Showing: $G(v_i, v_j) = 0$ when $i \neq j$ and $v_k$ are eigenvalues

Question

Suppose $G : V^2 \to \mathbb F$ is a bilinear form and $L : V \to V$ is a linear map. Furthermore, suppose:

1. $V$ is a finite-dimensional real vector space;
2. $G$ is symmetric positive-definite, i.e. $G(v, v) > 0$ for nonzero $v \in V$; and
3. $L$ is self-adjoint with respect to $G$, i.e. $G(v, Lw) = G(Lv, w)$.

I want to show that there is some basis $\{v_i\}$ of eigenvectors for $L$ so that $G(v_i, v_j) = 0$ when $i \neq j$. I have only been able to show this for a specific case:

Suppose $\{v_i\}$ are eigenvectors for $L$. Let's then say that $Lv_i = \lambda_i v_i$ for scalars $\lambda_i$. Then, $$ G(v_i, Lv_j) = G(Lv_i, v_j) , $$ so $$ G(v_i, \lambda_j v_j) = G(\lambda_i v_i, v_j) , $$ and finally $$ \lambda_j G(v_i, v_j) = \lambda_i G(v_i, v_j) . $$ Now if $\lambda_i \neq \lambda_j$, then yes, we can conclude that $G(v_i, v_j) = 0$. However, I cannot claim this when $\lambda_i = \lambda_j$.

Any hints or solutions?

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n.b. In the original question there is the phrase: "Show that $G$ has a basis $\{v_i\}$ of eigenvectors for $L$..." I have taken this to mean $\{v_i\}$ is a basis for $V$, but please correct me if I am wrong.