Prove $\lambda$ is eigenvalue of $A$ iff $\lambda^*$ is eigenvalue of $A^\dagger$ with kernel I don't know how to prove the statement &quot; $\lambda$ is an eigenvalue of $A$ iff $\lambda^*$ is an eigenvalue of $A^\dagger$&quot; 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. 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
$$\text{dim}(\text{ker}(A^\dagger - \lambda^* I)) = n - \dim(\text{Img}(A^\dagger - \lambda^* I)) &gt; 0$$
which would imply that...
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
$$\text{dim}(\text{ker}(A^\dagger - \lambda^* I)) = n - \dim(\text{Img}(A^\dagger - \lambda^* I)) &gt; 0$$
which would imply that...