I know that this type of question has been asked on here before but I am still having a hard understanding what is going on. The text that I am learning from is "Linear Algebra Done Right by Sheldon Axler" and I don't think it covers this topic very well. Does anyone have any suggestions? I am studying for my linear comp at the end of January.
Given the matrix
$$A = \left(\begin{matrix}0&-1&-1\\-3&-1&-2\\7&5&6\end{matrix}\right).$$
Find the Jordan form $J$ and an invertible matrix $Q$ such that $A = QJQ^{-1}$.
I know that I want to start by finding the eigenvalues of this matrix. I ended up getting det$(A – \lambda I) = (\lambda + 2)(\lambda62 + 3\lambda – 2) \implies \lambda = -2, \frac{-3 -\sqrt{17}}{2}, \frac{\sqrt{17} – 3}{2}$.
Now, from what I have read on this website, I know now that I want to find the null space of $A – \lambda I$ for each eigenvalue. Then my invertible matrix $Q$ will have entries whose columns are these vectors. Why does one do this? Is there a "nicer" way to go about doing this other than direct computation?
This problem is from a previous linear comp at my university, so I can't imagine that they would have wanted me to the direct calculation.
Hopefully, my questions and explanation of the problem were clear. Thanks for any help in advance.
Best Answer
Eigenvalues are $2,2,1$, which simplifies the calculation a lot.