I think this is false. Let $A = \begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix}$ be a 2x2 matrix, in the canonical basis of $\mathbb R^2$. Then A has a double eigenvalue b=1. If $v=\begin{pmatrix}1\\1\end{pmatrix}$, then $\langle v, Av \rangle < 0$.
The point is that the matrix can have all its eigenvalues strictly positive, but it does not follow that it is positive definite.
Your source talks about matrices over the complex numbers and $M$ is supposed to be Hermitian, that is, equal to its “conjugate transpose”.
I'll denote by $M^*$ the conjugate transpose of the (generic) matrix $M$. A couple of definitions; let $M$ be a square matrix:
The source also talks about the spectral theorem, which says that every normal matrix $M$ can be written as
$$
M=\sum_{k=1}^r \lambda_k P_k\tag{*}
$$
where $\lambda_1,\dots,\lambda_r$ are the distinct eigenvalues of $M$ and $P_1,\dots,P_r$ are the projections matrices to the eigenspaces; in particular they satisfy $P_k^2=P_k$ and $P_k^*=P_k$. Moreover $P_kP_l$ is the null matrix when $k\ne l$.
Clearly any Hermitian matrix is normal and it's easy to deduce that if $M$ is Hermitian then its eigenvalues are real: indeed, using its spectral decomposition (*), we have, from $M=M^*$,
$$
\sum_{k=1}^r \lambda_kP_k=\sum_{k=1}^r \lambda_k^*P_k
$$
and upon multiplying this by $P_l$ we get $\lambda_l=\lambda_l^*$.
The statement that any Hermitian matrix has real eigenvalues can be proved also without the spectral theorem, by directly using the definition: let $v$ be an eigenvector relative to the eigenvalue $\lambda$; then
$$
\lambda(v^*v)=v^*(\lambda v)=v^*Av=v^*A^*v=(Av)^*v=(\lambda v)^*v=\lambda^*v^*v
$$
Since $v^*v\ne0$, we get $\lambda=\lambda^*$.
Another important fact about normal matrices is that they can be diagonalized with a unitary matrix; this is actually an equivalent condition.
A square matrix $M$ is normal if and only if there exists a unitary matrix $U$ (that is, $U^{-1}=U^*$) such that $D=U^{-1}MU$ is diagonal.
Now, suppose $M$ is (Hermitian) and positive definite, that is, for $v\ne0$, $v^*Mv>0$. Consider $e_k$, the $k$-th vector of the canonical basis and consider $v_k=Ue_k$; then
$$
0<v_k^*Mv_k=e_k^*U^*UDU^*Ue_k=e_k^*De_k
$$
and, clearly, $e_k^*De_k$ is the entry at place $(k,k)$ in $D$. Since $D$ has on its diagonal the eigenvalues of $M$, we are done.
Conversely, if $D$ has all its diagonal entries positive, then, for every $v\ne0$ we have
$$
v^*Mv=(Uv)^*D(Uv)>0
$$
because clearly $D$ is positive definite.
Best Answer
Assume that for all $k=1,\ldots,n$, $\mathrm{tr}(A^k) = 0$ where $A$ is a $n\times n$ matrix.
We consider the eigenvalues in $\mathbb C$.
Suppose $A$ is not nilpotent, so $A$ has some non-zero eigenvalues $\lambda_1,\ldots,\lambda_r$.
Let $n_i$ the multiplicity of $\lambda_i$ then $$\left\{\begin{array}{ccc}n_1\lambda_1+\cdots+n_r\lambda_r&=&0 \\ \vdots & & \vdots \\ n_1\lambda_1^r+\cdots+n_r\lambda_r^r&=&0\end{array}\right.$$ So we have $$\left(\begin{array}{cccc}\lambda_1&\lambda_2&\cdots&\lambda_r\\\lambda_1^2 & \lambda_2^2 & \cdots & \lambda_r^2 \\ \vdots & \vdots & \vdots & \vdots \\ \lambda_1^r & \lambda_2^r & \cdots & \lambda_r^r\end{array}\right)\left(\begin{array}{c}n_1 \\ n_2 \\ \vdots \\ n_r \end{array}\right)=\left(\begin{array}{c}0 \\ 0\\ \vdots \\ 0\end{array}\right)$$ But $$\mathrm{det}\left(\begin{array}{cccc}\lambda_1&\lambda_2&\cdots&\lambda_r\\\lambda_1^2 & \lambda_2^2 & \cdots & \lambda_r^2 \\ \vdots & \vdots & \vdots & \vdots \\ \lambda_1^r & \lambda_2^r & \cdots & \lambda_r^r\end{array}\right)=\lambda_1\cdots\lambda_r\,\mathrm{det}\left(\begin{array}{cccc} 1 & 1 & \cdots & 1 \\ \lambda_1&\lambda_2&\cdots&\lambda_r\\\lambda_1^2 & \lambda_2^2 & \cdots & \lambda_r^2 \\ \vdots & \vdots & \vdots & \vdots \\ \lambda_1^{r-1} & \lambda_2^{r-1} & \cdots & \lambda_r^{r-1}\end{array}\right)\neq 0$$ (Vandermonde)
So the system has a unique solution which is $n_1=\ldots=n_r=0$. Contradiction.