Consider all linear systems with exactly one eigenvalue equal to 0. Which of these systems are conjugate?

Question

This question considers non-hyperbolic matrices (2x2), which is not covered in our textbook. Basically the problem I think tries to get us to think about conjugacy patterns of these cases, but I am stuck.

If A= ((a, b), (c, d)) a 2x2 has 1 eigenvalue = 0, then I believe c or b must be 0.

So A = ((a, 0), (c, d)) or ((a, b), (0, d)), but I'm not sure where to go from there.

Why did you believe this? Counterexample $$ \boldsymbol A = \begin{bmatrix}1&1\1&1 \end{bmatrix}. $$ — xbh, Oct 30 '18 at 03:13

score 1 · Answer 1 · answered Nov 24 '18 at 18:32

1

This means that the matrix has rank 1 or one row is a multiple of another row and one row is non zero. Formally$$k_1(a,b)+k_2(c,d)=0$$and $$a^2+b^2+c^2+d^2\ne 0$$

answered Nov 24 '18 at 18:32

Mostafa Ayaz

31,924

Consider all linear systems with exactly one eigenvalue equal to 0. Which of these systems are conjugate?

1 Answers1