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Is there a simple expression for the eigenvector of $D \mathbf a \mathbf a^H$ ?

Assuming $\lVert \mathbf a \rVert = 1$ and $D$ is positive definite and diagonal.

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By inspection, the vector $x = Da$ satisfies $Daa^Hx = Daa^HDa = (a^HDa)Da = (a^HDa)x$.

Hence, $Da$ is an eigenvector with eigenvalue $a^HDa$.

The other $\text{dim}(a)-1$ eigenvalues will be $0$, and the corresponding eigenvectors are any vectors orthogonal to $a$.

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