Let $X$ be a Kähler manifold and consider the dual operator $\Lambda$ of the Lefschetz operator $L$. Is the following a true statement?

A closed $(1,1)$-form $\eta$ is harmonic if and only if $\Lambda\eta = \text{constant}$.

This is supposedly proven in (https://people.math.harvard.edu/~siu/siu_reprints/dmv_book.pdf) at Remark (1.4) part (ii), but the author states in the proof that "because the product of two harmonic forms is harmonic" which certainly isn't always true. I'm wondering if this is a consequence of some of the Kähler identities, but could not come up with a complete proof. If $\eta$ is harmonic, then $\Delta\eta = 0$. Since $\Lambda$ commutes with $\Delta$ we have $$\Delta(\Lambda \eta)=0$$ i.e. $\Lambda \eta$ is harmonic. Since $\Lambda \eta$ is a $0$-form it must be constant. The converse seems harder.

Let $X$ be a Kähler manifold and consider the dual operator $\Lambda$ of the Lefschetz operator $L$. Is the following a true statement?

A closed $(1,1)$-form $\eta$ is harmonic if and only if $\Lambda\eta = \text{constant}$.

This is supposedly proven in Siu - Lectures on Hermitian–Einstein metrics for stable bundles and Kähler–Einstein metrics at Remark (1.4) part (ii), but the author states in the proof that "because the product of two harmonic forms is harmonic" which certainly isn't always true. I'm wondering if this is a consequence of some of the Kähler identities, but could not come up with a complete proof. If $\eta$ is harmonic, then $\Delta\eta = 0$. Since $\Lambda$ commutes with $\Delta$ we have $$\Delta(\Lambda \eta)=0$$ i.e. $\Lambda \eta$ is harmonic. Since $\Lambda \eta$ is a $0$-form it must be constant. The converse seems harder.

Let $X$ be a Kähler manifold and consider the dual operator $\Lambda$ of the Lefschetz operator $L$. Is the following a true statement?

A closed $(1,1)$-form $\eta$ is harmonic if and only if $\Lambda\eta = \text{constant}$.

This is supposedly proven in (https://people.math.harvard.edu/~siu/siu_reprints/dmv_book.pdf) at Remark (1.4) part (ii), but the author states in the proof that "because the product of two harmonic forms is harmonic" which certainly isn't always true. I'm wondering if this is a consequence of some of the Kähler identities, but could not come up with a proof.

Let $X$ be a Kähler manifold and consider the dual operator $\Lambda$ of the Lefschetz operator $L$. Is the following a true statement?

A closed $(1,1)$-form $\eta$ is harmonic if and only if $\Lambda\eta = \text{constant}$.

This is supposedly proven in (https://people.math.harvard.edu/~siu/siu_reprints/dmv_book.pdf) at Remark (1.4) part (ii), but the author states in the proof that "because the product of two harmonic forms is harmonic" which certainly isn't always true. I'm wondering if this is a consequence of some of the Kähler identities, but could not come up with a complete proof. If $\eta$ is harmonic, then $\Delta\eta = 0$. Since $\Lambda$ commutes with $\Delta$ we have $$\Delta(\Lambda \eta)=0$$ i.e. $\Lambda \eta$ is harmonic. Since $\Lambda \eta$ is a $0$-form it must be constant. The converse seems harder.