Let $ g $ be an invertible $ n\times n $ complex matrix. Show that $ g $ can be written as $$ g=su=us ,$$ where $ s $ is diagonalizable and all eigenvalues of $ u $ are equal to $ 1 $.

My Attempt:

Since $ g $ is an invertible complex matrix, then we assume that $ g $ is similar to $ \operatorname{diag}\{ J_1(\lambda_1), J_2(\lambda_2),\cdots, J_k(\lambda_k) \}=h^{-1}gh, \ h\in GL(n, \mathbb C) $ where $ J_i(\lambda_i) $ is the $ i $-th Jordan block with eigenvalue $ \lambda_i>0 $. If $ k=n $, i.e., $ g $ is diagonalizable, then take $ s=g, \enspace u=I_n $ the identity matrix and we are done. If not, it suffices to consider a non-diagonal Jordan block, say, $ \left(J_1(\lambda_1)\right)_{m_1\times m_1} $ with $ 2\le m_1\le n $, since diagonalization and eigenvalues are invariant under similar transformations.

Let's first try the easiest case which is $ m_1=2 $. If we want to find a $ s $ which is diagonalizable, then we can find invertible matrix $ Q $, s.t. $ Q^{-1}sQ=\operatorname{diag}\{ \tau_1, \tau_1 \}_{2\times 2},\ 0\ne\tau_1\in\mathbb C $. Then we can write: $$\begin{align} Q^{-1}J_1(\lambda_1)&=Q^{-1}su\\ &=(Q^{-1}sQ)Q^{-1}u\\ &=\tau_1Q^{-1}u .\end{align}$$ Hence we have $$ \begin{pmatrix}\lambda_1/\tau_1 & 1/\tau_1\\ 0&\lambda_1/\tau_1 \end{pmatrix}=u .$$ Since all eigenvalues of $ u $ are $ 1 $, we deduce that $ \lambda_1=\tau_1 $. Thus $$ u=\begin{pmatrix} 1&1/\lambda_1\\ 0&1 \end{pmatrix}. $$ And it should be clear that $$ s=\begin{pmatrix}\lambda_1&0\\0&\lambda_1\end{pmatrix} .$$ In general, we conclude that $$ s=\operatorname{diag}\{ \underbrace{\lambda_1,\cdots, \lambda_1}_{m_1},\underbrace{\lambda_2,\cdots, \lambda_2}_{m_2},\cdots, \underbrace{\lambda_k,\cdots, \lambda_k}_{m_k} \} ,$$ and $$ u=\operatorname{diag}\{ \frac{1}{\lambda_1}J_1(\lambda_1), \frac{1}{\lambda_2}J_2(\lambda_2),\cdots, \frac{1}{\lambda_k}J_k(\lambda_k) \} $$ up to similarity.

Is it right to give a proof like that?

  • $\begingroup$ It looks to me like your answer is generally correct, except that you switched between assuming $g$ was in Jordan form and assuming it was similar to a direct sum of Jordan blocks. $\endgroup$ – jgon Mar 26 at 14:39

Use the Chevalley Jordan decomposition ( cf. wiki).

$A=D+N$ where $DN=ND$, $D$ is diagonalizable over $\mathbb{C}$, $spectrum(D)=spectrum(A)$ and $N$ is nilpotent. Then $D$ is invertible and

$A=D(I_n+D^{-1}N)$. Take $s=D,u=I+D^{-1}N$.

  • $\begingroup$ Hi! Can you explain the reason why $ spectrum(D)=spectrum(A) $? $\endgroup$ – user549397 Mar 29 at 1:17
  • $\begingroup$ Since $DN=ND$, $D,N$ are simultaneously triangularizable. $\endgroup$ – loup blanc Mar 29 at 13:11
  • $\begingroup$ Hum...... still can't get it...maybe a link to proof is helpful. $\endgroup$ – user549397 Mar 29 at 15:18

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