Question: I write:

\begin{align} \lambda_1 & = {\sqrt[3]{\frac{1}{2}\bigg(9 + i \sqrt{687}\bigg)}\above 1.5 pt 3^{2/3}} + {4 \above 1.5pt {\sqrt[3]{\frac{3}{2} \bigg (9 + i \sqrt{687}\bigg)}}}\\ \lambda_2 & = -{\bigg(1-i\sqrt{3}\bigg)\sqrt[3]{\frac{1}{2}\bigg(9 + i \sqrt{687}\bigg)}\above 1.5 pt 2\times 3^{2/3}} - {2(1+i\sqrt{3})\above 1.5pt {\sqrt[3]{\frac{3}{2} \bigg (9 + i \sqrt{687}\bigg)}}} \\ \lambda_3 & = -{\bigg(1+i\sqrt{3}\bigg)\sqrt[3]{\frac{1}{2}\bigg(9 + i \sqrt{687}\bigg)}\above 1.5 pt 2\times 3^{2/3}} - {2(1-i\sqrt{3})\above 1.5pt {\sqrt[3]{\frac{3}{2} \bigg (9 + i \sqrt{687}\bigg)}}}\\ \end{align} Is it obviously true that $$|\lambda_1|^2+|\lambda_2|^2+|\lambda_3|^2 =8$$ ?

This question is driven entirely by curiosity and fun. Unless I am missing the obvious it seems if you were to work this out by hand it would be demonstrably difficult ? Here is the origin of the problem and my attempted solution:

Let $A(n)$ be a finite square $n \times n$ matrix with entries $a_{i,j}=1$ if $i+j$ is a perfect power; otherwise equals to $0$. Let $\chi_{A(n)}(X)$ be the characteristic polynomial of $A(n)$. By the fundamental theorem of algebra $\chi_{A(n)}(X)$ has $n$ roots (=eigenvalues). Denote the $n$ eigenvalues by $\lambda_i$ with $1 \leq i \leq n$. Note $A(n)=A(n)^t$ where $^t$ is the matrix transpose and in particular $A(n)$ is a real symmetric matrix consequently it is normal. I bring Schur's Inequality into play:

$$\sum_{i=1}^n |\lambda_i|^2 = \sum_{i=1}^n \sum_{j=1}^n |a_{ij}|^2$$

Immediately this tells me that the number of $1$'s in $A(n)$ is completely determined by its eigenvalues. Now consider $A(6)$.

$$A(6)= \text{ }\begin{pmatrix} 0&0&1&0&0&0\\ 0&1&0&0&0&1\\ 1&0&0&0&1&1\\ 0&0&0&1&1&0\\ 0&0&1&1&0&0\\ 0&1&1&0&0&0\\ \end{pmatrix}$$

I have that $$\chi_{A(6)}(X)=X^6-2X^5-4X^4+8X^3+2X^2-4X-1$$ and that can be factored into $$(X^3-4X-1)(X^2-X-1)(X-1)$$ $A(6)$ has 6 eigenvalues which can be written (courtesy of WOLFRAM ALPHA)

\begin{align} \lambda_1 & = {\sqrt[3]{\frac{1}{2}\bigg(9 + i \sqrt{687}\bigg)}\above 1.5 pt 3^{2/3}} + {4 \above 1.5pt {\sqrt[3]{\frac{3}{2} \bigg (9 + i \sqrt{687}\bigg)}}}\\ \lambda_2 & = -{\bigg(1-i\sqrt{3}\bigg)\sqrt[3]{\frac{1}{2}\bigg(9 + i \sqrt{687}\bigg)}\above 1.5 pt 2\times 3^{2/3}} - {2(1+i\sqrt{3})\above 1.5pt {\sqrt[3]{\frac{3}{2} \bigg (9 + i \sqrt{687}\bigg)}}} \\ \lambda_3 & = -{\bigg(1+i\sqrt{3}\bigg)\sqrt[3]{\frac{1}{2}\bigg(9 + i \sqrt{687}\bigg)}\above 1.5 pt 2\times 3^{2/3}} - {2(1-i\sqrt{3})\above 1.5pt {\sqrt[3]{\frac{3}{2} \bigg (9 + i \sqrt{687}\bigg)}}}\\ \lambda_4 & = -{1\above 1.5 pt \phi}\\ \lambda_5 & = \phi\\ \lambda_6 & = 1 \end{align}

where $\phi= {1+ \sqrt{5}\above 1.5pt 2}$ and is called the golden ratio. Note $\sum_{j=1}^6 a_{ij}=12$ and according to our first equation I can write

$$1+|\lambda_1|^2+|\lambda_2|^2+|\lambda_3|^2+ |-{1\above 1.5 pt \phi}|^2 +|\phi|^2 =12$$

