Linear algebra and matrices M3P9

Question Sheet 5 - Answers and solutions

1. We observe that A is the direct sum of two blocks, hence we find

eigenvectors of each block. The eigenvectors of ( \begin{matrix}5 & 4 \\ −2 & −1 \end{matrix} ) are the col-

umn vectors (1, −1)^{t} and (2, −1)^{t}. If we let P = ( \begin{matrix}1 & 2 & 0 \\ −1 & −1 & 0 \\ 0 & 0 & 1 \end{matrix} ), then

P^{−1}AP = ( \begin{matrix}1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{matrix} ) and P^{−1}BP = ( \begin{matrix}2 & 0 & 0 \\ 0 & 5 & 3 \\ 0 & −6 & −4 \end{matrix} ). It remains to find

an invertible matrix of size 2 which diagonalizes ( \begin{matrix}5 & 3 \\ −6 & −4 \end{matrix} ). Since the

corresponding block of P^{−1}AP is a scalar matrix is will remain the same after conjugation by any invertible matrix, and both matrices will become

diagonal. The eigenvectors of ( \begin{matrix}5 & 3 \\ −6 & −4 \end{matrix} ) are (1, −1)^{t} and (2, −1)^{t}, hence

we need the following matrix: Q = ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & −1 & −2 \end{matrix} ). Set R = P Q, then

R^{−1}AR = ( \begin{matrix}1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{matrix} ) and R^{−1}BR = ( \begin{matrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & −1 \end{matrix} ).

2. One sees that < e_{1}, e_{1} >= 4 ≠ 0. Set f_{1} = (1, 0, 0). Then < f_{1}, (a, b, c)^{t} >= 4a+2b−c, hence < f_{1} >^{⊥} contains f_{2} = e_{1}−2e_{2}, < f_{2}, f_{2} >= −4. (By the way this is an example of a vector v such that < v, v >< 0.) Next, < f_{1}, f_{2} >^{⊥} is generated by (a, b, c) such that 4a+2b−c = 2b+5c = 0, for example, by f_{3} = (−3, 5, −2). The Gram matrix of <, > in the basis f_{1}, f_{2}, f_{3} is diagonal.

3. The most obvious method is to find a basis in which the Gram matrix is diagonal. Since the initial matrix is non-singular, the number of 1’s plus the number of -1’s equals 5. One can start for example from f_{1} = e_{1} + e_{2}. Then < f_{1}, f_{1} >= 2, and the orthogonal complement to the span of f_{1} is generated by e_{1} − e_{2}, e_{1} − e_{3}, e_{4}, e_{2} − e_{5}. Set f_{2} = e_{1} − e_{2}, then < f_{2}, f_{2} >= −2. We also find that the orthogonal complement to the subspace generated by f_{1} and f_{2} is generated by e_{1} − e_{3}, e_{4}, e_{2} − e_{5}. However, these vectors are orthogonal to themselves, and we cannot take either of them as a basis vector. Let f_{3} = e_{1} − e_{3} − e_{4}, then < f_{3}, f_{3} >= 2, and the orthogonal space to the subspace generated by f_{1}, f_{2}, f_{3} is generated by f_{4} = e_{1} − e_{3} + e_{4} and e_{2} − e_{5}. We have < f_{4}, f_{4} >= −2. Continuing in the same way one chooses f_{5} = e_{1} −e_{2} −e_{3} +e_{4} +e_{5}, then < f_{5}, f_{5} >= 2. Summing up, we have found a basis in which the Gram matrix is diagonal with 2, −2, 2, −2, 2 on the main diagonal. Hence the signature is 1.

4. One can use the diagonalization process as in the lectures, but there is a shorter proof based on the observation that Ker(A) is one-dimensional. Hence the rank of A is 4, and the canonical form is the direct sum of two

standard blocks ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ), and a zero matrix of size 1.