Let $M_{n}$ denote the set of all $n\times n$-matrices over $\mathbb{C}$ and let us consider that $M_n$ is equipped with the Hilbert-Schmidt norm. Then we may identify $M_n\cong\mathbb{C^{n^2}}$. For any $A\in M_{n}$ we choose the distinct eigenvalues $\lambda_1,\dots,\lambda_n$ of $A$ with the eigenvectors $v_1,\dots,v_n$ and a circle $S=\{z\in \mathbb{C}:|z|=r\}$, where $r>0$, not meeting all of $\lambda_i$'s. Then we can define the matrix $Q\in M_{n}$ by $$Q= \frac{1}{2\pi i}\oint_S (zI-A)^{-1}dz.$$ (Note that the right-hand side means a contour integration of the matrix-valued function $z\mapsto (zI-A)^{-1}\in M_{n}(\cong \mathbb{C}^{n^2})$ defined on $S$, where the $I$ is the identity matrix, and that the matrix $zI-A$ is nonsingular for all $z\in S$). Then I would like to show the following properties:
- $QA=AQ$.
- $QQ=Q$.
If they are proved then we have, by the $(1)$, a decomposition $V=QV\oplus (I-Q)V$, where $V$ is the vector space $\mathbb{C}^{n}$. Moreover, since both of the subspaces $V_+=QV$ and $V_-=(I-Q)V$ are invariant for the linear transformation $A$ defined on $V$ by the (2), $A$ can be decomposed by the formula $A=A_+\oplus A_-$. Then all eigenvalues of $A_+$ reside inside of $S$ and the eigenvalues of $A_-$ does outside of $S$. Iterating this procedure, we have a decomposition $V=V_1\oplus \dots \oplus V_n$ with $V_i$ spanned by the $v_i$, which is the eigendecomposition of a vector space! (I heard that it is precisely a spectral decomposition, but I do not know the operator theory, unfortunately).
My attempt. I tried, anyway, the case of $n=2$ (though it is a tentative and concrete try). Suppose that $A$ has the form $$A= \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$ then $$(zI-A)^{-1}=\frac{1}{\text{det}(zI-A)} \begin{pmatrix} z-d & b \\ c & z-a \end{pmatrix}.$$ Since $\text{det}(zI-A)=(z-\lambda_1)(z-\lambda_2)$, if the eigenvalue lying inside of $S$ is the only $\lambda_1$ then the residue theorem says $$Q=\frac{1}{2\pi i} \begin{pmatrix} \oint_S \frac{z-d}{(z-\lambda_1)(z-\lambda_2)}dz & \oint_S \frac{b}{(z-\lambda_1)(z-\lambda_2)}dz \\ \oint_S \frac{c}{(z-\lambda_1)(z-\lambda_2)}dz & \oint_S \frac{z-a}{(z-\lambda_1)(z-\lambda_2)}dz \end{pmatrix} =\frac{1}{\lambda_1-\lambda_2} \begin{pmatrix} \lambda_1-d & b \\ c & \lambda_1-a \end{pmatrix}.$$ Similarly we see that if both of $\lambda_1$ and $\lambda_2$ lie inside of $S$ then $Q$ is determined by $$Q=\frac{1}{2\pi i} \begin{pmatrix} \oint_S \frac{z-d}{(z-\lambda_1)(z-\lambda_2)}dz & \oint_S \frac{b}{(z-\lambda_1)(z-\lambda_2)}dz \\ \oint_S \frac{c}{(z-\lambda_1)(z-\lambda_2)}dz & \oint_S \frac{z-a}{(z-\lambda_1)(z-\lambda_2)}dz \end{pmatrix} = \begin{pmatrix} \frac{\lambda_1-d}{\lambda_1-\lambda_2}+\frac{\lambda_2-d}{\lambda_2-\lambda_1} & \frac{b}{\lambda_1-\lambda_2}+\frac{b}{\lambda_2-\lambda_1} \\ \frac{c}{\lambda_1-\lambda_2}+\frac{c}{\lambda_2-\lambda_1} &\frac{\lambda_1-a}{\lambda_1-\lambda_2}+\frac{\lambda_2-a}{\lambda_2-\lambda_1} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. $$ It satisfies the commutativity (1) as $$AQ=\frac{1}{\lambda_1-\lambda_2} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \lambda_1-d & b \\ c & \lambda_1-a \end{pmatrix} = \begin{pmatrix} a\lambda_1-ad-cb & b\lambda_1 \\ c\lambda_1 & d\lambda_1-ad-cd \end{pmatrix} $$ and $$QA=\frac{1}{\lambda_1-\lambda_2} \begin{pmatrix} \lambda_1-d & b \\ c & \lambda_1-a \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a\lambda_1-ad-cb & b\lambda_1 \\ c\lambda_1 & d\lambda_1-ad-cd \end{pmatrix}. $$ Subsequently, to verify the idempotency (2), I calculate $$QQ=\frac{1}{(\lambda_1-\lambda_2)^2} \begin{pmatrix} \lambda_1-d & b \\ c & \lambda_1-a \end{pmatrix} \begin{pmatrix} \lambda_1-d & b \\ c & \lambda_1-a \end{pmatrix} =\frac{1}{(\lambda_1-\lambda_2)^2} \begin{pmatrix} (\lambda_1-d)^2+bc & b(\lambda_1-d)+b(\lambda_1-a) \\ c(\lambda_1-d)+c(\lambda_1-a) & (\lambda_1-a)^2+bc \end{pmatrix}, $$ but we have $\lambda_1+\lambda_2=a+d$, from which it follows that $(\lambda_1-d)+(\lambda_1-a) =\lambda_1-\lambda_2$. Additionally, observe from $\lambda_1+\lambda_2=a+d$ and $\lambda_1\lambda_2=ad-bc$ that $$(\lambda_1-\lambda_2)(\lambda_1-d) =\lambda_1^2-d\lambda_1+d\lambda_2-\lambda_1\lambda_2 =\lambda_1^2+(-2\lambda_1+a+d)d-ad+bc \\ =\lambda_1^2-2d\lambda_1+d^2+bc =(\lambda_1-d)^2+bc $$ and thus we conculude that $$QQ =\frac{1}{(\lambda_1-\lambda_2)^2} \begin{pmatrix} (\lambda_1-\lambda_2)(\lambda_1-d) & b(\lambda_1-\lambda_2) \\ c(\lambda_1-\lambda_2) & (\lambda_1-\lambda_2)(\lambda_1-a) \end{pmatrix} =Q. $$
My question. It is difficult that the above way is generalized, I think. How do I prove them for general $n$?