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>There's a more fundamental way... I'd wager to say there's nothing more fundamental about it. You are still finding eigenvalues and eigenvectors of a linear o
by romwell 8y ago
>There's a more fundamental way...
I'd wager to say there's nothing more fundamental about it. You are still finding eigenvalues and eigenvectors of a linear operator. The matrix is exactly the same.
- nimish 8y agoIt's more fundamental since you're working with the underlying vector space and linear operator vs representing it as a matrix over R^n, running a decomposition, then converting back. No need to change domains, but yes they are isomorphic (which is the point).