I'm looking for a matrix that is of rank 1 but has specified anti-diagonal sums. To give a few examples:
It is easy to make a rank 3 matrix which has anti-diagonals that sum to 1:
\begin{bmatrix}
1 & 0.5 & 0.33\\
0.5 & 0.33 & 0.5\\
0.33 & 0.5 & 1
\end{bmatrix}
here 1 = 0.5+0.5 = 0.33 + 0.33 + 0.33. This can easily be done for any MxM matrices.
So far, I've only been able to find rank 1 matrices for the 2×2 and 3×3 cases (I have not attempted the 4×4 case yet). For instance,
\begin{bmatrix}
1 & 1.618 & 1\\
-0.618 & -1 & -0.618\\
1 & 1.618 & 1
\end{bmatrix}
does the trick, and can be written as an outer product: $[1, (1-\sqrt{5})/2, 1]^T*[1, (1+\sqrt{5})/2, 1]$. My technique was basically to set up a system of equations for the components of vectors x and y, i.e.: $x_1 + y_1 = 1$, $x_1\cdot y_2 + x_2\cdot y_1 = 1$, and so on, and then substitute – but it generalizes poorly to larger matrices.
Are there any principled or smart approaches to figure it out for, e.g., a 10×10 matrix with arbitrary anti-diagonal sum values?
EDIT: here is the 2×2 matrix, written as an outer-product of $[1, 0.5-i\sqrt{3}/2]^T*[1, 0.5+i\sqrt{3}/2]$:
\begin{bmatrix}
1 & 0.5 + 0.866i \\
0.5 – 0.866i & 1
\end{bmatrix}
Best Answer
Your contruction can be expressed in terms of polynomial multiplications.
I will index coordinates vectors in $\mathbb{C}^n$ and matrices in $\mathbb{C}^{n \times n}$ from $0$, so that for example $u = \begin{bmatrix} u_0 \\ u_1 \\ \vdots \\ u_{n - 1} \end{bmatrix}$.
Let $A = uv^\top$ be a rank $1$ matrix. The $k$-th antidiagonal sum $w_k = \sum_{i + j = k} A_{ij} = \sum_{i + j = k} u_i v_j$. Conside the polynomials $U(t) = \sum_{i = 0}^{n - 1} u_i t^i$ and $V(t) = \sum_{j = 0}^{n - 1} v_j t^j$. The coefficients of the product polynomial $W(t) = U(t) V(t) = \sum_{k = 0}^{2n - 2} \left(\sum_{i + j = k} u_i v_j\right) t^k$ are exactly our antidiagonal sums.
Thus, to find the required rank $1$ matrix we need to factorize the polynomial $\sum_{k = 0}^{2n - 2} t^k = \frac{t^{2n - 1} - 1}{t - 1}$. Its roots are exactly the roots of unity of degree $2n - 1$, except $1$: $\sum_{k = 0}^{2n - 2} t^k = \prod_{i = 1}^{2n - 2} (t - \exp(\frac{2\pi i}{2n - 1}))$
Any partition of these roots of unity into two subsets of equal size gives a factorization of $\sum_{k = 0}^{2n - 2} t^k$ into two polynomials of degree $n-1$ and the corresponding rank $1$ matrix.
Your $2 \times 2$ example (and its transpose) comes from the unique factorization $x^2 + x + 1 = (x - \frac{1 + i\sqrt{3}}{2})(x - \frac{1 - i\sqrt{3}}{2})$, and the $3 \times 3$ example comes from the unique factorization of $x^4 + x^3 + x^2 + x + 1$ over $\mathbb{R}$.
In general, we can choose, for example, a partition into roots of unity with positive imaginary part and the negative imaginary part, so the polynomial $$U(t) = \prod_{i = 1}^{n - 1} (t - \exp(\frac{2\pi i}{2n - 1}))$$ $$= \sum_{k = 0}^{n - 1} (-1)^{k} t^{n - 1 - k} e_{k}(\exp(\frac{2\pi}{2n - 1}), \exp(\frac{4\pi}{2n - 1}), \dots, \exp(\frac{2\pi (n - 1)}{2n - 1}))$$ where $e_k$ are elementary symmetric polynomials, so we can take vector $u$ with coordinates given by $u_{n - 1 - k} = (-1)^k e_{k}(\exp(\frac{2\pi}{2n - 1}), \exp(\frac{4\pi}{2n - 1}), \dots, \exp(\frac{2\pi (n - 1)}{2n - 1}))$, and $v$ will be complex conjugate of $u$.
If $n$ is odd, it is also possible to have an answer over $\mathbb{R}$, since there is an even number of conjugate pairs of roots.