Jordan form and basis for 5 x 5 matrix

Question

Is anybody able to explain the solution to part c of this question above? I don't understand how the characteristic polynomial was determined, or how the basis was found.

Answer 1

First of all, from examining simple linear combinations of $A$’s columns, we can quickly see that its rank is at most two. In addition, its first two rows are obviously linearly independent, hence its rank is exactly two. So, we know that it has zero as an eigenvalue with geometric multiplicity $3$, and therefore that $x^3$ divides its characteristic polynomial.

Since the matrix is also traceless, instead of computing the characteristic polynomial directly as $\det(\lambda I-A)$, the solution makes a leap of faith (or uses insider knowledge) and guesses that $A$ might be nilpotent. Since we know that the geometric multiplicity of $0$ is three, this only requires checking at most $A^2$ and $A^3$. I’m not sure that this really saves that much work over the general method, though. As we see in the solution, $A^2\ne0$ but $A^3=0$, so the matrix is indeed nilpotent. Nilpotent matrices only have $0$ as an eigenvalue, which gives us the characteristic polynomial $x^5$, with minimal polynomial $x^3$. It looks like this text uses $\det(A-\lambda I)$ instead of $\det(\lambda I-A)$, which is why the book solution has $(-x)^5$ instead. The latter equivalent definition guarantees a monic polynomial for any order matrix.

The solution proceeds conventionally from there. There’s only one nontrivial Jordan block, so a simple-looking vector that’s not in $\ker(A^2)$ is chosen to start the chain. After that, the remaining two basis vectors are found by inspection. The first two columns of $A$ are equal, so $\mathbf e_1-\mathbf e_2$ is a null vector of $A$, as is $\mathbf e_5$ because the last column is zero. You should verify that the three vectors in the Jordan chain together with these two are linearly independent.