Linear algebra and matrices M3P9

Question Sheet 4 - Answers and solutions

1. 3 marks. By linearity it is enough to prove this for f(t) = t^{m}, for all m = 0, 1, 2, … Since J_{λ,n} = λ.Id + J_{0,n}, we have

J_{λ,n}^{m} = λ^{m}.Id + mλ^{m−1}.J_{0,n} + m^{2}/2.λ^{m−2}.J_{0,n}^{2} + …

Now (as observed in the lectures) the k-th power of J_{0,n} is the matrix all of whose entries are zeros except for j = i + k, when the entries equal 1. This proves what we need.

2. (1) 4 marks. Let us find the Jordan normal form of A. The only eigenvalue is c, hence the Jordan blocks are the J_{c,n_{i}}, where ∑_{i} n_{i} = n. If A is diagonalizable, its Jordan normal form is diagonal, that is, all n_{i} = 1. Then A is similar to c.Id, but P^{−1}(c.Id)P = cId for any invertible P, hence A must equal c.Id. So the answer is: A is diagonalizable iff A is diagonal.

(2) 4 marks. The answer is: “A is diagonalizable iff a = 0.” Indeed, A is diagonalizable iff its Jordan normal form is diagonal. Since f_{A}(t) = (t − λ)^{2}(t − μ) there are two possibilities for the Jordan normal form: the diagonal matrix with λ, λ, μ on the main diagonal, or the direct sum of the Jordan block J_{λ,2} and the one-element matrix μ. In the first case m_{A}(t) = (t − λ)(t − μ), whereas in the second case m_{A}(t) = (t − λ)^{2}(t − μ). Therefore, A is diagonalizable iff (A − λ.Id)(A − μ.Id) = 0. This computation gives a = 0.

3. 3 marks. The Jordan normal form of A is the direct sum of J_{3,2} and J_{1,1}. Indeed, f_{A}(t) = (t − 3)^{2}(t − 1), and as in the previous question there are two possibilities for the Jordan normal form. We exclude the diagonal matrix by checking that (t − 3)(t − 1) does not kill A, exactly in the same way as before.

3 marks. The Jordan normal form of B is the direct sum of two copies of J_{0,2}. One checks that 1 is the only eigenvalue of B, hence it makes sense to replace A by A′ = A − Id. It is actually easy to find that A^{′2} = 0, hence the sizes of Jordan blocks of A′ do not exceed 2. It is easy to see that rk(A′) = 2, and since the rank is invariant, there must be two blocks of size 2 (and no blocks of size 1 otherwise the rank is less than 2).

4. 3 marks.

First proof: show that −tr(A) is the coefficient by t^{n} in the characteristic polynomial of A, and use the fact that f_{A}(t) only depoends on the similarity class of A.

Second proof: let B = P^{−1}A, then we need to prove that tr(BP) = tr(P B). By this can be checked by a direct computation use the formula for the matrix coefficients of the product of two matrices. Both traces equal ∑_{i,j=1}^{n} b_{ij}p_{ji}.