# Is This Sum Obviously Equal to $8$.

Question: I write:

\begin{align} \lambda_1 & = {\sqrt{\frac{1}{2}\bigg(9 + i \sqrt{687}\bigg)}\above 1.5 pt 3^{2/3}} + {4 \above 1.5pt {\sqrt{\frac{3}{2} \bigg (9 + i \sqrt{687}\bigg)}}}\\ \lambda_2 & = -{\bigg(1-i\sqrt{3}\bigg)\sqrt{\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{\frac{3}{2} \bigg (9 + i \sqrt{687}\bigg)}}} \\ \lambda_3 & = -{\bigg(1+i\sqrt{3}\bigg)\sqrt{\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{\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{\frac{1}{2}\bigg(9 + i \sqrt{687}\bigg)}\above 1.5 pt 3^{2/3}} + {4 \above 1.5pt {\sqrt{\frac{3}{2} \bigg (9 + i \sqrt{687}\bigg)}}}\\ \lambda_2 & = -{\bigg(1-i\sqrt{3}\bigg)\sqrt{\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{\frac{3}{2} \bigg (9 + i \sqrt{687}\bigg)}}} \\ \lambda_3 & = -{\bigg(1+i\sqrt{3}\bigg)\sqrt{\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{\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 $8$.

• $\lambda_1,\lambda_2,\lambda_3$ are the (real) roots of $x^3-4x-1=0\,$, so by Vieta's relations $\lambda_1^2+\lambda_2^2+\lambda_3^2\,$ $=(\lambda_1+\lambda_2+\lambda_3)^2-2(\lambda_1\lambda_2+\lambda_1\lambda_3+\lambda_2\lambda_3)=0-2\cdot(-4)=8\,$.
– dxiv
Jun 2, 2017 at 16:59
• Seems like you could replace $\sqrt{\frac{1}{2}(9 + i \sqrt{687})}$ with $a$. Then your numerators have $a$ and your denominators have $\sqrt{3}a$. But ... @dxiv 's comment is far better. :)
– John
Jun 2, 2017 at 17:02
• @dxiv As in Vieta's Formula ? Jun 2, 2017 at 17:03
• @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.
– dxiv
Jun 2, 2017 at 17:07
• @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 Jun 2, 2017 at 17:09

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 $a = \sqrt{\frac32(9+i\sqrt{687})}$ 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 $|a|^2 = \sqrt{\frac{3^2}{2^2}(9^2 + 687)} = \sqrt{1728} = 12$, 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 $|a|^2$.

• Curiously $|a|^2 = \sqrt{\frac{3^2}{2^2}(9^2 + 687)} = \sqrt{1728} = 12$ which are exactly the number of $1$'s in $A(6)$. 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 $\alpha = \sqrt{\frac{1}{18}\left(9 + i \sqrt{687}\right)} = \frac{\sqrt{\frac{1}{2}\left(9 + i \sqrt{687}\right)}}{3^{2/3}}$ 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.