 Solution

A real number $\lambda$ is an eigenvalue of $A$ if there is a nonzero vector $v$ with $Av=\lambda v$; such a $v$ is an eigenvector for $\lambda$. The nonzero condition is essential.

For distinct eigenvalues ${\lambda }_1,\dots ,{\lambda }_r$, choose corresponding nonzero eigenvectors $v_1,\dots ,v_r$. Prove their linear independence by induction on $r$. The claim for one vector is immediate. Given a relation ${\sum }_{i=1}^r{a}_i{v}_i=0$, apply $A-{\lambda }_{r}I$ to getBy the induction hypothesis, every coefficient here is zero. Distinctness gives $a_i=0$ for $i<r$, and the original relation then gives $a_r=0$ since $v_r\ne 0$. Thus the eigenvectors are linearly independent.

With $n$ distinct real eigenvalues, the $n$ real eigenvectors form a basis of ${\mathbb{R}}^n$. Put them into the columns of $P$. Then $P$ is invertible andThis proves diagonalizability directly.