In SVD: $$M = UDV^T$$ How do you prove that columns of $U$ are eigenvector of $MM^T$ and columns of $V$ are eigenvectors of $M^TM$?
Proof of left and right singular vector in SVD
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matrices
eigenvalues-eigenvectors
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1 Answers
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For $U$: compute $$ MM^T = (UDV^T)(UDV^T)^T = UD^2U^T $$ Which is to say that $U$ diagonalizes $MM^T$. If that doesn't convince you, rewrite this as $$ (MM^T)U = UD^2 $$ which is to say that $$ (MM^T)\pmatrix{u_1 & \cdots & u_n} = \pmatrix{u_1&\cdots & u_n} \pmatrix{\sigma_1^2 \\ & \ddots \\ && \sigma_n^2} $$ The columns of the matrix on the left are $MM^T u_i$, but the columns of the matrix on the right are $\sigma_i^2 u_i$. The conclusion follows.