Linear algebra and matrices M3P9

Question Sheet 2 - Answers and solutions

1. 3 marks. It is immediate that (1, −1, 0, 0), (0, 1, −1, 0), (0, 0, 1, −1) are in Ker(A). Thus we have three linearly independent eigenvectors with eigenvalue 0. One observes that (1, 1, 1, 1) is an eigenvector with eigenvalue 4, so we have a basis of eigenvectors. Therefore we have P^{−1}AP = X where

P = ( \begin{matrix}1 & 1 & 0 & 0 \\ 1 & −1 & 1 & 0 \\ 1 & 0 & −1 & 1 \\ 1 & 0 & 0 & −1 \end{matrix} ) , X = ( \begin{matrix}4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} )

2. 2 marks. Scalar matrices (that is, A = c.Id, c ∈ k).

3. 8 marks, two for each matrix. We have m_{A_{1}}(t) = (t − λ)^{4};

m_{A_{2}}(t) = (t − λ)^{2}(t − μ)^{2} if λ ≠ μ, and m_{A_{2}}(t) = (t − λ)^{2} if λ = μ; m_{A_{3}}(t) = (t − λ)^{3}(t − μ) if λ ≠ μ, and m_{A_{3}}(t) = (t − λ)^{3} if λ = μ; m_{A_{4}}(t) = (t − λ)^{2}(t − μ) if λ ≠ μ, and m_{A_{4}}(t) = (t − λ)^{2} if λ = μ.

4. 7 marks, one for each case. By definition m_{A}(t) is a non-constant polynomial with real coefficients and leading coefficient 1, dividing t^{4} − 1, thus we have the following possibilities: m_{1}(t) = t − 1, m_{2}(t) = t + 1, m_{3}(t) = t^{2} + 1, m_{4} = t^{2} − 1, m_{5}(t) = (t − 1)(t^{2} + 1), m_{6}(t) = (t + 1)(t^{2} + 1), m_{7}(t) = (t − 1)(t + 1)(t^{2} + 1). Examples of square matrices of size 4 with such minimal polynomials: A_{1} = Id, A_{2} = −Id, A_{3} is the direct sum of two

copies of M := ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ), A_{4} = diag(−1, 1, 1, 1), A_{5} is the direct sum of

Id of size 2 and M, A_{6} is the direct sum of −Id of size 2 and M, A_{7} is the direct sum of diag(−1, 1) and M.

Question: what are possible sizes of a real matrix A such that m_{A}(t) = m_{i}(t), i = 1, …, 7?