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The following matrix:

$\begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$

is an indefinite matrix? Or it is that is a matrix semidefinite negative and semidefinite positive?

I am confused because some texts indicate that the indefinite matrices have non-zero eigenvalues of discordant sign, while others do not specify whether the zero eigenvalues are allowed.

Who should I give a reason? Where can I read a definition to be framed in the house and always follow that?

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    $\begingroup$ Of course the matrix is defined. It is just neither sort of semidefinite. It is called indefinite (some positive, some negative eigenvalues). $\endgroup$ – Ted Shifrin Jan 16 '17 at 23:19
  • $\begingroup$ For a symmetric matrix $A$, being positive semidefinite means that $x^TAx\ge0$, for every $x$. Change into $\le$ for semidefinite negative. A necessary and sufficient condition for $A$ to be positive semidefinite is that its eigenvalues are $\ge0$. $\endgroup$ – egreg Jan 16 '17 at 23:31
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We have $$\begin{pmatrix}{1}&{0}&{0}\end{pmatrix}\begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}{1}\\{0}\\{0}\end{pmatrix}=-1<0$$$$\begin{pmatrix}{0}&{0}&{1}\end{pmatrix}\begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}{0}\\{0}\\{1}\end{pmatrix}=1>0$$ so, the matrix is indefinite.

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  • $\begingroup$ So if a matrix has an eigenvalue zero isn't said to be semi-definite, could be indefinite. Thank you! $\endgroup$ – TeM Jan 16 '17 at 23:26
  • $\begingroup$ Yes, could be indefinite as in this case. Note that I've not used eigenvalues, but the "first" definition of indefinite. $\endgroup$ – Fernando Revilla Jan 16 '17 at 23:29
  • $\begingroup$ The definition of indefinite matrix is not unique? Oh God, tell me so! $\endgroup$ – TeM Jan 16 '17 at 23:31
  • $\begingroup$ No problem, there are equivalent definitions. Have a look here: mathworld.wolfram.com/IndefiniteQuadraticForm.html $\endgroup$ – Fernando Revilla Jan 16 '17 at 23:34
  • $\begingroup$ Perfect. Thanks again, very nice! $\endgroup$ – TeM Jan 16 '17 at 23:36

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