# An inequality with determinant

Let $A$ be an invertible $n \times n$ matrix such that all the elements of $A$ and $A^{-1}$ are integers. Prove that if all the eigenvalues of $A$ are real numbers then: $$|\det(A + A^{-1})| \ge 2^n$$ I've already known that if $A$ is a diagonalizable matrix with eigenvalues $\lambda_1$, $\lambda_2$,... $\lambda_n$ then we can prove that $|\det(A + A^{-1})| = |(\lambda_1 + \lambda_1^{-1})...(\lambda_n + \lambda_n^{-1})| \ge 2^n$ but I still cannot solve the problem, could somebody please give me some help? Thank you.

$A$ may not be diagonalizable, but it is triangularizable over $\mathbb C$. Since you know that all the eigenvalues of $A$ are real numbers, you get the same eigenvalues when considering $A$ as a matrix with complex entries, thus $A$ is similar to some $$T=\left( \begin{array}{ccccc} \lambda_1 \\ & \lambda_2 & & \huge *\\ & & \ddots \\ & \large 0 & & \lambda_{n-1} \\ & & & & \lambda_n \end{array} \right)$$ that is $A=PTP^{-1}$ where $P\in Gl_n(\mathbb C)$. $T^{-1}$ is upper triangular as well, with the form $$T^{-1}=\left( \begin{array}{ccccc} \lambda_1^{-1} \\ & \lambda_2 ^{-1} & & \huge **\\ & & \ddots \\ & \large 0 & & \lambda_{n-1} ^{-1} \\ & & & & \lambda_n^{-1} \end{array} \right)$$ so that $A^{-1}=PT^{-1}P^{-1}$, which yields $|\det(A + A^{-1})| = |(\lambda_1 + \lambda_1^{-1})...(\lambda_n + \lambda_n^{-1})|$.

Recall that $\det(A)=\prod_{i=1}^n \lambda_i$, where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$.

Hence, $\det(A^{-1})=\dfrac{1}{\prod_{i=1}^n \lambda_i}$ and $\det(A^2+Id)=\det(A+iId)\det(A-iId)=\prod_{j=1}^n (\lambda_j+i)\prod_{j=1}^n (\lambda_j-i)=\prod_{j=1}^n (\lambda_j^2+1)$.

Since $\lambda_i\in\mathbb{R}$ then $\lambda_i$ and $\lambda_i^{-1}$ have the same sign. So $|\lambda_i+\lambda_i^{-1}|=|\lambda_i|+|\lambda_i^{-1}|$.

Since $|\lambda_i|^2-2|\lambda_i|+1\geq 0$ then $|\lambda_i|^2+1\geq 2|\lambda_i|$. So $|\lambda_i|+|\lambda_i^{-1}|\geq 2$.

Finally, $|\det(A+A^{-1})|=|\det(A^{-1})\det(A^2+Id)|=\left|\prod_{i=1}^n \dfrac{\lambda_i^2+1}{\lambda_i}\right|=\prod_{i=1}^n \left|\lambda_i+\dfrac{1}{\lambda_i}\right|\geq 2^n$.

\begin{eqnarray*} \lambda+ \frac{1}{\lambda} =\left( \sqrt{\lambda} - \frac{1}{\sqrt{\lambda}} \right)^2+2 \geq 2. \end{eqnarray*}

• How does this help if $A$ is not diagonalizable? – Berci Nov 26 '17 at 12:34