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?
Best Answer
Whether two matrices commute has to do with their eigenvectors, not their eigenvalues. At any rate, seeing that the matrices commute is easier than you're making it.
I provided an explanation in an earlier answer.
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 $DN = ND$.