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Let $A\in\mathbb{R}^{9\times 9}$, and $I_3\in\mathbb{R}^{3\times 3}$ is the identity matrix. Now I am going to find a matrix $\Lambda\in\mathbb{R}^{3\times 3}$ and $x\in\mathbb{R}^9$ such that $$(A-\Lambda\otimes I) x=0$$ where $\otimes$ is the Kronecker product. It looks like a generalized eigenvalue problem compared to the regular one $(A-\lambda I)x=0$. Has anyone ever seen this kind of problem?

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Obviously $x=0$ will do, but I suppose you don't want that. There's a nonzero $x$ that works if and only if $\det(A - \Lambda \otimes I) = 0$. This will be a polynomial in the $\lambda_{ij}$ of total degree 9, with degree $3$ in each of the $\lambda_{ij}$: the terms of total degree 9 are $\det(\Lambda)^3$. So you should be able to specify random values for all but one $\lambda_{ij}$, and solve a cubic equation in the remaining $\lambda_{ij}$.

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    A good idea. But we are going to compute 9 variables from only one equation. So instead of randomly specifying the values of some $\lambda_{ij}$, I think it is better to add more constraints. Actually I have some constraints on $x$. BTW, the $\lambda_{ij}$ here are actually Lagrange multipliers, and $x$ is the one I am looking for. The constraint of $x$ is like this: $x=\mathrm{vec}V$ where $V$ should be an orthogonal matrix. Do you think we can get better results by adding this constraint on $x$?2011-10-11
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    I'm not sure what you mean by vec$V$. If you mean $x = V v$ where $v$ is a given vector and $V$ can be any orthogonal matrix, that just says $\|x\| = \|v\|$. And since the condition $(A - \Lambda \otimes I)x = 0$ is homogeneous in $x$, that's not a significant restriction.2011-10-11
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    Sorry I didn't make it clear. $\mathrm{vec}$ is an operator that reshapre a matrix to a column vector by stacking the matrix's column vectors one after another. $x=\mathrm{vec}V$ means $x$ is the column vector that contains $V$'s orthogonal column vectors.2011-10-12