Timeline for Arithmetic-geometric mean of positive matrices
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 2, 2011 at 18:19 | comment | added | Suvrit | Maybe the easiest way to prove this is to note that the geometric mean is the unique positive definite solution to the Riccati equation: $XA^{-1}X=B$. | |
| Aug 2, 2011 at 17:42 | vote | accept | Russel | ||
| Aug 2, 2011 at 7:34 | comment | added | Federico Poloni | @Gerald: this result is quite well-known; it is easier to prove using other characterizations of the geometric mean. For instance, using the fact that $Af(BA)=f(AB)A$ for all holomorphic $f$ (and in particular for the square root function), you can rewrite the definition as $A \sharp B = A(A^{-1}B)^{1/2}$, and from this form congruence invariance follows easily. | |
| Aug 1, 2011 at 21:46 | comment | added | Gerald Edgar | I worked out an example, and $S(A\sharp B)S^*=(SAS^*) \sharp (SBS^*)$ came out correct. So we just need a proof of it. Note that here $P$ is not unitary, but it is nonsingular, so we can solve backward for the result at the end. | |
| Aug 1, 2011 at 19:32 | comment | added | Federico Poloni | Sorry for the sloppiness, by "invariant" I mean that $S(A\sharp B)S^*=(SAS^*) \sharp (SBS^*)$. | |
| Aug 1, 2011 at 19:30 | comment | added | Federico Poloni | The square root is not, but the geometric mean is invariant under conjugation $A\mapsto SAS^*$ for each nonsingular $S$. | |
| Aug 1, 2011 at 18:58 | comment | added | Russel | I don't think there is "no loss of generality" to assume $A=I, B=D$. $(P^*AP)^{1/2}\ne P^*A^{1/2}P$ generally. | |
| Aug 1, 2011 at 18:09 | history | edited | Suvrit | CC BY-SA 3.0 |
fixed silly typo
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| Aug 1, 2011 at 18:08 | comment | added | Suvrit | @Tsuyoshi: thanks for catching the typo; fixing it now. | |
| Aug 1, 2011 at 18:01 | comment | added | Tsuyoshi Ito | This seems like a nice technique to remember. Typo: U should diagonalize $S^T Q^T BQS$ instead of $S^T Q^T BSQ$. | |
| Aug 1, 2011 at 17:22 | comment | added | Suvrit | @Gerald: No commutability was assumed; I had first put a comment, but now I have edited it into the answer. | |
| Aug 1, 2011 at 17:05 | history | edited | Suvrit | CC BY-SA 3.0 |
added extra proof, removed comment.
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| Aug 1, 2011 at 1:13 | comment | added | Gerald Edgar | I don't understand the "no loss of generality" here. Of course if $A,B$ commute, then we can reduce to the diagonal case. | |
| Jul 31, 2011 at 23:05 | history | answered | Suvrit | CC BY-SA 3.0 |