Shift matrix

Let L and U be the n × n lower and upper shift matrices, respectively. The following properties hold for both U and L. Let us therefore only list the properties for U: • det(U) = 0 • tr(U) = 0 • rank(U) = n − 1 • The characteristic polynomials of U is • : p_U(\lambda) = (-1)^n\lambda^n. • Un = 0. This follows from the previous property by the Cayley–Hamilton theorem. • The permanent of U is 0. The following properties show how U and L are related: {{unordered list : N(U) = \operatorname{span}\left\{ (1, 0, \ldots, 0)^\mathsf{T} \right\}, : N(L) = \operatorname{span}\left\{ (0, \ldots, 0, 1)^\mathsf{T} \right\}. : UL = I - \operatorname{diag}(0, \ldots, 0, 1), : LU = I - \operatorname{diag}(1, 0, \ldots, 0). These matrices are both idempotent, symmetric, and have the same rank as U and L }} If N is any nilpotent matrix, then N is similar to a block diagonal matrix of the form : \begin{pmatrix} S_1 & 0 & \ldots & 0 \\ 0 & S_2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & S_r \end{pmatrix} where each of the blocks S1, S2, ..., Sr is a shift matrix (possibly of different sizes). ==Examples==

: S = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{pmatrix}; \quad A = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 & 1 \\ 1 & 2 & 3 & 2 & 1 \\ 1 & 2 & 2 & 2 & 1 \\ 1 & 1 & 1 & 1 & 1 \end{pmatrix}. Then, : SA = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 2 & 1 \\ 1 & 2 & 3 & 2 & 1 \\ 1 & 2 & 2 & 2 & 1 \end{pmatrix}; \quad AS = \begin{pmatrix} 1 & 1 & 1 & 1 & 0 \\ 2 & 2 & 2 & 1 & 0 \\ 2 & 3 & 2 & 1 & 0 \\ 2 & 2 & 2 & 1 & 0 \\ 1 & 1 & 1 & 1 & 0 \end{pmatrix}. Clearly there are many possible permutations. For example, S^\mathsf{T} A S is equal to the matrix A shifted up and left along the main diagonal. : S^\mathsf{T}AS=\begin{pmatrix} 2 & 2 & 2 & 1 & 0 \\ 2 & 3 & 2 & 1 & 0 \\ 2 & 2 & 2 & 1 & 0 \\ 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}. ==See also==