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?
