Is Every Diagonalizable Endomorphism Defined by a Simple Polynomial?

[geoffrey159]

Hello, I am studying reduction of endomorphisms and I came across a theorem that I can't understand completely. It states that:

Theorem: Let $E$ be a finite dimensional $K$ vector space, $K$ sub-field of $\mathbb{C}$, and $f$ be an endomorphism of $E$. Then, $f$ is diagonalizable if and only if there exists a polynomial $P$ of $K[X]$, such that $P(f) = 0$, and $P$ can be written as a product of polynomials of degree 1, and its roots have order of multiplicity 1.

I understand the proof I have for the sufficient condition, but the proof for the necessary condition is hard to follow for me so I tried an alternate way. I would like you to tell me if this is correct please:

$\Leftarrow$ ) Assume that there exists distinct $\lambda_i$'s for $i = 1...p$, and a polynomial $P$ in the form $P = a \prod_{i=1}^p (X - \lambda_i)$ such that $P(f) = 0$.
I want to show that $E = \bigoplus_{\lambda \in \text{Sp}(f) } E_{f,\lambda}$, which is a necessary and sufficient condition of diagonalizability.
We have that for any $x\in E-\{0\}$, $P(f)(x) = 0$. So there exists $i \in \{ 1...p \}$ such that $f(x) = \lambda_i x$, and $x$ belongs to the eigenspace $E_{f,\lambda_i}$. Therefore $x\in \bigoplus_{\lambda \in \text{Sp}(f) } E_{f,\lambda}$, and $E \subset \bigoplus_{\lambda \in \text{Sp}(f) } E_{f,\lambda}$. The other inclusion is trivial. So $f$ diagonalizable.

 

[Samy_A]

We have that for any $x\in E-\{0\}$, $P(f)(x) = 0$. So there exists $i \in \{ 1...p \}$ such that $f(x) = \lambda_i x$, and $x$ belongs to the eigenspace $E_{f,\lambda_i}$.

I may be missing the obvious, but I don't understand this step. How can it be that each element of E is an eigenvector?

 

[geoffrey159]

If $P(f) = 0$, then $P(f)(x) = 0$ for all $x\in E$.

1. If $x = 0$ then $x \in E \cap \bigoplus_{\lambda \in \text{Sp}(f) } E_{f,\lambda}$
2. If $x\neq 0$, then $0 = P(f)(x) = a\prod_{i = 1}^p (f(x) - \lambda_i x)$. So at least one term in the product is equal to 0. Therefore, there exists $i$ such that $f(x) - \lambda_i x = 0$, which means that $x$ belongs to the eigenspace associated to $f$ and eigenvalue $\lambda_i$.

 

[Samy_A]

I'm very confused, and probably wasting your time, but let us take a trivial example.
$E={\mathbb C}²$, and $f$ the endomorphism represented (with the canonical basis) by the matrix $\begin{pmatrix}1 & 0 \\ 0 & 2 \end{pmatrix}$.
So $f(a,b)=(a,2b)$, and $P(X)=(X-1)(X-2)$.
Take $x=(1,1)$. $f(x)=(1,2)$, so $x$ is not an eigenvector of f.
But $P(f(x))=f(f(x))-3f(x)+2x=f(1,2)-3(1,2)+2(1,1)=(1,4)-(3,6)+(2,2)=(0,0)$.

 

[geoffrey159]

Oooops, sorry it's me who's wasting your time, I realize I am completely confused with the notations. I need to re-do this. Sorry

 

[geoffrey159]

proofreading

 

[Samy_A]

Developping the polynomial $Q(f)$ as a sum, and setting $n = \text{Card(Sp}(f))$, we see that for any vector $x\in E$, the family $(x,f(x),f^2(x),...,f^n(x))$ is linearly dependent in $E$. Therefore $\text{dim}(E) \le n$.

I'm somewhat troubled by this step, but maybe I'm wrong about the notation.
If I understood it correctly, ${Sp}(f)$ is the set of all eigenvalues of $f$.

