_1 & -\pi_2 & \cdots & -\pi_{k-1}\\
-\pi_1 & 1-\pi_2 & \cdots & -\pi_{k-1}\\
\vdots & &  \vdots&\\
-\pi_1 & -\pi_2 & \cdots & 1-\pi_{k-1}
\end{bmatrix}$$
Now use the identity $I_{k-1}$ to split the matrix above as
$$
B :=\begin{bmatrix}
1-\pi_1 & -\pi_2 & \cdots & -\pi_{k-1}\\
-\pi_1 & 1-\pi_2 & \cdots & -\pi_{k-1}\\
\vdots & &  \vdots&\\
-\pi_1 & -\pi_2 & \cdots & 1-\pi_{k-1}
\end{bmatrix} = \begin{bmatrix}
-\pi_1 & -\pi_2 & \cdots & -\pi_{k-1}\\
-\pi_1 & -\pi_2 & \cdots & -\pi_{k-1}\\
\vdots & &  \vdots&\\
-\pi_1 & -\pi_2 & \cdots & -\pi_{k-1}
\end{bmatrix} + I_{k-1}$$
The lines of the matrix in the RHS are all equal to the first one, so $\lambda = 0$ is an eigenvalue of algebraic multiplicity of $k-2$, while
$$
\begin{bmatrix}
-\pi_1 & -\pi_2 & \cdots & -\pi_{k-1}\\
-\pi_1 & -\pi_2 & \cdots & -\pi_{k-1}\\
\vdots & &  \vdots&\\
-\pi_1 & -\pi_2 & \cdots & -\pi_{k-1}
\end{bmatrix}\begin{bmatrix} 1\\1\\ \vdots \\1\end{bmatrix} = \begin{bmatrix} -(\pi_1+\pi_2+\ldots+\pi_{k-1})\\-(\pi_1+\pi_2+\ldots+\pi_{k-1})\\ \vdots\\-(\pi_1+\pi_2+\ldots+\pi_{k-1})\end{bmatrix}
$$
indicates that $\lambda = -(\pi_1+\pi_2+\ldots+\pi_{k-1}) = \pi_k-1$ is also an eigenvalue. As adding the identity increases each eigenvalue by one, the unshifted matrix $B$ will have eigenvalues $\lambda = \pi_k$ (multiplicity $1$) and $\lambda = 1$ (multiplicity $k-2$), so its determinant becomes
$$
\det \begin{bmatrix}
1-\pi_1 & -\pi_2 & \cdots & -\pi_{k-1}\\
-\pi_1 & 1-\pi_2 & \cdots & -\pi_{k-1}\\
\vdots & &  \vdots&\\
-\pi_1 & -\pi_2 & \cdots & 1-\pi_{k-1}
\end{bmatrix} = \prod_{i} \lambda_i = \pi_k
$$
which shows that $\det(A_k) = \pi_1\pi_2\ldots\pi_k \neq 0$ for $\pi_i \neq 0$.
A: Since this explicitly involves probabilities, it's fitting to do a proof using tools from probability.  OP's matrix is a Covariance Matrix for a particular set of Bernoulis.  Once we show this set to be linearly independent, the result follows.
Let $U$ be a Uniform random variable taking on values in $(0,1]$ and partition these values into intervals and associated indicator random variables
$\Big\{\big(0,\pi_1\big],\big(\pi_1, \pi_1 + \pi_2\big],...,\big(\sum_{i=1}^k\pi_i, \sum_{i=1}^{k+1}\pi_i =1\big]\Big\}$
$\Big\{X_1,X_2,...,X_{k+1}\Big\}$
these random variables are dependent since for any $\omega$ exactly one of them is 1 and the others are identically zero.
Consider the centered random variable $Y_i:= X_i - \pi_i$
Note that $X_1+...+X_{k}+X_{k+1} = 1\implies Y_1+...+Y_{k}+Y_{k+1}=0$
Thus using all $k+1$ random variables gives a linearly dependent set.  So instead we focus on $Y_i$ for $i\in\{1,..,k\}$ and associated random vector
$\mathbf y: =
\begin{bmatrix}
Y_1 \\
Y_2\\
\vdots \\
Y_{k}\\
\end{bmatrix}=
\begin{bmatrix}
X_1 \\
X_2\\
\vdots \\
X_{k}\\
\end{bmatrix}-
\begin{bmatrix}
\pi_1 \\
\pi_2\\
\vdots \\
\pi_{k}\\
\end{bmatrix}$
these $Y_i$ are linearly independent, i.e for $\mathbf v\in \mathbf R^k$ we have $\big(\sum_{i=1}^{k}v_iY_i\big) = 0 \implies \mathbf v = \mathbf 0$
supposing this wasn't the case: then some there is some $i$ such that WP1
$Y_i = \big(\sum_{1\leq j\lt i} \alpha_j Y_j\big)+\big(\sum_{i\lt j \leq k} \alpha_j Y_j\big)$
yet with probability $\pi_i\gt 0$ the LHS is $1-\pi_i$ and the RHS is $c$, and with probability $\pi_{k+1}\gt 0$ the LHS is $-\pi_i$ and the RHS is $c$, thus $1-\pi_i = -\pi_i\implies 1=0$, a contradiction.
Finally, consider the Covariance Matrix
$\Sigma = E\big[\mathbf y\mathbf y^T\big] = 
\begin{bmatrix}
\pi_1(1-\pi_1) & -\pi_1\pi_2 & -\pi_1\pi_3& \cdots & -\pi_1\pi_{k}\\
-\pi_1\pi_2 & \pi_2(1-\pi_2) & -\pi_2\pi_3 & \cdots & -\pi_2\pi_{k}\\
\vdots & & &  \vdots&\\
-\pi_{k}\pi_1 & -\pi_{k}\pi_2 & -\pi_{k}\pi_3 & \cdots & \pi_{k}(1-\pi_{k})
\end{bmatrix}$
for any $\mathbf v \in \mathbf R^{k}-\{\mathbf 0\}$
$\mathbf v^T\Sigma\mathbf v$
$=\sum_{i=1}^{k}\sum_{j=1}^{k}v_i v_jE\big[Y_iY_j\big]$
$=E\big[\sum_{i=1}^{k}\sum_{j=1}^{k}v_i v_jY_iY_j\big]$
$=E\big[\big(\sum_{i=1}^{k}v_iY_i\big)^2\big]$
$\gt 0$
Thus $\Sigma\succ \mathbf 0$ and the matrix $A_{k-1}$ is a leading principal minor and hence it is PD as well (which implies non-singular).
