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Let $n \in \mathbb N^{*}$, $A \in GL_n (\mathbb C)$ and $N \in M_n (\mathbb C)$ with $N$ Nilpotent such as $AN=NA$. Prove that there exists $B \in M_n (\mathbb C) $ such as $B^{2} = A + N$

Since $A$ and $N$ are triangularisable and are commuting matrices, $A$ and $N$ are simultaneously triangularisable. There exists $P \in GL_n (\mathbb C)$ such as

$A=P\left( \begin{array}{ccccc} \lambda_1 \\ & \ddots & & \huge*\\ & & \ddots \\ & \huge0 & & \lambda_n \end{array} \right)P^{-1}\:\:\:$ with $(\lambda_1,\dots,\lambda_n)$ the eigenvalues of $A$. $A$ is invertible thus $(\lambda_1,\dots,\lambda_n) \ne(0,\dots,0)$

and $N=P\left( \begin{array}{ccccc} 0 \\ & \ddots & & \huge{*'}\\ & & \ddots \\ & \huge0 & & 0 \end{array} \right)P^{-1}$

Let's set $T=\left( \begin{array}{ccccc} \lambda_1 \\ & \ddots & & \huge*\\ & & \ddots \\ & \huge0 & & \lambda_n \end{array} \right)+\left( \begin{array}{ccccc} 0 \\ & \ddots & & \huge{*'}\\ & & \ddots \\ & \huge0 & & 0 \end{array} \right)=\left( \begin{array}{ccccc} \lambda_1 \\ & \ddots & & \huge*''\\ & & \ddots \\ & \huge0 & & \lambda_n \end{array} \right)$

Let's define $ L \in M_n (\mathbb C)$ by $\forall (i,j) \in [\![1;n]\!]^{2}, L_{i,j}=\left\{ \begin{array}{lll} 0 & \mbox{if } i>j \\ \alpha_i & \mbox{if } i=j \\ \frac{1}{\alpha_i+\alpha_j} \times(T_i,j - \sum_{k=i+1}^{j-1} L_{i,k} L_{k,j}) &\mbox{else} \end{array} \right.$, with $(\alpha_1,\dots,\alpha_n) \:\:$square roots of $(\lambda_1,\dots,\lambda_n)$.

The diagonal of L is well defined because for all $i$, $\alpha_i \ne 0$ since $\lambda_i \ne 0$. The upper part of L is defined step by step, the diagonal then the first subdiagonal, then the second subdiagonal, and so on; (for $j-i$ going from $0$ to $n-1$, with $(i,j) \in [\![1;n]\!]^{2}$ and $j\geq i$).

$L$ is an upper triangular matrix so $L^{2}$ is also an upper triangular matrix. Let $(i,j) \in [\![1;n]\!]^{2}$ and $j> i$

$L^{2}_{i,i}= \alpha_i^{2} = \lambda_i =T_{i,i}\\ L^{2}_{i,j}= \sum_{k=0}^{n}L_{i,k}L_{k,j}=\sum_{k=i}^{j}L_{i,k}L_{k,j}= L_{i,i}L_{i,j}+ \sum_{k=i+1}^{j-1}L_{i,k}L_{k,j} + L_{i,j}L_{j,j} = L_{i,j} ( \alpha_i+\alpha_j) + \sum_{k=i+1}^{j-1}L_{i,k}L_{k,j}=\frac{1}{\alpha_i+\alpha_j} \times(T_i,j - \sum_{k=i+1}^{j-1} L_{i,k} L_{k,j}) \times ( \alpha_i+\alpha_j) + \sum_{k=i+1}^{j-1}L_{i,k}L_{k,j}=T_{i,j}$ ie $L^{2}=T$

let's pose $B=PLP^{-1}$ we have $ B^{2}= PL^{2}P^{-1}= PTP^{-1}= A+N$

Several questions: Is the definition of $L$ correct? Is there another method, which does not use the matrix L ?

turo
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1 Answers1

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To answer half of your question, here is an alternative approach. First, if we define $M = A^{-1}N$, then we find that $M$ is nilpotent and that $$ A^{-1}(A + N) = I + M. $$ Now, there are various ways to see that the matrix $I+M$ is invertible. For one, we can consider the eigenvalues of $I + M$ (which must all be equal to $1$, since the eigenvalues of $M$ must all be zero). Another approach is to consider the Neumann-series $$ (I + M)^{-1} = \sum_{k=0}^\infty (-1)^k M^k, $$ for which all but finitely many terms are finite. Because $I + M$ is invertible, it follows that $A + N = A(I + M)$ is also invertible.

By various means (confer this post and the linked posts therein), it can be shown that there exists a matrix $B$ such that $B^2 = A + N$ (because $A + N$ is invertible), which was what we wanted.

Ben Grossmann
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