If \(N\) is a \(k \times k\) elementary nilpotent matrix, i.e., \(N^k = 0\) but \(N^{k-1} \neq 0\), show that \(N^t\) is similar to \(N\). Now use the Jordan form to prove that every complex \(n \times n\) matrix is similar to its transpose.
Solution
1. An elementary nilpotent matrix is similar to its transpose.
If \(N\) is \(k\times k\), nilpotent, and \(N^{k-1}\ne0\), then its nilpotency index is \(k\). Hence its Jordan form consists of one block \(J_k(0)\) [@HK71]. It therefore suffices to compare \(J_k(0)\) with its transpose.
Let \(R\) be the permutation matrix reversing the standard basis: \[ Re_j=e_{k+1-j}. \] Then \(R^{-1}=R\), and direct inspection gives \[ R J_k(0)R=J_k(0)^t. \] Thus \(J_k(0)^t\sim J_k(0)\), and hence \(N^t\sim N\).
2. Every Jordan block is similar to its transpose.
For any \(c\in\mathbb C\), \[ J_k(c)=cI+J_k(0). \] Using the same reversal matrix, \[ R J_k(c)R =cI+R J_k(0)R =J_k(c)^t. \] So each Jordan block is similar to its transpose.
3. Apply the blockwise similarity to an arbitrary complex matrix.
Let \(A=PJP^{-1}\) be a Jordan form. Then \[ A^t=(P^{-1})^tJ^tP^t, \] so \(A^t\) is similar to \(J^t\). By step 2, each block of \(J^t\) is similar to the corresponding block of \(J\), hence their direct sums are similar: \(J^t\sim J\). Therefore \[ A^t\sim J^t\sim J\sim A. \] Thus every complex square matrix is similar to its transpose.