Understanding proof that symmetric matrix is diagonalizable This text gives a short proof that a symmetric matrix is diagonalizable.

If the symmetric matrix $A$ is not diagonalizable then it must have
  generalized eigenvalues of order $2$ or higher. That is, for some repeated
  eigenvalue $λ_i$ there exists  $v\ne 0$ such that
  $$(A − λ_iI)^2
v = 0, \ \ (A − λ_iI)v \ne 0$$
But note that
$$0 = v^∗
(A − λ_iI)
^2v = v^∗
(A − λ_iI)(A − λ_iI)v \ne  0,$$
  which is a contradiction. Therefore, as there exists no generalized eigenvectors of order $2$ or higher, $A$ must be diagonalizable.

The text defines $v^*=\bar{v}^T$. I am having trouble understanding the proof starting from "But note that...".
In particular, 
(a).  I don't understand what role $v^*$ is playing here.
(b). I also don't understand how they know that $v^∗(A − λ_iI)(A − λ_iI)v \ne  0$ in order to make this contradiction. 
 A: To flesh out my comment: that text is just wrong because it fails to say real symmetric matrix. A similar result does hold for Hermitian (complex) matrices, of which real symmetric matrices are a special case (since complex conjugation does not change real matrices); one could prove the result for real symmetric matrices by deducing it from the more general result for Hermitian matrices (and this could be derived from an even more general result for normal matrices if one likes), but one needs to state clearly what one is doing.
So a proof for Hermitian matrices is possible, for instance as follows. Eigenvalues of such matrices must be real since if $v$ is an eigenvector for$~\lambda$ then
$$
  \lambda\left\|v\right\|=\langle v\mid Av\rangle=\langle Av\mid v\rangle
 =\overline\lambda\left\|v\right\|
$$
(wth the middle equality because $A$ is Hermitian; here $\langle\cdot\mid\cdot\rangle$ is the complex inner product, linear on the right),
and $\left\|v\right\|\neq0$. A linear operator on a complex vector space of finite nonzero dimension always has at least one eigenvalue (and hence eigenvectors), so if one can find a stable complementary subspace to the $1$-dimensional space generated by an eigenvector$~v$, the diagonalisability will follow by induction on the dimension (restricting to such a complement one gets a linear operator on a lower dimensional space). But one has this for a Hermitian operator: if $w$ is a vector orthogonal to such an eigenvector $v$, then
$$
  \langle v\mid Aw\rangle = \langle Av\mid w\rangle
= \overline\lambda\langle v\mid w\rangle = 0
$$
so $Aw$ is again orthogonal to $v$.
I prefer this to trying to rule out generalised eigenvectors that are not eigenvectors as you text tries to do (not very successfully, since there is at least a typo of a missing $v$, and even when corrected as in the question the argument is not very clear). Once one has eigenvectors for real eigenvalues, repeated restriction to the orthogonal complements will give a basis of eigenvectors explicitly; there is no need to use generalised eigenspaces at all.