I can show that

$$|-{1\above 1.5 pt \phi}|^2 +|\phi|^2=3$$

Cancelling out terms and regrouping yields

$$|\lambda_1|^2+|\lambda_2|^2+|\lambda_3|^2 =8$$

and we are done.

Outside of hand calculations which I am too lazy to do to I am not aware of any other approach that will obviously and for that matter quickly show the sum equals $8$.

  • 11
    $\begingroup$ $\lambda_1,\lambda_2,\lambda_3$ are the (real) roots of $x^3-4x-1=0\,$, so by Vieta's relations $\lambda_1^2+\lambda_2^2+\lambda_3^2\,$ $=(\lambda_1+\lambda_2+\lambda_3)^2-2(\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3)=0-2\cdot(-4)=8\,$. $\endgroup$
    – dxiv
    Jun 2, 2017 at 16:59
  • 1
    $\begingroup$ Seems like you could replace $\sqrt[3]{\frac{1}{2}(9 + i \sqrt{687})}$ with $a$. Then your numerators have $a$ and your denominators have $\sqrt[3]{3}a$. But ... @dxiv 's comment is far better. :) $\endgroup$
    – John
    Jun 2, 2017 at 17:02
  • 2
    $\begingroup$ @dxiv As in Vieta's Formula ? $\endgroup$
    – Anthony
    Jun 2, 2017 at 17:03
  • 1
    $\begingroup$ @AnthonyHernandez Right. That's assuming you already know the polynomial, of course. Otherwise, if you are only given the expressions for $\,\lambda_{1,2,3}\,$ with no clue of where they come from, then you have to do the calculations the hard way, or (with a leap of imagination) "guess" which cubic they are the roots of. $\endgroup$
    – dxiv
    Jun 2, 2017 at 17:07
  • 2
    $\begingroup$ @dxiv Having done this without scrolling down to see the polynomial, it definitely is hard! I reconstructed the polynomial by trying to invert Cardano's method for cubics, so it is possible $\endgroup$
    – B. Mehta
    Jun 2, 2017 at 17:09

2 Answers 2


Let's assume we have never seen the polynomial $x^3 - 4x - 1$ before. We can compute the value of $|\lambda_1|^2 + |\lambda_2|^2 + |\lambda_3|^2$ in following manner.

Let $a = \sqrt[3]{\frac32(9+i\sqrt{687})}$ and $\omega = e^{2\pi/3 i}$, a primitive cubic root of unity. We have $$\lambda_1 = \frac{a}{3} + \frac{4}{a},\quad \lambda_2 = \frac{a}{3}\omega + \frac{4}{a}\omega^2,\quad \lambda_3 = \frac{a}{3}\omega^2 + \frac{4}{a}\omega^4$$ This can be summarized as $$\lambda_k = \frac{a}{3} \omega^{k-1} + \frac{4}{a}\omega^{2(k-1)}\tag{*1}$$ This leads to $$|\lambda_1|^2 + |\lambda_2|^2 + |\lambda_3|^2 = \sum_{k=1}^3 \lambda_k \bar{\lambda}_k = \sum_{k=1}^3 \left(\frac{a}{3} + \frac{4}{a}\omega^{k-1}\right)\left( \frac{\bar{a}}{3} + \frac{4}{\bar{a}}\omega^{-(k-1)}\right)$$

Using the fact $\sum_{k=1}^3 \omega^{\ell(k-1)} = \begin{cases}3, & \ell \equiv 0 \pmod 3\\0, & \ell \not\equiv 0\pmod 3\end{cases}$, we can simplify above as $$|\lambda_1|^2 + |\lambda_2|^2 + |\lambda_3|^2 = 3 \left(\left|\frac{a}{3}\right|^2 + \left|\frac{4}{a}\right|^2\right)\tag{*2}$$

