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
$\def\id{\mathrm{id}}$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.