Do tests for positive definiteness for nonsymmetric matrices exist? More specifically I am working with bidiagonal upper/lower triangular matrices with positive eigenvalues and I need to check to see if they are positive definite or not.
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1if $(Ax,x)\ge 0$ for all $x\in V,$ then $A=A*$ provided you are working over complex numbers. – leshik May 10 '13 at 03:04
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1You can't start the question with "Do they exist". What are "they"? Make the question self-contained, and remember that a title is just an elevator pitch line to draw people and let them know what the question is about, in particular the title is not part of the question's text. – Asaf Karagila May 10 '13 at 03:18
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1Welcome to Math.SE. Thank you for your question. It will help us to better answer it if you give a more complete context for your question and share what you've tried so far. Also, you may wish to follow the customary style used here; see the FAQ for details. – vadim123 May 10 '13 at 03:24
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Suppose that $A$ is real. Note that $x^TAx=(x^TAx)^T=x^TA^Tx$, being a scalar. Hence $x^TAx>0$ if and only if $2x^TAx=x^TAx+x^TA^Tx=x^T(A+A^T)x>0$. Thus $A$ is positive definite if and only if $A+A^T$, a symmetric matrix, is positive definite. You can test this using the Cholesky decomposition.
If $A$ is complex, replace $x^TAx$ with $x^\star Ax$ and $A^T$ with $A^\star$.
vadim123
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1Thank you very much, I used Sylvester's criterion on $A^\dagger + A$ to show that it was not positive definite. – lowrybh May 10 '13 at 04:01
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3In the complex case $A$ "positive definite" ($x^T Ax \gt 0$ for all $x$) implies that $A$ must be Hermitian. – hardmath Jan 19 '15 at 14:12