Posted in Blog - news

## Chúc mừng năm mới Ất Mùi

Chúc các bạn dồi dào sức khỏe và thành công trong năm mới

Posted in Geometric topology

## Exercise of diff topo 1

BT của Munkres,

BT1, Cho $M$ là một đa tạp chiều $m$ có biên khác rỗng. Gọi $M_0 = M \times 0$$M_1 = M \times 1$ là hai mảnh của $M$. “Double” của $M$, ký hiệu là $D(M)$, là không gian tô pô tạo bởi $M_0 \cup M_1$ bằng cách gắn đồng nhất $(x,0)$$(x,1)$ với mỗi $x \in Bd(M)$ (biên của M). Chứng minh rằng $D(M)$ là một đa tạp không có biên $m$ chiều.

Posted in Olympiad exercises

## Trace of powers of a nilpotent matrix

This note, I copied from Yoshi on MathStackExchange:

Let ${A}$ be an ${n\times n}$ complex nilpotent matrix. Then we know that because all eigenvalues of ${A}$ must be 0, it follows that ${\text{tr}(A^n)=0}$ for all positive integers ${n}$.

What I would like to show is the converse, that is, if ${\text{tr}(A^n)=0}$ for all positive integers ${n}$, then ${A}$ is nilpotent. I tried to show that 0 must be an eigenvalue of ${A}$, 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 ${\det(A)=0}$.

May I know of the approach to show that ${A}$ is nilpotent?

The answer of Alvarez:

If the eigenvalues of ${A}$ are ${\lambda_1}$, ${\dots}$, ${\lambda_n}$, then the eigenvalues of ${A^k}$ are ${\lambda_1^k}$, ${\dots}$, ${\lambda_n^k}$. It follows that if all powers of ${A}$ have zero trace, then ${\lambda_1^k+\dots+\lambda_n^k=0, \forall k\geq 1}$.

Using [Newton’s identities] to express the elementary symmetric functions of the ${\lambda_i}$‘s in terms of their power sums, we see that all the coefficients of the characteristic polynomial of ${A}$ (except that of greatest degree, of course) are zero. This means that ${A}$ is nilpotent.
The answer of JBC:

Assume that for all ${k=1,\ldots,n}$, ${\mathrm{tr}(A^k) = 0}$ where ${A}$ is a ${n\times n}$ matrix. We consider the eigenvalues in ${\mathbb{C}}$.

Suppose ${A}$ is not nilpotent, so ${A}$ has some non-zero eigenvalues ${\lambda_1,\ldots,\lambda_r}$.
Let ${n_i}$ the multiplicity of ${\lambda_i}$ then

$\displaystyle \left\{\begin{array}{ccc}n_1\lambda_1+\cdots+n_r\lambda_r&=&0 \\ \vdots & & \vdots \\ n_1\lambda_1^r+\cdots+n_r\lambda_r^r&=&0\end{array}\right.$

So we have

$\displaystyle \left(\begin{array}{cccc}\lambda_1&\lambda_2&\cdots&\lambda_r\\\lambda_1^2 & \lambda_2^2 & \cdots & \lambda_r^2 \\ \vdots & \vdots & \vdots & \vdots \\ \lambda_1^r & \lambda_2^r & \cdots & \lambda_r^r\end{array}\right)\left(\begin{array}{c}n_1 \\ n_2 \\ \vdots \\ n_r \end{array}\right)=\left(\begin{array}{c}0 \\ 0\\ \vdots \\ 0\end{array}\right)$

But

$\displaystyle \mathrm{det}\left(\begin{array}{cccc}\lambda_1&\lambda_2&\cdots&\lambda_r\\\lambda_1^2 & \lambda_2^2 & \cdots & \lambda_r^2 \\ \vdots & \vdots & \vdots & \vdots \\ \lambda_1^r & \lambda_2^r & \cdots & \lambda_r^r\end{array}\right)=\lambda_1\cdots\lambda_r\,\mathrm{det}\left(\begin{array}{cccc} 1 & 1 & \cdots & 1 \\ \lambda_1&\lambda_2&\cdots&\lambda_r\\\lambda_1^2 & \lambda_2^2 & \cdots & \lambda_r^2 \\ \vdots & \vdots & \vdots & \vdots \\ \lambda_1^{r-1} & \lambda_2^{r-1} & \cdots & \lambda_r^{r-1}\end{array}\right)\neq 0$

(Vandermonde)

So the system has a unique solution which is ${n_1=\ldots=n_r=0}$. Contradiction.
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

$\displaystyle f(z) = \sum_{k\ge 0} z^k \text{tr}(A^k) = \sum_{i=1}^n \frac{1}{1 - z \lambda_i}$

where ${\lambda_i}$ are the eigenvalues of ${A}$. As a meromorphic function, ${f(z)}$ has poles at the reciprocals of all of the nonzero eigenvalues of ${A}$. Hence if ${f(z) = n}$ 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 ${\text{tr}(A^k) = 0}$ for ${1 \le k \le n}$. Newton’s identities are in fact equivalent to the identity

$\displaystyle f(z) = n - \frac{z p'(z)}{p(z)}$

where ${p(z) = \prod_{i=1}^n (1 - z \lambda_i)}$. To prove this identity it suffices to observe that

$\displaystyle \log p(z) = \sum_{i=1}^n \log (1 - z \lambda_i)$

and differentiating both sides gives

$\displaystyle \frac{p'(z)}{p(z)} = \sum_{i=1}^n \frac{- \lambda_i}{1 - z \lambda_i}.$

(The argument using Newton’s identities is also valid over any field of characteristic zero.)