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Matrix determinant inequality

Suppose that all eigenvalues of A\in \mathcal{M}_n(\mathbb{R}) are positive real numbers. Show that

    \[\det \left( A + A^{-1} \right) \geq 2^n\]


Let the eigenvalues of A be \lamda_i , 1 \leq 1 \leq n. Consider the Jordan normal form of A;  this Jordan form is an upper triangular matrix that has the eigenvalues of A in the main diagonal. Let this matrix be called J.  Furthermore \det A = \det J , as a matrix and its Jordan normal form are similar. As J is upper triangular, its inverse is given by an upper triangular matrix whose diagonal entries are the inverses of the diagonal entries of J. That is,

    \[J +J^{-1} = \begin{pmatrix} \lambda_1+\lambda^{-1}_1 & & & & \\ 0 & \lambda_2+\lambda_2^{-1} & & U & \\ 0& 0 & \lambda_3+\lambda_3^{-1} & & \\ \vdots& \vdots & \vdots &\ddots &x \\ 0& 0 &0 &\cdots &\lambda_n +\lambda_n^{-1} \end{pmatrix}\]

is an upper-triangular matrix, and its determinant can be computed simply as the product of the elements of the diagonal. Hence,

    \[\det \left ( A + A^{-1} \right ) = \prod_{i=1}^{n} \left ( \lambda_i + \frac{1}{\lambda_i} \right ) \geq 2^n\]

due to the inequality x+\frac{1}{x} \geq 2 for all x>0.
Note: The main point here is that if \lambda is an eigenvalue of A then \frac{1}{\lambda} is an eigenvalue of A^{-1}. Taking into account that

    \[\det A = \prod_{i} \lambda_i\]

yields a shorter solution to the problem. No need for upper triangular matrices.

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