# Decompose an invertible matrix into an exchangeable product of diagonalizable matrix and a matrix with all the eigenvalues equal to $1$

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?

• 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. – jgon Mar 26 '19 at 14:39

$$A=D+N$$ 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 $$s=D,u=I+D^{-1}N$$.
• Hi! Can you explain the reason why $spectrum(D)=spectrum(A)$? – Bach Mar 29 '19 at 1:17
• Since $DN=ND$, $D,N$ are simultaneously triangularizable. – user91684 Mar 29 '19 at 13:11