Since $|a|^2 = \sqrt[3]{\frac{3^2}{2^2}(9^2 + 687)} = \sqrt[3]{1728} = 12$, we find

$$|\lambda_1|^2 + |\lambda_2|^2 + |\lambda_3|^2 = 3 \left(\frac{12}{3^2} + \frac{4^2}{12}\right) = 8$$

In above computation, the critical piece is the representation of $(\lambda_1,\lambda_2,\lambda_3)$ in $(*1)$.
We can view triplet $(\lambda_1, \lambda_2, \lambda_3)$ as a DFT (discrete fourier transform) of the triplet $\left( 0, \frac{a}{3}, \frac{4}{a}\right)$. Equality $(*2)$ is the result when one apply Plancherel theorem to this particular DFT. That's the underlying reason why $|\lambda_1|^2 + |\lambda_2|^2 + |\lambda_3|^2$ has such a simple expression in $|a|^2$.

  • $\begingroup$ Curiously $|a|^2 = \sqrt[3]{\frac{3^2}{2^2}(9^2 + 687)} = \sqrt[3]{1728} = 12$ which are exactly the number of $1$'s in $A(6)$. $\endgroup$
    – Anthony
    Jun 2, 2017 at 19:16

It's worth pointing out that showing those are the roots to the polynomial $t^3 - 4t - 1=0$ isn't enough to conclude that $|\lambda_1|^2 + |\lambda_2|^2+|\lambda_3|^2=8$, only that $\lambda_1^2 + \lambda_2^2+\lambda_3^2=8$, and we need to check that the roots you've given are actually real numbers (even if they don't look like it). The simplest way I can see is to use the intermediate value theorem, checking that there is a sign change between $-\infty$, $-1$, $0$, $\infty$, giving at least three real roots. Of course this means there are exactly three real roots, and so the modulus signs made no difference.

Let's write $\alpha = \sqrt[3]{\frac{1}{18}\left(9 + i \sqrt{687}\right)} = \frac{\sqrt[3]{\frac{1}{2}\left(9 + i \sqrt{687}\right)}}{3^{2/3}}$ for convenience, and $\omega$ for the cube root of unity, $\omega = e^{2\pi i/3}$. Then your roots are $$\begin{align} \lambda_1 &= \alpha+\frac{4}{3\alpha} \\ \lambda_2 &= \alpha \omega^2 + \frac{4}{3\alpha \omega^2} \\ \lambda_3 &= \alpha \omega + \frac{4}{3\alpha \omega} \\ \end{align}$$

To recap Cardano's method, to solve a depressed cubic $t^3 = at+b$, we make the substitution $t = u+v$, so $u^3 + v^3 + 3uv(u+v) = a(u+v)+b$, and choose $u^3 + v^3 = b$, $3uv=a$. We can solve this by writing $v = \frac{a}{3u}$, so $u^3 + \frac{a^3}{27u^3} = b$, producing $u^6 + \frac{a^3}{27} = b u^3$, a quadratic equation in $u^3$.

The value of $\alpha$ given above seems like the cube root of a solution to a quadratic, so let's guess $u=\alpha$ and $v = \frac{4}{3\alpha}$, so we compute $3uv = 4$, and so $a = 4$. Then, $x = u^3$ should satisfy $x^2 -bx +\frac{64}{27}=0$. We find $x = \frac1{18}(9+i \sqrt{687})$, so $(18x-9)^2 = -687$, and expanding and cancelling gives $x^2 - x + \frac{64}{27}=0$, hence $b=1$.

Therefore, $\lambda_1$ satisfies $t^3 = 4t+1$. You can do the same thing for $\lambda_2$ and $\lambda_3$, or just observe that when the solution via Cardano's method is found, we have the choice of three cube roots of $u^3$ to take, corresponding to $\alpha$, $\alpha \omega$ and $\alpha \omega^2$. (The choice of two square roots from the quadratic doesn't give extra solutions, since it would give the $\frac{4}{3\alpha}$ term instead, and you'd get the same answer.)

So, $\lambda_1, \lambda_2, \lambda_3$ are the roots to $t^3 = 4t+1$, and as others have said, Vieta's formulas (or more precisely, Newton's identities) finish off the answer.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .