# Dunford decomposition proof : why such a form?

I'm working on the proof of the Dunford decomposition theorem :

All matrices $$A\in M_n(K)$$ such their characteristic polynomials split can be written in the form $$A=D+N$$ where $$D$$ is diagonalizable and $$N$$ is nilpotent.

The proof :

Let $$A \in M_n(K)$$ a linear operator with spectrum $$\sigma(A)=\{\lambda_1,...\lambda_r\}$$. The characteristic polynomial of $$A$$ can be written : $$X_A(t)=\prod\limits_{i=1}^{r}(t-\lambda_i)^{m_i} \quad m_i\text{ is the algebraic multiplicity of }\lambda_i$$ Then, we know, thanks to the primary reduction theorem, that : $$\ker(X_A(A))=K_n=N_1 \oplus \cdots \oplus N_r$$ where $$N_i=N_{\lambda_i}(A)=\ker(A-\lambda_iI_n)^{m_i}$$

If we call $$B_i$$ a basis of $$N_i$$ far all $$i \in \{1,...,r\}$$, then $$B=B_1 \cup \cdots \cup B_r$$ is a basis of $$K^n$$. Here is the thing that I don't understant : why, in this basis, the matrix of $$A$$ will have this form :

$$\begin{pmatrix} A_1 & 0 & \dots & 0 \\ 0 & A_2 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & A_r \end{pmatrix}$$

Where $$A_i, i\in \{1,...,r\}$$ is a matrix with size $$m_i\times m_i$$ ???

• You want $m_i$ the algebraic multiplicities, not geometric multiplicities. – user10354138 Jun 5 '19 at 20:28
• ooh yes thank you... – Dicordi Jun 5 '19 at 20:31
• Note that $A$ maps $N_i$ to $N_i$, since $A$ commutes with any polynomial in $A$, in particular $(A-\lambda_i)^{m_i}$. – user10354138 Jun 5 '19 at 20:36

It has this form because each $$N_i$$ is stable under $$A$$, so the basis vectors of $$B_i$$ are sent to combinations of vectors of $$B_i$$ : their components on $$B_j, j\neq i$$ are therefore $$0$$.
More generally, when $$E$$ is a vector space, $$f$$ an endomorphism and $$F,W$$ two stable subspaces such that $$E=F\oplus W$$, then the matrix of $$f$$ can be written as $$\begin{pmatrix} A_1 & 0 \\ 0 & A_2 \end{pmatrix}$$ in the basis $$B_1\cup B_2$$ if $$B_1$$ is a basis of $$F,B_2$$ of $$W$$, and $$A_1$$ is the matrix of the restriction-corestriction of $$f$$ to $$F$$ in the basis $$B_1$$, $$A_2$$ similarly but with $$W$$ and $$B_2$$
• Yes thanks, this is because $(f-\lambda_i I_d)^{m_i}(f-\lambda_i I_d)(v)=-(f-\lambda_i I_d)^{m_i}(\lambda_i v)+(f-\lambda I_d)^{m_i}(f(v))=0$ So $N_i$ Is $f-$invariant ! – Dicordi Jun 6 '19 at 8:55