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Month: February 2015
Một bổ đề về KerA và KerAA^t
Exercise of diff topo 1
BT của Munkres,
BT1, Cho là một đa tạp chiều
có biên khác rỗng. Gọi
và
là hai mảnh của
. “Double” của
, ký hiệu là
, là không gian tô pô tạo bởi
bằng cách gắn đồng nhất
và
với mỗi
(biên của M). Chứng minh rằng
là một đa tạp không có biên
chiều.
Trace of powers of a nilpotent matrix
File: Trace of powers of a nilpotent matrix
This note, I copied from Yoshi on MathStackExchange:
\quad
Let be an
complex nilpotent matrix. Then we know that because all eigenvalues of
must be 0, it follows that
for all positive integers
.
What I would like to show is the converse, that is, if for all positive integers
, then
is nilpotent. I tried to show that 0 must be an eigenvalue of
, then try to show that all other eigenvalues must be equal to 0. However, I am stuck at the point where I need to show that
.
May I know of the approach to show that is nilpotent?
\quad
The answer of Alvarez:
If the eigenvalues of are
,
,
, then the eigenvalues of
are
,
,
. It follows that if all powers of
have zero trace, then
.
Using [Newton’s identities] to express the elementary symmetric functions of the ‘s in terms of their power sums, we see that all the coefficients of the characteristic polynomial of
(except that of greatest degree, of course) are zero. This means that
is nilpotent.
\quad
The answer of JBC:
Assume that for all ,
where
is a
matrix. We consider the eigenvalues in
.
Suppose is not nilpotent, so
has some non-zero eigenvalues
.
Let the multiplicity of
then
So we have
But
(Vandermonde)
So the system has a unique solution which is . Contradiction.
\quad
The answer of Qiaochu Yuan: Here is an argument that does not involve Newton’s identities, although it is still closely related to symmetric functions. Write
where are the eigenvalues of
. As a meromorphic function,
has poles at the reciprocals of all of the nonzero eigenvalues of
. Hence if
identically, then there are no such nonzero eigenvalues.
The argument using Newton’s identities, however, proves the stronger statement that we only need to require for
. Newton’s identities are in fact equivalent to the identity
where . To prove this identity it suffices to observe that
and differentiating both sides gives
(The argument using Newton’s identities is also valid over any field of characteristic zero.)