If $A$ is an $n$-by-$n$ matrix with complex entries, (i.e., $A\in M_n(\mathbb{C})$,) $A$ must have $n$ eigenvalues, counting algebraic multiples. But it is not always true that $A$ has $n$ linearly independent eigenvectors. So, what necessary and sufficient condition may be add, to ensure that $A$ has $n$ linearly independent eigenvectors?

Of course, the simpler the better.

I have another related question:

I can't figure out why, intuitively, that algebraic multiple doesn't mean more than one linearly independent eigenvectors. I mean in my tuition, a multiple appear because there is a subspace with dimension>1 being scaled "evenly" in every direction. If the multiple is 2, how come I may fail to find 2 linearly independent vectors in this subspace?

  • $\begingroup$ The necessary and sufficient condition is that $\mathbf{A}$ commutes with $\mathbf{A}^{\dagger}$. $\endgroup$ – chaohuang Sep 10 '12 at 4:44

Iff $\,A\,$ is diagonalizable iff the minimal polynomial of A is a product of different linear factors...

There you have two (equivalent, of course) conditions

  • $\begingroup$ Thank, you. You're right but these are the answers I know (or can find in textbook.) What I really want to know is that, WHAT exactly, is the intrinsic differences, as linear transformations, between a diagonalizable matrix and a non-diagonalizable matrix. $\endgroup$ – xzhu Sep 10 '12 at 3:20
  • $\begingroup$ I realize this doesn't best describe what I'm wondering, so I might as well start a new question. $\endgroup$ – xzhu Sep 10 '12 at 3:23
  • $\begingroup$ @Voldemort: the intrinsic difference is the presence of nontrivial Jordan blocks. See en.wikipedia.org/wiki/Jordan_normal_form . Another relevant keyword here is "semisimple module over $\mathbb{C}[x]$" (see en.wikipedia.org/wiki/Semisimple_module). $\endgroup$ – Qiaochu Yuan Sep 10 '12 at 4:24

For example, gcd$(f(X), f'(X)) = 1$, where $f(X)$ is the minimal polynomial of $A$. gcd$(f(X), f'(X))$ can be calculated by the Euclid's algorithm.

  • $\begingroup$ This is sufficient but not necessary. $\endgroup$ – Qiaochu Yuan Sep 10 '12 at 3:02
  • $\begingroup$ @QiaochuYuan I edited my answer. $\endgroup$ – Makoto Kato Sep 10 '12 at 3:28
  • $\begingroup$ Please clear the downvotes not for me but for readers who might think this answer is not correct because of them. $\endgroup$ – Makoto Kato Sep 10 '12 at 3:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.