1

I have consulted some answers on SE already, such as these two:

If $A^m=I$ then A is Diagonalizable

Show that if $A^{n}=I$ then $A$ is diagonalizable.

However, I am still confused. I understand that if we can show that $p_A (\lambda) = \lambda^n - 1$, then setting it to zero, we will have $(\lambda - 1)(\lambda^{n-1} + \cdots)=0$, so we have $n$ distinct roots. (or even simpler, root of unity, so it follows)

But how do we get $p_A (\lambda) = \lambda^n - 1$ in the first place? Cayley-Hamilton says that $A$ should satisfy its own characteristic polynomial. But in this case we don't have the characteristic polynomial to begin with?

Community
  • 1
3x89g2
  • 7,542
  • The characteristic polynomial of $A$ does not necessarily have pairwise different roots; take for example $A = I$. The claim is that the minimal polynomial of $A$ has pairwise different roots (which is equivalent to $A$ being diagonalizable). – Jendrik Stelzner Oct 03 '16 at 17:39
  • The minimal polynomial is not necessarily $\lambda^n-1$ but it divides it (otherwise some combination of the two would result in a polynomial that also kills $A$ and has lower degree, contradicting minimality of the minimal polynomial). And if $\lambda^n-1$ has distinct roots, so has any of its factors. – Peter Franek Oct 03 '16 at 17:45
  • 1
    @PeterFranek So this problem has nothing to do with Cayley-Hamilton? – 3x89g2 Oct 03 '16 at 17:59
  • @Misakov Right, I don't think it is Cayley-Hamilton. It is a combination of the fact that $\lambda^n-1$ kills $A$ and hence the minimal polynomial is a factor of $\lambda^n-1$. – Peter Franek Oct 03 '16 at 18:02
  • @Misakov You can see it also directly from jordan forms: the eigenvalues are only one's and if you get a nontrivial block $(1,1;0,1)$ or anything like that, then no power of that will ever result in a diagonal block. – Peter Franek Oct 03 '16 at 18:04

2 Answers2

2

To show that $A$ is diagonalizable, it's sufficient to show that its minimal polynomial has distinct roots. Since $A$ satisfies $A^m-I$, its minimal polynomial must divide $\lambda^m-1$, which has distinct roots. So the minimal polynomial has distinct roots, meaning that $A$ is diagonalizable.

For a proof of the first sentence, see for example Minimal polynomial and diagonalizable matrix

Community
  • 1
211792
  • 2,656
0

Here is a different argument that does not use characteristic or minimal polynomials.

We have that $A$ is similar to its Jordan form, i.e., $A=SJS^{-1}$. Then $$ J^m=(S^{-1}JS)^m=A^m=I. $$

Now, to have $J^m=I$, we need that each Jordan block $J_k(\lambda)$ satisfies $J_k(\lambda)^m=I_k$. This can only happen when $k=1$ (see below), which forces $J$ to be diagonal, and so $A$ is diagonalizable.

Here is the proof that $J_k(\lambda)^m\ne I$ if $k\geq2$. The case $\lambda=0$ is trivial because the diagonal entries will always be zero. When $\lambda\ne0$, we can prove by induction on $m$ that $[J_k(\lambda)^m]_{12}$ (the $1,2$ entry of $J_k(\lambda)^m$) is $m\lambda^{m-1}$. When $m=1$, we have $J_k(\lambda)_{12}=1=m\lambda^{m-1}$. Assume as inductive hypothesis that $[J_k(\lambda)^m]_{12}=m\lambda^{m-1}$ and $[J_k(\lambda)^m]_{rr}=\lambda^m$ for all $r$. Then \begin{align} [J_k(\lambda)^{m+1}]_{12}&=[J_k(\lambda)\,J_k(\lambda)^{m}]_{12} =\sum_{r=1}^n[J_k(\lambda)]_{1r}\,[J_k(\lambda)^m]_{r2}\\ \ \\ &=[J_k(\lambda)]_{11}\,[J_k(\lambda)^m]_{12}+[J_k(\lambda)]_{12}\,[J_k(\lambda)^m]_{22}\\ \ \\ &=\lambda\,m\lambda^{m-1}+1\,\lambda^m=(m+1)\lambda^m. \end{align}

Martin Argerami
  • 205,756