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

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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