# Jordan Normal Form matrix decomposition into the sum of parts that commute.

I'm learning about Jordan Normal Form matrices, and I've read that we can decompose a Jordan Normal Form matrix into the sum of two parts $$J=N_J+D_J$$where $D_J$ is the diagonal part (i.e. the eigenvalues) and $N_J$ is the superdiagonal part (hence is nilpotent). It is then stated that $D_J$ and $N_J$ commute ($N_JD_J=D_JN_J$).

Why do $D_J$ and $N_J$ always commute? Initially I thought this would relate to the eigenvalues of the parts, but $N_J$ is nilpotent so has eigenvalues of zero only, whereas $D_J$ does not. Can someone point me in the right direction?

Let $J_k(\lambda)$ denote the Jordan block of size $k$ associated with $\lambda$. Let $A_1 \oplus A_2 \oplus \cdots \oplus A_m$ denote the block-diagonal matrix $$\pmatrix{A_1 \\ & A_2 \\&& \ddots \\ &&& A_m}$$ We note that for two block diagonal matrices partitioned in the same fashion, we have $$(A_1 \oplus A_2 \oplus \cdots \oplus A_m)+ (B_1 \oplus B_2 \oplus \cdots \oplus B_m) = (A_1 + B_1) \oplus (A_2 + B_2) \oplus \cdots \oplus (A_m + B_m) \\ (A_1 \oplus A_2 \oplus \cdots \oplus A_m) (B_1 \oplus B_2 \oplus \cdots \oplus B_m) = (A_1 B_1) \oplus (A_2B_2) \oplus \cdots \oplus (A_m B_m)$$ Now, the Jordan form of the matrix $A$ can be written as $$J = J_{k_1}(\lambda_1) \oplus J_{k_2}(\lambda_2) \oplus \cdots \oplus J_{k_m}(\lambda_m)$$ Let $I_k$ denote the identity matrix of size $k$. We define $$D = (\lambda_1 I_{k_1}) \oplus (\lambda_2 I_{k_2}) \oplus \cdots \oplus (\lambda_m I_{k_m})\\ N = J_{k_1}(0) \oplus J_{k_2}(0)\oplus \cdots \oplus J_{k_m}(0)$$ Verify that $J = D + N$, and that $DN = ND$.
• Thanks, that's great! So, for each block we have each $(\lambda_i I_{k_i})$ and $J_{k_i}$ commuting (since it one is always the identity element) and it follows that the product must commute. An additional question: it is also stated that $N$ and $D$ commute with $J$. For $D$ and $J$ this makes sense, but I can't see how $N$ and $J$ commute using your answer above. How does this work? – Zac Aug 25 '15 at 15:39
• Remember, $J = D + N$. We then have $$NJ = \\ N(D + N) = \\ ND + N^2=\\ DN + N^2=\\ (D+N)N =\\ JN$$ – Omnomnomnom Aug 25 '15 at 15:47
This is simply because if $D=\begin{bmatrix}d_1&0&\dots&0\\0&d_2&\dots&0\\\dots&\dots&\dots&\dots\\0&0&\dots &d_n\end{bmatrix}$, left multiplying any matrix $A$ by $D$ results in multiplying row no $k$ of $A$ by $d_k$ ($\;k=1,\dots, n$), and right multiplying results in multiplying column no $k$ by $d_k$, and that for the nilpotent part of a Jordan matrix, it has the same effect, as there is only $1$ non zero element per row and per column.