Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Definite matrix
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
== Extension for non-Hermitian square matrices == The definition of positive definite can be generalized by designating any complex matrix <math>M</math> (e.g. real non-symmetric) as positive definite if <math>\mathcal{R_e} \left\{\mathbf{z}^* M \mathbf{z}\right\} > 0</math> for all non-zero complex vectors <math>\mathbf{z},</math> where <math>\mathcal{R_e}\{c\}</math> denotes the real part of a [[complex number]] <math>c.</math><ref name="mathw">{{cite web |last = Weisstein |first = Eric W. |url = http://mathworld.wolfram.com/PositiveDefiniteMatrix.html |title = Positive definite matrix |website = MathWorld |publisher = Wolfram Research |access-date= 2012-07-26 }}</ref> Only the Hermitian part <math display="inline">\frac{1}{2}\left(M + M^*\right)</math> determines whether the matrix is positive definite, and is assessed in the narrower sense above. Similarly, if <math>\mathbf{x}</math> and <math>M</math> are real, we have <math>\mathbf{x}^\mathsf{T} M \mathbf{x} > 0</math> for all real nonzero vectors <math>\mathbf{x}</math> if and only if the symmetric part <math display="inline">\frac{1}{2}\left(M + M^\mathsf{T} \right)</math> is positive definite in the narrower sense. It is immediately clear that <math display="inline">\mathbf{x}^\mathsf{T} M \mathbf{x} = \sum_{ij} x_i M_{ij} x_j</math>is insensitive to transposition of <math>M.</math> A non-symmetric real matrix with only positive eigenvalues may have a symmetric part with negative eigenvalues, in which case it will not be positive (semi)definite. For example, the matrix <math display=inline>M = \left[\begin{smallmatrix} 4 & 9 \\ 1 & 4 \end{smallmatrix}\right]</math> has positive eigenvalues 1 and 7, yet <math>\mathbf{x}^\mathsf{T} M \mathbf{x} = -2 </math> with the choice <math>\mathbf{x} = \left[\begin{smallmatrix} -1 \\ 1 \end{smallmatrix}\right]</math>. In summary, the distinguishing feature between the real and complex case is that, a [[Bounded operator|bounded]] positive operator on a complex Hilbert space is necessarily Hermitian, or self adjoint. The general claim can be argued using the [[polarization identity]]. That is no longer true in the real case.
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)