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* This is not always true for the [[matrix multiplication|product]]: given symmetric matrices <math>A</math> and <math>B</math>, then <math>AB</math> is symmetric if and only if <math>A</math> and <math>B</math> [[commutativity|commute]], i.e., if <math>AB=BA</math>.
* For any integer <math>n</math>, <math>A^n</math> is symmetric if <math>A</math> is symmetric.
* Rank of a symmetric matrix <math>A</math> is equal to the number of non-zero eigenvalues of <math>A</math>.
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<math display="block">A = LL^\textsf{T}.</math>
If the matrix is symmetric indefinite, it may be still decomposed as <math>PAP^\textsf{T} = LDL^\textsf{T}</math> where <math>P</math> is a permutation matrix (arising from the need to [[pivot element|pivot]]), <math>L</math> a lower unit triangular matrix, and <math>D</math> is a direct sum of symmetric <math>1 \times 1</math> and <math>2 \times 2</math> blocks, which is called Bunch–Kaufman decomposition <ref>{{cite book |author-
A general (complex) symmetric matrix may be [[defective matrix|defective]] and thus not be [[diagonalizable]]. If <math>A</math> is diagonalizable it may be decomposed as
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The transpose of a symmetrizable matrix is symmetrizable, since <math>A^{\mathrm T}=(DS)^{\mathrm T}=SD=D^{-1}(DSD)</math> and <math>DSD</math> is symmetric. A matrix <math>A=(a_{ij})</math> is symmetrizable if and only if the following conditions are met:
# <math>a_{ij} = 0</math> implies <math>a_{ji} = 0</math> for all <math>1 \le i \le j \le n.</math>
# <math>a_{i_1 i_2} a_{i_2 i_3} \dots a_{i_k i_1} = a_{i_2
== See also ==
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== References ==
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*{{citation|
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[[Category:Matrices (mathematics)]]
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