A = B+ I, B^2 = 0

Let {A = \begin{pmatrix} a + 1& -a\\ a & -a + 1\\ \end{pmatrix}}. What can we say about {A}?

Remark: {A} has some interesting properties: {A = B + I} where {B^2 = O}.

  1. {A^n = nB+I, n \in \mathbb{Z}}.
    Indeed, we can prove it by induction.
    {A^2 = (B+I)(B+I) = B^2 + 2B + I = 2B+I},  {(B+I)(-B+I) = I \Rightarrow A^{-1} = -B + I, A^{-2} = (-B + I)(-B+I) = -2B+I\dots} Suppose that {A^k = kB+I}, we have {A^{k+1} = (kB+ I)(B+I) = kB^2 + (k+1)B + I = (k+1)B + I}. Moreover, suppose that {A^{-k} = -B+I}, we have {A^{-k-1} = (-kB+I)(-B+I) = -(k+1)B + I}.
  2. {A^{-1} = - B+I \Leftrightarrow B + I = - A^{-1} + 2I \Leftrightarrow A = -A^{-1} + 2I \Leftrightarrow A + A^{-1} = 2I}.
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Trace of powers of a nilpotent matrix

File: Trace of powers of a nilpotent matrix

This note, I copied from Yoshi on MathStackExchange:
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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?
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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.
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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.
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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.)