I was reading this pdf file about Cartan-Diedonne theorem and it was stated that the minimal number needed to obtain an isometry $f$ is $$\dim V − \dim\big(\ker (f − \text{id})\big).$$
Unfortunately, no prove has been added, though below it states that this remark will be proven later.
I would be glad, if someone provide me any links for that proof.
$$ \newcommand{\u}{ {\mathbf u}} \newcommand{\v}{ {\mathbf v}} $$
Here's a quick sketch of a proof:
Treat $V$ as $\Bbb R^n$ (or whatever field you're using) to make things concrete.
Call $f$ by the name $T$ (because I like it better).
Let $W$ be the subspace on which $T$ is the identity (i.e., ker$(T-I)$), and $dim W = n-k$. Pick a basis $\v_1, \ldots, \v_{n-k}$ for $W$.
Pick a basis $\u_1, \ldots, \u_k$ for $W^\perp$; let $$ S: \Bbb R^k \times R^{n-k} \to \Bbb R^n : (x_1, \ldots, x_k; y_1 \ldots y_{n-k}) \mapsto \sum_i x_i u\u_i + \sum_j y_i \v_i $$
$S$ is an isomorphism; let $H: R^n \to R^n : z \mapsto S^{-1} T Sz$. The matrix of $H$ with respect to the standard basis then has the form $$ \pmatrix{B & | & 0 \\ 0 & | & I} $$ So we can look at the isometry $Q$ on $\Bbb R^k$ defined by $x \mapsto Bx$, and know that $Q$ has no fixed points, and if we can write this using no more than $k$ reflections, we're done.
But that's exactly what the usual proof that every rotation (which is what $Q$ is, except perhaps for a reflection) can be written as a product of Givens rotations (search for "Givens" in this reference). There's a small detail --- the matrix $T$ might have determinant $-1$; in that case, you might need to add a reflection to one of the Givens rotations, but I leave that detail to you.