Linear algebra and matrices M3P9

Question Sheet 6 - Answers and solutions

1. (1) False. Counterexample: A = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ), B = ( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ).

(2) False. Counterexample: A = B = ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ).

(3) True. Proof: (AB − BA)^{t} = (AB)^{t} − (BA)^{t} = B^{t}A^{t} − A^{t}B^{t} = (−B)(−A) − (−A)(−B) = −(AB − BA).

2. The eigenvalues are λ_{1} = 3 + \sqrt{2}, λ_{2} = 3 − \sqrt{2}. The corresponding eigenvectors are (1, 1 + \sqrt{2})^{t} and (1, 1 − \sqrt{2})^{t}. The length of the first one is (4 + 2\sqrt{2})^{1/2}, and the length of the second one is (4 − 2\sqrt{2})^{1/2}. The matrix P whose columns are

((4+2\sqrt{2})^{−1/2}, (1+\sqrt{2})(4+2\sqrt{2})^{−1/2})^{t} and ((4−2\sqrt{2})^{−1/2}, (1−\sqrt{2})(4−2\sqrt{2})^{−1/2})^{t}

is orthogonal, and

P^{t}A_{1}P = P^{−1}A_{1}P = ( \begin{matrix}3 + \sqrt{2} & 0 \\ 0 & 3 − \sqrt{2} \end{matrix} ) .

Let us observe that A_{2} + Id is an all-one matrix, whose eigenvalues are easily seen to be 5 with eigenvector (1, 1, 1, 1, 1)^{t}, and 0 with eigenspace of dimension 4 generated by e_{i} − e_{j}, i ≠ j. Hence the same vectors are eigenvectors of A_{2} with eigenvalues 4 and -1, respectively. To construct an orthonormal basis take f_{1} = 5^{−1/2}(1, 1, 1, 1, 1)^{t}, f_{2} = 2^{−1/2}(1, −1, 0, 0, 0), f_{3} = 6^{−1/2}(1, 1, −2, 0, 0), f_{4} = 12^{−1/2}(1, 1, 1, −3, 0), f_{5} = 20^{−1/2}(1, 1, 1, 1, −4). Let Q be the matrix with such columns, then

Q^{t}A_{2}Q = Q^{−1}A_{2}Q = ( \begin{matrix}4 & 0 & 0 & 0 & 0 \\ 0 & −1 & 0 & 0 & 0 \\ 0 & 0 & −1 & 0 & 0 \\ 0 & 0 & 0 & −1 & 0 \\ 0 & 0 & 0 & 0 & −1 \end{matrix} ) .

3. We denote complex conjugation by the bar. Let z = (z_{1}, …, z_{n})^{t}, z_{i} ∈ C, be the complex eigenvector, that is, a non-zero vector with complex coordinates such that Az = λ.z. Then \overline{z}^{t}.A.z = λ.\overline{z}^{t}.z = λ ∑_{i=1}^{n} |z_{i}|^{2}. Since A is real we have \overline{A} = A, we also have A^{t} = −A. Hence the same expression equals \overline{(A^{t}.z)}^{t}.z = \overline{−(A.z)}^{t}.z = −\overline{λ}.\overline{z}^{t}.z = −\overline{λ} ∑_{i=1}^{n} |z_{i}|^{2}. Since ∑_{i=1}^{n} |z_{i}|^{2} ≠ 0, we conclude that λ = −\overline{λ}, which is what we had to prove.

4. In the same spirit we have < v_{1}, v_{2} >= v_{1}^{t}.(A.v_{2}) = λ_{2}.v_{1}^{t}.v_{2}, and this same expression equals (A.v_{1})^{t}.v_{2} = λ_{1}.v_{1}^{t}.v_{2}. Since λ_{1} ≠ λ_{2} we must have v_{1}^{t}.v_{2} = 0, hence, by the first formula, we must also have < v_{1}, v_{2} >= 0.

5. That the form is bilinear is obvious, and its symmetricity follows from the fact that tr(AB) = tr(BA) (sheet 4). The vector space of matrices has a natural basis consisting of E_{i,j} which has 1 at the place (i, j), and 0 elsewhere. Then the Gram matrix of the form tr(AB) in this basis is the direct sum of

n(n − 1)/2 blocks ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) (corresponding to E_{ij} and E_{ji} for i < j), and

the identity matrix of size n (corresponding to the subspace generated by the E_{ii}’s). The Gram matrix of the form tr(A)tr(B) is the direct sum of the zero matrix (corresponding to the subspace V_{1} generated by the E_{ij}, i ≠ j), and the all-one matrix (corresponding to the subspace V_{2} generated by the E_{ii}’s). We see that V = V_{1} ⊕ V_{2}, and V_{1} and V_{2} are orthogonal with respect to our form. Hence we can compute its rank as the sum of the ranks of its restrictions to V_{1} and V_{2}. The first one is zero if α = 0, and n(n − 1) otherwise. The second one is 0 if α = β = 0, 1 if α = 0, β ≠ 0, n if α ≠ 0, β = 0, in the remaining cases it equals n unless α + nβ = 0 when it equals n − 1.

The final answer is:

0 if α = β = 0,

1 if α = 0 and β ≠ 0,

in the remaining cases it equals n^{2} unless α + nβ = 0 when it equals n^{2} − 1.