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Let $\vec{x} \in \mathbb{R}^n$ be a real column vector and $\mathbf{X} = \vec{x} \otimes \vec{x} \triangleq \vec{x}\vec{x}^T \in \mathbb{R}^{n \times n}$ the Kronecker product of $\vec{x}$ with itself. As $\mathbf{X}$ is real symmetric its spectral decomposition is $\mathbf{X} = \sum_{i = 1}^n \lambda_i V_iV_i^T$,

Are there any result on how the individual eigenvalues (and/or eigenvectors) of $\mathbf{X}$ relate to the elements of $\vec{x}$ in this special case (i.e. self Kronecker product)?

(Note that although $\sum_i \lambda_i = \sum_i x_i^2$ here I'm specifically asking about the relationship between each $\lambda_i$ and elements of $\vec{x}$)

user400479
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1 Answers1

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Since that operator $xx^\top$ is rank-one, the eigenvalue $0$ occurs with multiplicity $n-1$, with eigenspace consisting of the orthogonal complement of $x$. The other eigenvalue is length-squared of $x$, with eigenvalue $x$ itself. These assertions are one-liners to verify, once observed...

Is this what is being asked?

paul garrett
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