Melodie, composed by Tchaikovsky, played by Daniel Shafran (cello).
Melodie, composed by Tchaikovsky, played by Daniel Shafran (cello).
Scriabin Etude, Op.42-5, played by Evgeny Kissin. So romantic! I’ll try to learn this some time soon. Just printed out the sheet music, actually. Maybe I’ll try it out this afternoon.
Rachmaninoff Cello Sonata Movement #3, Daniel Shafran cello, Yakov Flier piano.
Let A and B be 2-by-2 matrices with integer entries such that A, A+B, A+2B, A+3B, and A+4B are all invertible matrices whose inverses have integer entries. Show that A+5B is invertible and that its inverse has integer entries.
This problem has bugged me for a long time, but I solved it immediately after learning about unimodular matrices. Here is how I solved it. For any invertible matrix M, we have |M| * |M^(-1)| = 1. Thus, unless both numbers are +1 or -1, either the absolute value of |M| or |M^(-1)| is smaller than 1, but greater than 0. There are many characterizations of the determinant of a matrix, and one of them says that the determinant of a matrix is the sum of products of some elements from it. Therefore, an integer matrix cannot have a non-integer determinant. This shows that whenever M has integer entries and is invertible, then |M| is either +1 or -1. (This argument applies to matrices of any size, not just 2-by-2.)
Now going back to the original problem, the remark above states that |A+tB| is +1 or -1 for t = 0, 1, 2, 3, 4. By pigeonhole principle, at least for three different t’s, |A+tB| is equal. Since the matrices are of size 2-by-2, |A+tB| is a quadratic polynomial in t. However, since it achieves the same value at three different t’s, |A+tB| must be constant. This shows that |A+5B| is either +1 or -1.
We’re almost done. To finish, we have to argue that A+5B is an integer matrix, and it does have an integer inverse. This is not hard, but cannot be left off because in the first part, we only showed that “integer inverse -> determinant +1 or -1”, but not the other direction. The fact that A+5B is an integer matrix is pretty obvious, because both A and A+B had integer entries. The inverse of a 2-by-2 matrix [a b; c d] is given by [d -b; -c a] / (ad-bc). We know that ad-bc is either +1 or -1. Thus, if the entries were integers, then the inverse would have integer entries, too. (In fact, regardless of its dimension, if an integer matrix has determinant +1 or -1, then it does have an integer inverse. This is because the inverse of a matrix is the cofactor matrix divided by the determinant of the original matrix.) Finally we’re done.
Given the following matrix of 25 elements
11, 17, 25, 19, 16
24, 10, 13, 15, 3
12, 5, 14, 2, 18
23, 4, 1, 8, 22
6, 20, 7, 21, 9,
choose five of these elements, no two coming from the same row or column, in such a way that the minimum of these five elements is as large as possible. Prove that your answer is correct.
Eddie Higgins Trio, Into the Memory (기억 속으로)
Kapustin, Toccatina Op. 36, played by Marc-Andre Hamelin. Somehow the file I have is named as the 2nd movement of sonata #2, recorded from his 1999 Bremen concert. It could be the case that Hamelin actually played the toccatina instead of the original second movement… Who knows.
A gorgeous transcription of the main theme of Schindler’s List. I wonder if it’s possible to transcribe this again, to a piano solo piece…
Wow, why didn’t I know this piece before? I’m definitely learning this at some point soon. Chopin’s Barcarolle, op. 60.
It’s December already… George Winston’s Joy from December.