Diagonal and Similar Matrices

We know that $1$ is an eigenvalue with geometric multiplicity $r$. This means there exists $r$ linear independent eigenvectors $v_1,v_2, \dots, v_r$ for the eigenvalue $\lambda=1$, i.e. \[Av_i=v_i, \ \ \ \ i=1,2, \dots r.\] On the other hand we know $rank (A)=r$. By the Rank-Nullity Theorem (https://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem) we have \[\text{Rank}(A)+\text{Nullity}(A)=n \Rightarrow \text{Nullity}(A)=n-r. \] Note that by definition, $\text{Nullity}(A)$ is the dimension of the Null space of A: \[\text{Null}(A)=\{x\in \mathbb{R}^n: Ax=0\},\] which we know is an $(n-r)$ dimensional space. So there exists $n-r$ linearly independent vectors $w_1,w_2, \dots w_{n-r}$ that form a basis for the Null($A$). Note that each $w_i$ is an eigenvector for the eigenvalue $\lambda=0$, i.e. \[Aw_i=0, \ \ i=1,2, \dots n-r.\] Since, in general, the eigenvectors corresponding to distinct eigenvalues are linearly independent, the set of eigenvectors \[\{ v_1,v_1, \dots v_r, w_1, w_2, \dots w_{n-r} \}\] contains exactly $n  (=r+n-r)$ linearly independent vectors. Hence by Diagonalization Theorem, $A$ is diagonalizable.

Diagonalization Theorem. An n × n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.