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So, there is this question in which I am suppose to find $UDU^{T}$ factorization of $$A = \begin{bmatrix} -3&1&-1\\1&-3&1\\-1&1&-3 \end{bmatrix}$$ I don't understand what is the meaning of $UDU^{T}$ factorization. I don't want solution as such. I just want to know what the question is asking. Any help is appreciated.

EDIT

I just know that $xAx^{T}$ here represents a standard quadratic form. Is it related to something like that?

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2 Answers 2

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Are you familiar with the eigendecomposition?

$D$ will be your diagonal matrix with the eigenvalues in the diagonal and $U$ is your matrix of eigenvectors normalized so that each eigenvector has norm 1.

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It is in fact the same prominent Schur decomposition for square matrices. Check the link below for further reading:

https://en.wikipedia.org/wiki/Schur_decomposition

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  • $\begingroup$ Schur decomposition finds an upper triangular U, not a diagonal D... $\endgroup$ Commented Nov 23, 2018 at 22:02
  • $\begingroup$ Then what if your matrix is real and symmetric as it is the case now? $\endgroup$ Commented Nov 23, 2018 at 22:03

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