Am reading about kernel PCA. The set-up is a set of centered observations $\big\{x_k: k = 1,..., M\big\} \subset \mathbb{R}^N$ and we want to diagonalize the covariance matrix :
$$C = \frac{1}{M} \sum_{j=1}^M x_j x_j' $$
The text am following reads :
To do this, one has to solve the eigenvalue equation $$\lambda v = Cv$$ for eigenvalues $λ ≥ 0$ and $v ∈ \mathbb{R}^N\setminus{0}$. Since $Cv = \frac{1}{M} \sum_{j=1}^M \langle x_j, v \rangle x_j $, all solutions $v$ with $λ \neq 0$ must lie in the span of $x_1,..., x_M$. In that case the previous equation is equivalent to
$$λ\langle x_k, v\rangle = \langle x_k, Cv \rangle $$for all $k = 1,..., M$.
Question: Why is this last set of equations equivalent to the eigenvalue equation?
Clearly $\lambda v = Cv$ implies $$λ\langle x_k, v\rangle = \langle x_k, Cv \rangle $$for all $k = 1,..., M$ but how can I show the converse?
Any help is greatly appreciated.
I will assume that the vectors $x_1, \dots, x_M$ are linearly independent. Given that $\lambda \langle x_k, v \rangle = \langle x_k, Cv \rangle$ for all integers $1 \leq k \leq M,$ we have that $\langle x_k, \lambda v \rangle = \langle x_k, Cv \rangle,$ from which it follows that $\langle x_k, Cv - \lambda v \rangle = 0$ for all integers $1 \leq k \leq M.$ (Both equalities here hold because our inner product is taken over $\mathbb R.$)
Considering that the inner product $\langle \cdot, Cv - \lambda v \rangle$ is a linear functional, it follows that $\langle \cdot, Cv - \lambda v \rangle$ is uniquely determined by the values that it takes on a basis. Consequently, if we restrict $\langle \cdot, Cv - \lambda v \rangle$ to $\operatorname{span}_{\mathbb R}\{x_1, \dots, x_M \},$ then in view of the fact that $\langle x_k, Cv - \lambda v \rangle = 0$ for all integers $1 \leq k \leq M,$ it follows that $\langle \cdot, Cv - \lambda v \rangle$ is the zero functional on $\operatorname{span}_{\mathbb R}\{x_1, \dots, x_M \}$ so that $\langle Cv - \lambda v, Cv - \lambda v \rangle = 0.$ But then, considering that the inner product is positive-definite, we must have that $Cv = \lambda v,$ as desired.