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If I have an $n \times n$ positive definite symmetric matrix $A$, with eigenvalues $\lambda_{1}>\lambda_{2}\cdots>\lambda_{n}$, can I claim that the highest value which matrix $A^{-1}$ can have will be $\lambda_{n}^{-1}$ (where $\lambda_{n}$ is the minimum eigenvalue of the matrix $A$)

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  • $\begingroup$ Indeed, positive definiteness tells you (with the right definition, that is) that the eigenvalues are strictement positif. And the rest is trivial. $\endgroup$ – knsam Apr 11 '16 at 4:24
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Yes. Hint. Let $B=A^{-1}$.

  1. Clearly, $B$ is positive definite. By using Sylvester's criterion, show that for every off-diagonal entry $b_{ij}$, we have $|b_{ij}|\le\max\{b_{ii},b_{jj}\}\le\max_kb_{kk}=\max_ke_k^TBe_k$, where $\{e_1,\ldots,e_n\}$ is the standard basis of $\mathbb R^n$.
  2. By using the fact that every real symmetric matrix is orthogonally diagonalisable, show that $\max_ke_k^TBe_k\le\lambda_\max(B)=\lambda_n^{-1}$. Obviously, the upper bound $\lambda_n^{-1}$ is attainable by a positive diagonal $B$.
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