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Matrix $A$ is $$ \begin{pmatrix} 1 & 1 & 1 \\ 1 & w^2 & w \\ 1 & w & w^2 \\ \end{pmatrix} $$

Where $1,w,w^2$ are cube roots of unity.

My problem is to obtain the result $|\lambda_1| + |\lambda_2| +|\lambda_3| \le 9$, where, $\lambda_1 , \lambda_2, \lambda_3$ are eigenvalues of matrix $A^2$.

My approach was, to use the fact that eigenvalues of $A^2$ are squares of eigenvalues of $A$. But I couldn't find a way to relate them through an in equation.

Bérénice
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praveenchand
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  • Hint: Compute $A^4$. – Batominovski Aug 14 '16 at 01:18
  • I assume you mean $\omega$ is a primitive root of unity? If so: $1+\omega +\omega^2 = 0$. Using this, calculate $B= A^2$. If I didn't mess up(!), the answer is $3$ times a permutation matrix $P$ of order $2$; $P$ has eigenvalues $1,1,-1$. This is really @Batominovski's hint, I think... – peter a g Aug 14 '16 at 03:19

1 Answers1

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In general, fix $n\in\mathbb{N}$ and pick a primitive $n$-th root of unity $\omega\in \mathbb{C}$. Define $$\mathbf{A}:=\left[\omega^{ij}\right]_{i,j\in\{0,1,2,\ldots,n-1\}}\,.$$ Then, we have $$\mathbf{A}^2=\left[\sum_{\mu=0}^{n-1}\,\omega^{(i+j)\mu}\right]_{i,j\in\{0,1,2,\ldots,n-1\}}\,.$$ Ergo, $\mathbf{A}^2$ is given by the $n$-by-$n$ permutation matrix $$\mathbf{A}^2=n\,\left[\begin{array}{c|cccccc} 1&0&0&\cdots&0&0&0\\ \hline 0&0&0&\cdots&0&0&1\\ 0&0&0&\cdots&0&1&0\\ 0&0&0&\cdots&1&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 0&0&1&\cdots&0&0&0\\ 0&1&0&\cdots&0&0&0 \end{array}\right]\,.$$ Therefore, $$\mathbf{A}^4=n^2\,\mathbf{I}_n$$ where $\mathbf{I}_n$ is the $n$-by-$n$ identity matrix. Hence, all eigenvalues of $\mathbf{A}$ must belong to the set $\big\{+\sqrt{n},-\sqrt{n},+\sqrt{n}\text{i},-\sqrt{n}\text{i}\}$.

This means the eigenvalues of $\mathbf{A}^2$ lie in the set $\{+n,-n\}$. In fact, since $$\text{Trace}\left(\mathbf{A}^2\right)=\begin{cases}n\,,&\mbox{if }n\mbox{ is odd}\,,\\ 2n\,,&\mbox{if }n\mbox{ is even}\,,\end{cases}$$ we see that the eigenvalue $+n$ of $\mathbf{A}^2$ has multiplicity $\left\lceil\frac{n+1}{2}\right\rceil$, whereas the eigenvalue $-n$ of $\mathbf{A}^2$ has multiplicity $\left\lfloor\frac{n-1}{2}\right\rfloor$.

Batominovski
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