# Prove $\lambda$ is eigenvalue of $A$ iff $\lambda^*$ is eigenvalue of $A^\dagger$ with kernel

I don't know how to prove the statement " $$\lambda$$ is an eigenvalue of $$A$$ iff $$\lambda^*$$ is an eigenvalue of $$A^\dagger$$" being $$A$$ an operator in a complex, finite Hilbert space. And only using kernels, images, orthogonality, and inner product concepts to prove it (can't use dimensions).

I think it has to be something similar to the following one:

$$Ax=\lambda x \xrightarrow[]{} (A-\lambda I)x=0$$

Where $$x$$ is the eigenvector, and then:

$$\ker(A-\lambda I)=(Img(A-\lambda I)^{\dagger})^{\perp} = Img(A^{\dagger}-\lambda^* I)^{\perp}$$

But at that point I get stuck, any hints about how to continue?

Thank you so much!

• You're almost there. You don't need to find a relation between the kernels to conclude. Just think in term of "if this is empty/full, then this is..." Dec 16, 2020 at 19:32
• Is the dimension finite? Dec 16, 2020 at 19:40
• I can't use the concept of dimension to prove it :( Dec 16, 2020 at 19:47
• @GEdgar Yes (for densely defined operators) but an answer below uses a dimension argument. Dec 16, 2020 at 19:49
• @Clash Finite Hilbert space? So we can use dimension! Dec 16, 2020 at 19:54

You're almost there. Think about the dimension of the space $$\text{ker}(A - \lambda I) = \text{Img}(A^\dagger - \lambda^* I)^\perp$$ We know that it's dimension is nonzero, so in particular this means that the dimension of the $$\text{Img}(A^\dagger - \lambda^* I)$$ must be strictly less than $$n$$, which is the order of $$A$$ and $$A^\dagger$$. But from rank-nullity, this would imply that $$\text{dim}(\text{ker}(A^\dagger - \lambda^* I)) = n - \dim(\text{Img}(A^\dagger - \lambda^* I)) > 0$$ which would imply that...