Consider the following matrix:

$$\left[\begin{array}{cccc} 1+x_1y_1 & 1+x_1y_2 & \ldots & 1+x_1y_n \\ 1+x_2y_1 & 1+x_2y_2 & \ldots & 1+x_2y_n \\ 1+x_3y_1 & 1+x_3y_2 & \ldots & 1+x_3 y_n \\ \vdots & & & \\ 1+x_ny_1 & 1+x_ny_2 & \ldots & 1+x_n y_n \\ \end{array} \right]$$ with $x_k, y_k$ being real numbers. I suppose that there must be some neat way of computing the determinant of this matrix. Those ones are annoying to me.

  • $\begingroup$ This appears to be singular, after doing several matrices out by hand and doing some symbolic calcs in Octave $\endgroup$
    – costrom
    Commented May 16, 2016 at 20:30
  • 1
    $\begingroup$ @costrom this heavily depends on your choice of the $x_k$ and $y_k$ respectively $\endgroup$
    – user190080
    Commented May 16, 2016 at 20:34

3 Answers 3


The matrix can be expressed as the sum of two rank $1$ matrices, each of which is the outer product of vectors $$M=\mathbf{11}^T+\mathbf{xy}^T$$ where $\mathbf{1}=[1,1,1,\cdots,1]^T\in\mathbb{R}^{n\times 1}$, $\mathbf{x}=[x_1,x_2,\cdots,x_n]^T\in\mathbb{R}^{n\times 1}$ and $\mathbf{y}=[y_1,y_2,\cdots,y_n]^T\in\mathbb{R}^{n\times 1}$.

Using the subadditive property of the rank, where $\text{rank}(A+B)\leq\text{rank}(A)+\text{rank}(B)$, we have $$\text{rank}(M)\leq\text{rank}(\mathbf{11}^T)+\text{rank}(\mathbf{xy}^T)\Rightarrow \text{rank}(M)\leq 2$$ Thus, only when $n=2$ the matrix can be full rank, and so potentially have a non-zero determinant, which is easily evaluated as $(1+x_1y_1)(1+x_2y_2)-(1+x_2y_1)(1+x_1y_2)$. However for $n\geq3$, the matrix will be rank deficient, in which case the determinant will be $0$.

  • $\begingroup$ This is impressively simple. Well done! (+1) $\endgroup$ Commented May 16, 2016 at 21:08

Here are two approaches.

1) By inspecting the pattern of this matrix, one sees that it is of the form $A+x y^T$ where $A$ is the matrix of ones and $x,y$ are appropriate column vectors. The matrix determinant lemma then gives $\det(A+x y^T)=\det(A)+y^T \text{adj}(A)x$ where $\text{adj}(A)$ is the adjugate matrix of $A$. But the determinant of $A$ vanishes; moreover, if $n\geq 3$ then then so too does the the adjugate, so the determinant is equal to zero for $n\geq 3$.

2) If we imagine trying to compute this determinant by cofactors, we see that should be some degree polynomial of degree $2n$ in the variables $x_k,y_k$. Furthermore, said polynomial must be antisymmetric with respect to swaps of $x$- or $y$- variables. This suggests an answer of the form $\displaystyle A\prod_{i<j}(x_i-x_j)(y_i-y_j)$ where $A$ is some overall constant. To normalize this, we take $x_k=y_k=\omega^k$ where $\omega=e^{2\pi i/k}$. Then the matrix is circulant and for $n\geq 3$ has a zero eigenvalue, but the above polynomial doesn't vanish. Consequently both $A$ and the determinant itself must vanish if $n\geq 3$.


If the matrix is $M$, then


where $J$ is the vector of all ones. $X$ is an $n*n$ matrix that has $x_1,x_2,...,x_n$ on the diagonal and zeros elsewhere. $Y$ is an $n*n$ matrix that has $y_1,y_2,...,y_n$, with the same order, on every row.

$rank(J)=1, rank(Y)=1,rank(X)= n$

$rank(XY)\leqslant min(rank(X),rank(Y))=min(1,n)=1$



$rank(J+XY)\leqslant rank(J)+rank(XY)=2$

If $2<n$, then we can conclude

$rank(J+XY)\leqslant 2 < n$

So, the rank of $M=J+XY$ is less that $n$ and consequently, its determinant is zero. But the correctness depends on the assumption that $2<n$


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