📂Matrix Algebra

SN Decomposition of a Matrix

Theorem¹

For the matrix $A \in M_{n \times n}(\mathbb{C})$, there exists a unique decomposition satisfying the following.

$$ A = S + N $$

Here $S$ is a diagonal matrix, and $N$ is a nilpotent matrix.

Explanation

This is called the SN decomposition (decomposition) or the Jordan–Chevalley decomposition. Apart from being a complex square matrix, $A$ has no further restrictions. The diagonal matrix $S$ has the eigenvalues of $A$ on its diagonal entries.

Proof

Throughout, $A$ is regarded as a matrix and as a linear transformation.

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Let the distinct eigenvalues of $A$ be $\lambda_{1}, \lambda_{2}, \dots, \lambda_{k}$. Let $W_{\lambda_{i}}$ denote the generalized eigenspace corresponding to $\lambda_{i}$. Then $W_{\lambda_{i}}$ is a $A$-invariant invariant subspace, and the following holds.

$$ \mathbb{C}^{n} = W_{\lambda_{1}} \oplus W_{\lambda_{2}} \oplus \cdots \oplus W_{\lambda_{k}} $$

Let $A_{\lambda_{i}} = A|_{W_{\lambda_{i}}}$. Then the following holds.

$$ A = A_{\lambda_{1}} \oplus A_{\lambda_{2}} \oplus \cdots \oplus A_{\lambda_{k}} = \begin{bmatrix} A_{\lambda_{1}} & O & \cdots & O \\ O & A_{\lambda_{2}} & \cdots & O \\ \vdots & \vdots & \ddots & \vdots \\ O & O & \cdots & A_{\lambda_{k}} \end{bmatrix} $$

Now let $N_{\lambda_{i}} = A_{\lambda_{i}} - \lambda_{i} I$. Then $N_{\lambda_{i}}$ is a nilpotent transformation (matrix) on $W_{\lambda_{i}}$. Therefore $A$ equals the following.

$$ \begin{align*} A &= \begin{bmatrix} \lambda_{1} I + N_{\lambda_{1}} & O & \cdots & O \\ O & \lambda_{2} I + N_{\lambda_{2}} & \cdots & O \\ \vdots & \vdots & \ddots & \vdots \\ O & O & \cdots & \lambda_{k} I + N_{\lambda_{k}} \end{bmatrix} \\ &= \begin{bmatrix} \lambda_{1} I & O & \cdots & O \\ O & \lambda_{2} I & \cdots & O \\ \vdots & \vdots & \ddots & \vdots \\ O & O & \cdots & \lambda_{k} I \end{bmatrix} + \begin{bmatrix} N_{\lambda_{1}} & O & \cdots & O \\ O & N_{\lambda_{2}} & \cdots & O \\ \vdots & \vdots & \ddots & \vdots \\ O & O & \cdots & N_{\lambda_{k}} \end{bmatrix} \end{align*} $$

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