How can I show that a real symmetric matrix $A$ has only positive eigenvalues if and only if there exists a matrix B such that $A=B^TB$?
$A$ definite positive if and only if $A=B^TB$
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linear-algebra
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0The answer is there (https://en.wikipedia.org/wiki/Positive-definite_matrix) under the hidden form of being a "Gram matrix". – 2017-01-18
1 Answers
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comment: JeanMarie
Since $A$ is a real symmetric matrix it is orthogonally diagonalizable, that is, $$A = QDQ^T$$ with $D = \{ \lambda_i \}$ a diagonal matrix consisting of the eigenvalues from $A$. Now define $C = \{c_i\}$ also diagonal with $c_i = \sqrt{\lambda_i}$. Since it holds that $C^TC = D$, $A$ is equal to $$A = QDQ^T = QC^TCQ^T = (CQ)^T(CQ) = B^TB.$$ Note that $C$ is defined if and only if $\lambda_i > 0$, that is, $A$ has strictly positive eigenvalues.