It sounds to me like you have heard, and are trying to remember the statement of, the following fact about the *Hodge Laplacian* $d^*d+dd^*$ on forms.

Slogan: eigenforms come in pairs.

Sketch proof: If $(d^*d+dd^*)\eta=-\lambda\eta$, then $(d^*d+dd^*)(d\eta)=dd^*d\eta=d(d^*d+dd^*)\eta=-\lambda d\eta$.

Statement: Recall the Hodge decomposition $\Gamma(M,\Lambda^kM)=A_k\oplus B_k\oplus H_k=\operatorname{im}(d^*_{k+1})\oplus\operatorname{im}(d_{k-1})\oplus (\operatorname{ker}(d_k^*)\cap\operatorname{ker}(d_k))$, such that for all $k$, $d_k:A_k\to B_{k+1}$ is an isomorphism. Then for each $\lambda$, $d_k$ sends the $\lambda$-eigenspace of $A_k$ to the $\lambda$-eigenspace of $B_{k+1}$.

Special case: Each nonzero eigenvalue of the Laplace operator on functions is also an eigenvalue of the Laplace operator on 1-forms.

Also, if you take any sensible choice whatsoever of Laplacian on $\Gamma(M,\Lambda^kM)$, then the asymptotic distribution (à la Weyl) of the eigenvalues will be the same as the asymptotic distribution of the eigenvalues of the Laplacian on scalars (up to multiplication by the rank of $\Lambda^kM$). See, eg, Theorem 2.41 of Berline-Getzler-Vergne.