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Let $L$ be a lower-triangular matrix such that $LL^T=A$.

Then $B=L^TL$ has the same eigenvalues as $A$, but different eigenvectors.

1) Why are the eigenvalues the same?

2) Is there an analytical relationship between the eigenvectors of $A$ and $B$?

Jackson Loper
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    If $\lambda$ is an eigenvalue $LL^T$ then for some $v$, $LL^Tv=\lambda v$. Then $L^T(\lambda v) = L^T(LL^T)v = (L^TL)L^Tv$. This is similar to http://math.stackexchange.com/questions/821934/eigenvalues-of-ab-and-ba-matrices – ndruiven Dec 08 '14 at 16:16

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$$B = L^{-1} A L$$ So $B$ is similar to $A$ with the change of basis $L$. The eigenvectors of $B$ are thus $L^{-1}v$ where $v$ is an eigenvector of $A$. The corresponding eigenvalues are trivially equal:

$$BL^{-1} v = L^{-1} A L L^{-1} v = L^{-1} Av = L^{-1} \lambda v = \lambda L^{-1} v$$

AlexR
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