Rank 1 matrix with given anti-diagonal sums

Question

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 2x2 and 3x3 cases (I have not attempted the 4x4 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 10x10 matrix with arbitrary anti-diagonal sum values?

EDIT: here is the 2x2 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}

Answer 1

Let us consider the case $n=4$.

Working with matrix $U^TV$ (notation explained in (1)), I had the idea to pre-/ post mutiply it by $X^T$ and $X$ resp. giving the $1 \times 1$ matrix (= number) :

$$(X^TU)(V^TX)=\underbrace{\pmatrix{1&t&t^2&t^3}}_{X^T}\underbrace{\pmatrix{a\\b\\c\\d}}_{U}\underbrace{\pmatrix{e&f&g&h}}_{V^T}\underbrace{\pmatrix{1\\t\\t^2\\t^3}}_{X}\tag{1}$$

giving rise to the identity :

$$(a + bt + ct^2 +dt^3)(e+ft+gt^2+ht^3)=1+t+t^2+t^3+t^4+t^5+t^6\tag{1'}$$

because, in the expansion of the LHS, the coefficients are precisely those

$$ae,be+af,\cdots dg+ch,dh$$

we want to be equal to $1$ (or to any pre-assigned values) being the sums of the different antidiagonals.

From (1'), how can adequate values for

$$a,b,c,d,e,f,g,h \tag{2}$$

be obtained ?

In (1'), he roots of the RHS, are the seven "$7$th roots of unity" deprived of root $1$ :

$$w,w^2,w^3,w^4,w^5,w^6 \ \text{with} \ w:=e^{2i \pi/7}$$

Therefore we have to "constrain" the LHS of (1) to have the same roots. This can be done in many different ways, for example :

$$\underbrace{[(t-w)(t-w^2)(t-w^3)]}_{\text{first 3rd degree pol.}}\underbrace{[(t-w^4)(t-w^5)(t-w^6)]}_{\text{second 3rd degree pol.}}$$

Any gathering three by three of the roots of the RHS is possible. In this case ($n=4$), whatever this gathering, some coefficients will be complex. This is not necessarily the case (see case $n=5$ below).

Remark 1: The methodology above can be extended to any value of $n$. Here is for example a solution for the $n=5$ case (therefore based on the $2n-1=9$th roots of unity) with all-real coefficients rounded to their 4th decimal :

$$\left(\begin{array}{rrrrr} 1.0000 & 2.8794 & 3.8794 & 2.8794 &1.0000\\ -1.8794 & -5.4115 &-7.2909 &-5.4115 &-1.8794\\ 2.5321 & 7.2909 & 9.8229 & 7.2909& 2.5321\\ -1.8794 & -5.4115&-7.2909 &-5.4115& -1.8794\\ 1.0000& 2.8794 & 3.8794& 2.8794 &1.0000\end{array}\right)$$

(check ! ). Please note that this matrix has a central symmetry.

Remark 2 : in the case $n=3$, the solution you have found is related to the following factorization :

is based on the following factorization :

$$(1 + \phi t +t^2)(1 - \frac{1}{\phi}t+t^2)=1+t+t^2+t^3+t^4 \tag{3}$$

with the usual notation :

$$\phi=\tfrac12(1+\sqrt{5})$$

for the golen ratio.

(one can check identity (3) by expanding and using the fact that $\phi^2=\phi + 1$).

Answer 2

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.