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I am dealing with a greater problem where all matrices are $\mathbb{R}^{3 \times 3}$ and the main focus is a certain ${\mathbf X} \in [-2,2]^{3 \times 3}$ which can be non-symmetric. I spare you the full problem context, I just want to show that ${\mathbf X}$ has eigenvalues fulfilling $1 \leq |\lambda_k| \leq 2$. I know that this is true from numerical experiments.

I have sucessfully shown that for certain unitary matrices ${\mathbf A}$ and ${\mathbf B}$, the matrix ${\mathbf A} {\mathbf X} {\mathbf B}^\text{T}$ has eigenvalues $2$, $-1$, $-1$. Or, if you prefer the decomposition, $${\mathbf A} {\mathbf X} {\mathbf B}^\text{T} = {\mathbf U} \left(\begin{array}{rrr}2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array}\right) {\mathbf U}^\text{T}$$ with ${\mathbf U}$ unitary. Intuitively, I feel like this already implies my eigenvalue bounds: ${\mathbf A}$ and ${\mathbf B}$ "smear" the magnitude of eigenvalues over the interval $[1,2]$ and only a special case like ${\mathbf A} = {\mathbf B} = {\mathbf I}_3$ achieves the boundaries.

So my question is: Is it possible to prove my bounds $1 \leq |\lambda_k| \leq 2$ about ${\mathbf X}$ using only the provided information and suitable properties about unitary maps and eigenvalues?

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Consider the matrices \begin{align} A= \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 0 &-1 \\0& 1 &0 \end{matrix} \right) && B= \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 0 &1 \\0& 1 &0 \end{matrix} \right) && X= \left( \begin{matrix} 2 & 0 & 0 \\ 0 & 2 & \sqrt{3} \\0& \sqrt{3} &2 \end{matrix} \right) \end{align} So $A$ and $B$ are unitary. In particular, we can focus on the sub $2 \times 2$-matrices in the lower right corner, and forget about the space spanned by $e_1$. Then we have \begin{eqnarray} AXB^T &=& \left( \begin{matrix} 0 & -1 \\ 1 &0 \end{matrix} \right) \left( \begin{matrix} 2 & \sqrt{3} \\ \sqrt{3} &2 \end{matrix} \right) \left( \begin{matrix} 0 & 1 \\ 1 &0 \end{matrix} \right) \\ &=& \left( \begin{matrix} 0 & -1 \\ 1 &0 \end{matrix} \right) \left( \begin{matrix} \sqrt{3} & 2 \\2 &\sqrt{3} \end{matrix} \right)\\ &=&\left( \begin{matrix} -2 & -\sqrt{3} \\ \sqrt{3} &2 \end{matrix} \right) \end{eqnarray} So the eigenvalues of $AXB^T$ sum to $0$ and have product $-1$, so the eigenvalues of $AXB^T$ are $1$ and $-1$, with eigenvectors \begin{eqnarray} \left( \begin{matrix} -1 \\ \sqrt{3} \end{matrix}\right) &\textrm{ and }& \left( \begin{matrix} -\sqrt{3} \\ 1 \end{matrix}\right) \end{eqnarray} Thus $X$ satisfies the properties, but has eigenvalues $\lambda_1 = 2-\sqrt{3}$, $\lambda_2 = 2 $ and $ \lambda_3 = 2+\sqrt{3}$, and so $$ | \lambda_1 | < 1 < 2 < | \lambda_3|.$$

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  • $\begingroup$ Sorry for the late reply, I was on the road over new year's. Thank you very much, I verified your counterexample and it perfectly demonstrates that I have to involve more constraints from my problem ($X$ can't take the value that you legitimately assumed, it's actually very specific). $\endgroup$ – GDumphart Jan 4 '16 at 11:22
  • $\begingroup$ @GDumphart If $A$,$B$, and $X$ commute, you can diagonalize them w.r.t. to the same basis, and you are done. So you could look if it is enough for $A$ and $B$ to commute $\endgroup$ – Hetebrij Jan 4 '16 at 13:06
  • $\begingroup$ Thanks, unfortunately I couldn't utilize that. However, I made some progress with my problem and asked a new, self-contained question. If you're interested: math.stackexchange.com/questions/1600829/… $\endgroup$ – GDumphart Jan 5 '16 at 14:19

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