Linear algebra and matrices M3P9

Question Sheet 1 - Answers and solutions

1. (1) P = ( \begin{matrix}x^{−1} & 0 \\ 0 & 1 \end{matrix} )

(2) Id is not similar to any other matrix since P^{−1}P = Id.

(3) P = ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} )

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

2. t^{2} − 2at + (a^{2} + b^{2}), t^{4}, t^{4}, t^{4}. In the first case there are no real eigenvectors unless b = 0; if b = 0 then any non-zero vector is an eigenvector. The second matrix, call it A_{2}: the only eigenvector (up to multiplication by a constant factor) is e_{1}. For the third matrix, say A_{3}, every eigenvector is proportional either to e_{1} or to e_{3}; for the fourth matrix, say A_{4}, we have e_{1} and e_{4}.

3. A_{2} is similar neither to A_{3} nor to A_{4} because it has only one eigenvector (up to proportionality) whereas the other two matrices have two. We have A_{3}^{2} = 0, whereas A_{4}^{2} ≠ 0, hence A_{3} is not similar to A_{4}.

4. Every real polynomial of degree 3 has a real root.

5. In Exercise 1 all matrices except Id have either one eigenvector (up to proportionality), hence are not diagonalizable, or have two different eigenvalues, hence are diagonalizable. In Exercise 2 the first matrix has no eigenvalues (over the reals), and hence is not diagonalizable, unless b = 0 when it is already diagonal. The other matrices are not diagonalizable because they have not enough eigenvectors for this. (Recall that A is diagonalizable iff there exists a basis of the vector space V consisiting of eigenvectors of A.)