Linear algebra and matrices M3P9

Question Sheet 3 - Answers and solutions

1. Hint: Prove by direct calculation that A^{m} has the property that its (i, j)-entries are zeroes unless j ≥ i + m.

2. The first matrix is the nilpotent Jordan block J_{0,n}. The second matrix is

( \begin{matrix}a & 0 & b & 0 & … & 0 \\ 0 & a & 0 & b & … & 0 \\ … & & & & & … \\ 0 & 0 & 0 & 0 & … & a \end{matrix} )

3. m_{c}(t) = l.c.m.(m_{A}(t), m_{B}(t)).

4. The non-degeneracy of A implies that the function which associates to the column j the row i such that a_{ij} = 1 is a permutation, say σ, of {1, 2, …, n}. Then Ae_{j} = e_{σ(j)}. Any permutation has finite order, that is, σ^{m} = 1 for some m. Then A^{m} = Id, which implies that every eigenvalue of a is an m-th root of 1.

5. The set {1, 2, …, n} can be represented as the disjoint union of orbits of the permutation σ, say {1, 2, …, n} = I_{1} ∪ … ∪ I_{s}. Then the ambient space V is the direct sum of V_{j} :=< e_{i}|i ∈ I_{j} > (the linear span of e_{i} for i ∈ I_{j}). The linear transformation A preserves this decomposition, hence its matrix is the direct sum of the matrices of A restricted to each V_{i}. But on each particular V_{i} the matrix of A in a suitably rearranged basis e_{i}, i ∈ V_{j} is the matrix

( \begin{matrix}0 & 1 & 0 & 0 & … & 0 \\ 0 & 0 & 1 & 0 & … & 0 \\ 0 & 0 & 0 & 1 & … & 0 \\ … & & & & … & 1 \\ 1 & 0 & 0 & 0 & … & 0 \end{matrix} )

Hence its characteristic polynomial is t^{dim(V_{i})} −1 (check!), which has different roots. Thus the restriction of A to each V_{i} is diagonalizable (in the field of complex numbers), hence A is also diagonalizable.