If that is indeed the case, $n = \text{Card(Sp}(f))$ is the number of eigenvalues of $f$. You claim that $\text{dim}(E) \le n$. But that can't be true in general, because an endomorphism can have less distinct eigenvalues than the dimension of the vector space, and still be diagonalizable.

I also don't immediately see why the linearly dependence of the family $(x,f(x),f^2(x),...,f^n(x))$ for all x implies that $\text{dim}(E) \le n$.

 

[geoffrey159]

Lol, this is exactly the reason why I deleted my post :-) I need to work more on this. Thank you for your help

 

[Samy_A]

Lol, this is exactly the reason why I deleted my post :-) I need to work more on this. Thank you for your help

Ok, no problem. :)

 

[geoffrey159]

But I'm not giving up on this, I want to find the solution :-)

 

[geoffrey159]

Think I have it now.

In a previous post, we said that $P(f) = 0$ implies that the eigenvalues of $f$ are among the zeros of $P$.
Then it had to be true that $P(f) = 0 \iff Q(f) = 0$, where $Q = \prod_{\lambda \in \text{Sp}(f)} (X - \lambda)$.

Now I want to show that $E = \bigoplus_{\lambda\in\text{Sp}(f)} \text{Ker}(f-\lambda e)$. The inclusion $\supset$ is trivial, and we now want to show $\subset$. We need to prove that any $x\in E$ has a decomposition over the eigenspaces of $f$ in the form $x = x_1 + ... + x_n$, where $x_i \in \text{Ker}(f-\lambda_i e)$.

That would be done if we could find $n$ polynomials $Q_k$ such that the endomorphism $Q_k(f)$ sends any $x\in E$ in the $k$-th eigenspace and :
$0 = Q(f) = e - \sum_{k=1}^n Q_k(f)$.

So the following constraints must be satisfied:

1. $Q_k(f) \in \text{Ker}(f-\lambda_k e) \iff f(Q_k(f)) = \lambda_k Q_k(f) \iff X(f) \circ Q_k(f) = (\lambda_k e)(f)\circ Q_k(f) \iff ((X-\lambda_k ) Q_k)(f) =0$
2. Existence of a non-zero constant $\beta$ such that $Q = \beta \ (1 - \sum_{k=1}^n Q_k)$, and the $Q_k$'s have the same degree as $Q$.

So if $Q_k$ has the form $Q_k = \alpha (X-\mu) \prod_{i\neq k} (X-\lambda_i)$, where $\alpha,\mu$ are constants to be determined, then the first constraint is satisfied, and $Q_k$ has the same degree as $Q$. For the second constraint, we must satisfy that $1 - \sum_{k=1}^n Q_k$ has the same roots as $Q$. So it seems logic to determine $\alpha,\mu$ such that $Q_k(\lambda_i) = \delta_{ik} \iff \alpha = \frac{1}{(\lambda_k-\mu) \prod_{i\neq k} (\lambda_k - \lambda_i)}$ and $\mu$ is not an eigenvalue of $f$.

Finally, there exists a constant $\beta \neq 0$ such that $Q = \beta (1 - \sum_{k=1}^n \frac{X-\mu}{\lambda_k -\mu} \prod_{i\neq k} \frac{ (X-\lambda_i)}{ (\lambda_k - \lambda_i)})$, and for any $x\in E$, $P(f)(x) = 0 \iff Q(f)(x) = 0 \Rightarrow x \in \bigoplus_{\lambda\in \text{Sp}(f)} \text{Ker}(f-\lambda e)$, which proves that $f$ is diagonolizable.

Is it correct now ?

 

[Samy_A]

Looks correct to me. I'm a little lost in your point 1) concerning the polynomials $Q_k$, but the formula for these polynomials fits the bill.

 

[geoffrey159]

Finally! That made me sweat ;-) Thank you very much Samy for your patience