your argument also work for linear map...

$x(x-1)(x+1)$ is not the minimal polynomial. The minimal polynomial for $f$ has to be a divisor of $x(x-1)(x+1).$

Why is it applicable for linear maps also?

If your vector space is finite dimensional, then all linear maps can be represented by matrices

@SouravGhosh: that's not true. Take $T$ being the identity. Then $T$ is diagonalizable, however, $P(x)=(x-1)(x^2+x+1)$ is s.t. $p(I)=0$ but you can't express $p$ as a product of linear factor.

@SouravGhosh Never use "any" in mathematics. It is an ambiguous word in English meaning both "some" or "every", which are important to be distinguished in mathematics. They read it as "every", which gives a vacuous theorem, instead of the "some" that you intended.

@Surb Well. I have used "any" for "at least one" but it's not standard term( infact a bad term) as "any" means "forall". Ok correct form: if there is a $p(x) \in K[x]$..

$T\in\mathcal{L}(V) $ is diagonalizable if $\exists p\in \textrm{Ann}_T$ such that $p(x) =0$ with product of distinct linear factors.

@100xln2 Although you haven't defined it for diagonalizability for arbitrary linear maps, there is indeed a valid definition of "diagonalizable" that applies to linear maps.

The equality $f(f-I)(f+I)=0$ implies that the space is the direct sum of $\ker f,$ $\ker(f-I)$ and $\ker(f+I).$ Thus it is diagonalizable. No matrices are needed.

The statement in question is true only if the characteristic of the underlying field is not $2$.

Nobody's answering your question! A linear map from a f.d. vector space to itself is diagonalisable iff there's a basis s.t. the matrix of the map in this basis is diagonalisable. Equivalently: the matrix of the map wrt all bases is diagonalisable, or the matrix of the map is diagonal in some basis. If you haven't defined this in class that's bad of your teacher but you should've been able to come up with it yourself! In linear algebra often there's a property of matrices that's preserved under similarity and hence can be defined for maps. EG: determinant, trace, char poly, min poly, ...

@SouravGhosh you said that for the polynomial (x-1)(x^2+x+1)=0 the identity is a diagonizable solution but the polynom cant be expressed as a product of linear factors. This is true. But. This does only go one way: if the matrix is diagonizable =/=> there exists a polynomial which can be expressed as linear factors. BUT: if the polynomial of the matrix can be expressed in linear factor it indeed does imply that the matrix is diagonizable. Do i understand you correctly?

@100xln2 If you replace anhilating polynomial by a special anhilating polynomial ( minimal polynomial) then you get a necessary and sufficient condition.

There is a similar result for any linear map of finite order (in group theoretic terms): if $f^n=\rm{Id}$ for some $n\in\Bbb N$, $f$ will be diagonalisable.

@FShrike Not true. You need to be in a field of characteristic prime to $n$. Otherwise group theory would be a very dull subject indeed.

@ancientmathematician I didn’t mention the field, true, but I was assuming the ‘default’ ones for linear algebra, say $\Bbb R$ or $\Bbb C$.

@FShrike I know, I know. But I just think we've collectively made a bad choice in making this the default.