0
$\begingroup$

I have the following well known theorem:

Let $V$ be a vector space over a field $K$, $\dim_KV=n$ and let $\phi:V\to V$ be an endomorphism. It is $\phi$ diagonalizable iff $\text{minpoly}_\phi(t)=\prod_{i=1}^m (t-\lambda_i)$ with $\lambda_i\not=\lambda_j$ for $i\not=j$.

I know this question could be a duplicate, but I have a concrete question about my try to prove this $"\Rightarrow "$ direction.

My try: $\phi$ is diagonalizable, therefore the characteristic polynomial of $\phi$ splits into linear factors: $\chi_{\phi}(t)=\prod_{i=1}^m (t-\lambda_i)^{k_i}$ with $\lambda_i\not=\lambda_j$ for $i\not=j$ and $k_i\ge 1$ and the geometric multiplicity of the $\lambda_i's$ is the same as the algebraic multiplicity. Because of $\phi$ is diagonalizable, there exists a basis of $V$,$(v_1,...,v_n),$ of eigenvectors of $\phi$ related to the eigenvalues $\lambda_i$. The minimal polynomial divides the characteristic polynomial, therefore it is $\text{minpoly}_\phi(t)=\prod_{i=1}^m (t-\lambda_i)^{t_i}$ with $t_i\le k_i$ for every i. Let $g(t)=\prod_{i=1}^m (t-\lambda_i)$. Claim: $g=\text{minpoly}_\phi$.

First of all $g$ divides the characteristic polynomial of $\phi$, therefore I have to show that $g(\phi)=0$, this means $g(\phi)(v)=0$ for every $v\in V$.

Let $v\in V, v=\sum\limits_{k=1}^n\mu_kv_k$. It is $\def\id{\mathrm{id}}g(\phi)(v)=\sum\limits_{k=1}^n\mu_kg(\phi)(v_k)=\sum\limits_{k=1}^n\mu_k\prod\limits_{i=1}^m (\phi-\lambda_i\id)(v_k)=\sum\limits_{k=1}^n\mu_k\prod\limits_{i=1}^m (\phi(v_k)-\lambda_i\id(v_k))=0$, because for an $i\in\{1,..,m\}$ it is $i=k$. This shows $\Rightarrow$ (i hope).

My question is: Is the equation $\sum\limits_{k=1}^n\mu_k\prod\limits_{i=1}^m (\phi-\lambda_i\id)(v_k)=\sum\limits_{k=1}^n\mu_k\prod\limits_{i=1}^m (\phi(v_k)-\lambda_i\id(v_k))$ true? I think yes, but I'm not sure.

Could you give me a proof or explain me, how to construct a basis of eigenvectors of this $\Leftarrow$ direction? Regards.

Edit: Is there a mistake in the formulation of the theorem (see comments below)?I have found a proof for the direction $\Leftarrow$: http://www.maths.ed.ac.uk/~tl/minimal.pdf

$\endgroup$
  • $\begingroup$ The identity matrix is a counterexample to this theorem. $\endgroup$ – Titus Apr 10 '15 at 6:40
  • 1
    $\begingroup$ hm, no? the identity matrix is diagonalizable and $\text{minpoly}_{id}(t)=(t-1)$? I don't see the mistake.. Maybe I have a mistake in the formulation $\endgroup$ – user229528 Apr 10 '15 at 6:46
  • $\begingroup$ Please don't type <br/> all the time, just leave a blank line to start a new paragraph. This is much more readable. $\endgroup$ – Marc van Leeuwen Apr 10 '15 at 7:12
  • $\begingroup$ sorry! I will do/change this in feature. $\endgroup$ – user229528 Apr 10 '15 at 7:17
  • $\begingroup$ I don't get the RHS of your equation. How do you multiply a bunch of vectors? $\endgroup$ – darij grinberg Apr 10 '15 at 7:25
0
$\begingroup$

The equation $$ \sum\limits_{k=1}^n\mu_k\prod\limits_{i=1}^m (\phi-\lambda_i\id)(v_k) =\sum\limits_{k=1}^n\mu_k\prod\limits_{i=1}^m (\phi(v_k)-\lambda_i\id(v_k))$$ that you are asking about is nonsensical, so it is neither true not false. The left hand side should be written $$ \sum\limits_{k=1}^n\mu_k\left(\left(\prod\limits_{i=1}^m (\phi-\lambda_i\id)\right)(v_k)\right) $$ to express what it means: a product (composition) of polynomials in $\phi$, applied to $v_k$, then multiplied by the scalar $\mu_k$ and summed. [The inner large parentheses are obligatory, since function application binds more strongly than the large operator $\prod$ does (to the right); the outer parentheses are just for readability, since function application also binds more strongly than multiplication.] Your error is maybe misreading your own formula, thinking that the function application is taking place in each factor of the product (which is in fact a function composition), essentially interpreting $(\phi_1\circ\cdots\circ \phi_m)(v_k)$ as a "product of vectors" $\phi_1(v_k)\phi_2(v_k)\ldots\phi_m(v_k)$ that makes no sense.

Instead you can argue that the expression is$~0$ because the polynomials in$~\phi$ commute, so that you can move the factor $\phi-\lambda_i\id$ for which $\lambda_i$ is the eigenvalue of $v_k$ to the right, so that it acts on $v_k$ first; the result is the zero vector, and then the other factors will leave it zero because they are linear operators.

$\endgroup$
  • $\begingroup$ Ok thank you. Do you mean I yould write $\sum\limits_{k=1}^n\mu_k\prod\limits_{ i=1}^m (\phi-\lambda_i\id)(v_k)=\sum\limits_{k=1}^n\mu_k\prod\limits_{ i=1\; i\not= k }^m (\phi-\lambda_i\id)(\phi-\lambda_k\id)(v_k)= \sum\limits_{k=1}^n\mu_k\prod\limits_{ i=1\; i\not= k }^m (\phi-\lambda_i\id)(\phi(v_k)-\lambda_k\id (v_k))=0$ $\endgroup$ – user229528 Apr 10 '15 at 8:33
  • $\begingroup$ I'm sorry.. I have problems with LaTeX and formel editor's in general, it is new for me.. I try my best to learn $\endgroup$ – user229528 Apr 10 '15 at 8:36
  • $\begingroup$ Yes, I think that is exactly what I meant. Don't worry too much about LaTeX in comments, it's horrible, and occasionally the system will actually break correct LaTeX by inserting a space (invisible to you) in a random place because it thinks that this is getting too long for a line (which bug you can circumvent by placing some spaces yourself in harmless places). By the way, you can type \neq instead of \not=. $\endgroup$ – Marc van Leeuwen Apr 10 '15 at 8:39
  • $\begingroup$ ok thank's. everything is clear now. I really need a break now from this question.. Later I will vote up your answers. I really appreciate your help, thank you! This helps a lot! $\endgroup$ – user229528 Apr 10 '15 at 8:51

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.