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Suppose I have the following Toeplitz symmetric matrix

\begin{align} M=\begin{bmatrix} 1 & c & c & x \\ c & 1 & c & c \\ c & c & 1 & c \\ x & c & c & 1 \end{bmatrix} \end{align}

I want to write an algorithm that takes $c$ as input and calculates the range of $x$ for which matrix $M$ is positive semidefinite.

Currently, I do Gaussian elimination by hand and reduce the problem to checking the determinant of a $2 \times 2$ matrix. But how do I automate the process so I can write a function that takes $c$ and $n$ as inputs, where $n$ is the dimension of $M$, and returns the range of $x$. Thanks!

Rodrigo de Azevedo
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dynamic89
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3 Answers3

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For your specific example, done with pen and paper, $$M_4=\left( \begin{array}{cccc} 1 & c & c & x \\ c & 1 & c & c \\ c & c & 1 & c \\ x & c & c & 1 \end{array} \right)$$ $$\Delta_4=\left(4 c^3-5 c^2+1\right)+\left(4 c^2-4 c^3\right) x+\left(c^2-1\right) x^2$$ With a computer $$M_5=\left( \begin{array}{ccccc} 1 & c & c & c & x \\ c & 1 & c & c & c \\ c & c & 1 & c & c \\ c & c & c & 1 & c \\ x & c & c & c & 1 \end{array} \right)$$ $$\Delta_5=\left(-6 c^4+14 c^3-9 c^2+1\right)+6 \left(c^4-2 c^3+c^2\right) x+\left(-2 c^3+3 c^2-1\right) x^2$$

Claude Leibovici
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Essentially, you want to compute the diameter of a $1$-dimensional spectrahedron. This diameter can be found by solving two semidefinite programs

$$\begin{array}{ll} \text{minimize} & \pm x\\ \text{subject to} & \begin{bmatrix} 1 & c & c & x \\ c & 1 & c & c \\ c & c & 1 & c \\ x & c & c & 1 \end{bmatrix} \succeq \mathrm O_4\end{array}$$

Rodrigo de Azevedo
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Hint:

Definiteness can be assessed by the signs of the leading principal minors. The minors of order up to $n-1$ have the same Toeplitz structure and are functions of $c$ alone, let $N_k(c)$.

Now notice that the determinant of the matrix $M$ is a quadratic function of $x$, and an obvious root is $x=1$. In addition, the determinant evaluated at $x=c$ is $N_n$. And finally, the coefficient of the term $x^2$ is $N_{n-2}$, by removing the top-right and bottom-left elements.

Hence the determinant is

$$\det(M)=N_{n-2}(x-r)(x-1)$$

with $$N_{n-2}(c-r)(c-1)=N_n$$

and the second root is

$$r=c-\frac{N_n}{N_{n-2}(c-1)}.$$

  • @dynamic89: typo, read principal minors. –  Aug 10 '20 at 20:34
  • Thanks so much for your response. Why is it that the principal minors of $M$ up to order $n-1$ are only functions of c? What about ${\rm det} [[1,x]' [x,1]']$ and ${\rm det} [[1,c,x]' [c,1,c]' [x,c,1]']$? So for example when I write $p(x)=\sum_{r=0}^n x^r S_{n-r}$, where $S_{n-r}$ denotes the sum of all principal minors of $M$ of order $n-r$, ${\rm det} [[1,x]' [x,1]']$ and ${\rm det} [[1,c,x]' [c,1,c]' [x,c,1]']$ should be part of $S_2$ and $S_3$. Maybe I'm missing something? – dynamic89 Aug 10 '20 at 20:37
  • @dynamic89: sorry, I meant leading principal minors. –  Aug 10 '20 at 20:39
  • so are they all leading principal minors you talking about? – dynamic89 Aug 10 '20 at 20:43
  • @dynamic89: yep, can't you guess why ? How do you check definiteness ? –  Aug 10 '20 at 20:44
  • I mean you can expand determinant in terms of both principal minors and leading principal minors. I need to think more! – dynamic89 Aug 10 '20 at 20:46
  • @dynamic89: write the matrix with $x=1$ ! –  Aug 11 '20 at 15:57
  • sure but is there some intuition? or just trying a few numbers and find a root that way? – dynamic89 Aug 11 '20 at 15:59
  • @dynamic89: no, open your eyes. What makes a determinant zero ? –  Aug 11 '20 at 16:00
  • ahh i see! thanks! – dynamic89 Aug 11 '20 at 16:04