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 $A$). All the values of $A$ are real.
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1$\begingroup$ Do you mean the highest value in $A^{-1}$ on highest eigenvalue in $A^{-1}$? Also you should specify that the matrices are real or complex. They are probably real but its better to clarify. $\endgroup$– PushpendreCommented Apr 11, 2016 at 4:43
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$\begingroup$ @Pushpendre I mean the highest value in $A^{-1}$ $\endgroup$– Rohit ShuklaCommented Apr 11, 2016 at 4:44
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3$\begingroup$ Assume $A \in \mathbb{R}^{n \times n}$. Since $A$ is symmetric PD therefore $A^{-1}$ exists. The highest eigen value of $A^{-1}$ equals $\lambda_n^{-1}$. This means $\max ||A^{-1}x||$ over all $x \in \mathbb{R}^n$ such that $||x|| = 1$ equals $\lambda^{-1}$ . Consider $x$ to be a basis vector with 1 in the i-th row. Let $a_i$ be the ith column of $A^{-1}$. Clearly $0 \le ||a_i|| \le \lambda_n^{-1}$. This implies that the maximum element of the i-th element of $a_i$ is less than $\lambda_n^{-1}$. But $i$ was chosen arbitrarily. In other words $||Ax||_\infty \leq ||Ax||_2 \leq || A||_2 ||x||_2$. $\endgroup$– PushpendreCommented Apr 11, 2016 at 5:14
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3$\begingroup$ The largest value of an entry of $A^{−1}$ is at most $\lambda_{n}^{-1}$ but equality need not occur. For example, it can happen that $\lambda_{n}$ is irrational and $A$ has rational entries. $\endgroup$– Geoff RobinsonCommented Apr 11, 2016 at 6:37
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$\begingroup$ Obviously, the entries are bounded by the operator norm (whether the matrix is positive definite or not), which is $\lambda_n^{-1}$ in your setting. $\endgroup$– Christian RemlingCommented Apr 13, 2016 at 2:12
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1 Answer
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Let the (unique wlog) largest element of $A^{-1}$ be at position $(i,j)$. Since $A^{-1}$ is positive definite, it follows that \begin{equation*} 2e_i^TA^{-1}e_j \le e_i^TA^{-1}e_i + e_j^TA^{-1}e_j. \end{equation*} But the rhs is clearly upper bounded by $2\lambda_n^{-1}$, so the largest element cannot be larger than $\lambda_n^{-1}$.
