Timeline for Largest element in inverse of a positive definite symmetric matrix [closed]
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2016 at 7:51 | history | closed |
Christian Remling Alex Degtyarev Stefan Waldmann Wolfgang Mikhail Katz |
Not suitable for this site | |
| Apr 13, 2016 at 2:28 | review | Close votes | |||
| Apr 13, 2016 at 7:51 | |||||
| Apr 13, 2016 at 2:12 | comment | added | Christian Remling | 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. | |
| Apr 13, 2016 at 2:07 | answer | added | Suvrit | timeline score: 3 | |
| Apr 11, 2016 at 6:37 | comment | added | Geoff Robinson | 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. | |
| Apr 11, 2016 at 5:14 | comment | added | Pushpendre | 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 $||Ax||_\infty \leq ||Ax||_2 \leq || A||_2 ||x||_2$. | |
| Apr 11, 2016 at 4:44 | comment | added | Rohit Shukla | @Pushpendre I mean the highest value in $A^{-1}$ | |
| Apr 11, 2016 at 4:43 | history | edited | Rohit Shukla | CC BY-SA 3.0 |
added 33 characters in body
|
| Apr 11, 2016 at 4:43 | comment | added | Pushpendre | 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. | |
| Apr 11, 2016 at 4:26 | history | asked | Rohit Shukla | CC BY-SA 3.0 |