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Linear Algebra - eigendecomposition

Suppose that a matrix A has n linearly independent eigenvectors {v(1)v^{(1)}, . . . , v(n)v^{(n)}} with corresponding eigenvalues {λ1\lambda _1, . . . , λn\lambda _n}. We may concatenate all the eigenvectors to form a matrix V with one eigenvector per column: V = [v(1)v^{(1)} , . . . , v(n)v^{(n)}]. Similarly, we can get a vector containing eigenvalues, naming it as λ\lambda. The eigendecomposition of A is then given by A=Vdiag(λ)V1A = Vdiag(\lambda)V^{-1}

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Updated 2021-05-17

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