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Suppose a matrix $A\in\mathbb{R}^{n\times n}$ is given. Faced with a proof for $$x^TAx>0,$$ for a non-zero vector $x\in\mathbb{R}^{n}$, I was thinking to use the information of the spectrum of $A$ (note that the proof of the above is the proof of positive definiteness of $A$). Namely, it is known that the spectrum of $A$ is strictly positive. Does that mean that $A$ is positive definite? Or, does that mean that only for symmetric $A$, $A$ is positive definite if and only if the spectrum of $A$ is positive?

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  • $\begingroup$ I don't know that this question is appropriate for SciComp because it's really more of a proof-based math question usually seen on Math.SE, but it's easy to answer here. If you want me to migrate it, let me know. $\endgroup$ – Geoff Oxberry Aug 6 '12 at 8:57
  • $\begingroup$ What do you want prove? $x^T A x$ ? $\endgroup$ – shuhalo Aug 6 '12 at 17:07
  • $\begingroup$ I suppose you meant to write that you want to prove that $x^T Ax > 0$? $\endgroup$ – Wolfgang Bangerth Aug 7 '12 at 6:44
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Usually, the term "positive definite" refers to any square, Hermitian matrix $A$ such that $x^{H}Ax > 0$ for all nonzero vectors $x$, where the superscript $H$ denotes the Hermitian transpose. An equivalent definition states that a positive definite matrix $A$ is Hermitian and has strictly positive spectrum (this statement is well-posed because Hermitian matrices have real spectra); see Wikipedia.

This definition is sometimes extended to include matrices that are not Hermitian, but such extended definitions may be inappropriate for the applications you're considering. It's probably safer to require that $A$ be Hermitian.

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  • $\begingroup$ But your relation only involves the "if" direction. Does it hold that a Hermitian matrix $A$ is positive-definite if and only if it has a positive spectrum? $\endgroup$ – usero Aug 6 '12 at 9:24
  • $\begingroup$ Yes. I chose to use the verb "is" rather than "if and only if" for brevity, since I also indicated that I was stating a definition. Adding the phrase "if and only if" seems redundant. $\endgroup$ – Geoff Oxberry Aug 6 '12 at 9:55
  • $\begingroup$ I disagree: the statement "a complex matrix is pos. def. iff $\Re[x^H A x] > 0$" is not an extended definition. I would rather say that your $x^HAx>0$ for hermitian matrices is a special case. $\endgroup$ – Stefano M Aug 6 '12 at 14:22
  • $\begingroup$ According to the source you cite: "Confusingly, the discussion of positive definite matrices is often restricted to only Hermitian matrices, or symmetric matrices in the case of real matrices (Pease 1965, Johnson 1970, Marcus and Minc 1988, p. 182; Marcus and Minc 1992, p. 69; Golub and Van Loan 1996, p. 140)." These are standard sources; Artin's algebra textbook and Strang's linear algebra textbooks also state the same definition. You're free to disagree, but semantically, saying that the Hermitian case is "special" is the same as saying that the non-Hermitian case is "extended". $\endgroup$ – Geoff Oxberry Aug 6 '12 at 16:09
  • $\begingroup$ It is not my intention to start a war on definitions. The point for me is that $x^TAx \lesseqgtr 0$ for real matrices nicely maps to $\Re [ x^HAx] \lesseqgtr 0$ for the complex case. On the contrary $x^HAx \in \mathbb{R}$, $x^HAx \lesseqgtr 0$, since it implies Hermitian $A$, chokes the discussion on $x^TAx \lesseqgtr 0$ for real general matrices, which is actually the subject of this question. But you are right: my previous comment on extended/special was rather silly, it is like arguing about an half empty/half full glass... $\endgroup$ – Stefano M Aug 6 '12 at 17:16
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According to MathWorld a matrix $A \in \mathbb{R}^{n \times n}$ is positive definite iff $$ (x^T A x) > 0 $$ for all non zero vectors $x\in\mathbb{R}^n$. It is trivial to obtain that $$ x^T\,A\,x = x^T\, \left [ \frac12(A+A^T) \right ]\, x $$ and to recognize that positive definiteness is linked to the spectrum of the symmetric part of matrix $A$.

A general real square matrix $A$ is positive definite iff its symmetric part $\frac12(A+A^T)$ has all positive eigenvalues.

A positive spectrum for $A$ does not implies that $\frac12 (A+A^T)$ has positive spectrum, as it can seen for $$ A = \begin{pmatrix}a & 1\\ 0 &b\end{pmatrix} $$ If we assume $a>0$ and $b>0$ this matrix has positive spectrum, but $$ \mathrm{det} \frac12 (A+A^T) = ab-\frac14 $$ now if $ab < 1/4$, $\mathrm{det} \frac12(A+A^T)<0$ and $A$ is not positive definite.

Bottom line: for unsymmetric matrices $A$, if you are interested in the sign of $x^TAx$ you have to study the eigenvalues of $(A+A^T)/2$.

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