The definition of a positive semidefinite/definite matrix seems assuming the matrix is symmetric. Is my understanding correct? If it is symmetric, is there any reason behind that? Otherwise, can anyone give an example of a non-symmetric matrix that is positive semidefinite/definite?
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$M=\left[\begin{array}{llll}1 & 0 \ 1 & 0 \end{array}\right]$ is positive semidefinite over real scalars but not symmetric. Any positive semidefinite over complex scalars is symmetric. – Kavi Rama Murthy Aug 26 '21 at 00:14
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@Kavi Rama Murthy Then for $x^T=(x_1,x_2)$, does $x^TMx=x^2_1+x_1\cdot x_2\ge 0$ always hold? I am confused. – nlite Aug 30 '21 at 22:41
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PD- and PSD-ness only make sense for symmetric matrices. Note that the skew-symmetric part contributes nothing. – Rodrigo de Azevedo May 05 '22 at 16:12
2 Answers
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There seems to be disagreement in the literature for the definition of a positive definite matrix. For discussion purposes let us assume the matrix is real.
Here are possible two definitions:
- Let $M$ be a $n \times n$ matrix. $M$ is a positive definite matrix if and only if $M$ is symmetric AND $x^T M x > 0$ for all non-zero vectors of order $n$.
- Let $M$ be a $n \times n$ matrix. $M$ is a positive definite matrix if and only if $x^T M x > 0$ for all non-zero vectors of order $n$.
For example, let $M = \begin{pmatrix} 1 & 1\\ -1 & 1\\ \end{pmatrix}$ and $x = \left(a,b\right)$
Since $x^T M x = a^2+b^2$, $M$ is positive definite by the second definition, but not by the first because $M$ is not symmetric.
The first definition agrees with the following:
- Strang G., "Introduction To Applied Mathematics", Wellesley-Cambridge Press, 1986, pages 15 to 18.
- https://en.wikipedia.org/wiki/Definite_matrix
- https://proofwiki.org/wiki/Definition:Positive_Definite_Matrix
The second definition is supported by:
- https://mathworld.wolfram.com/PositiveDefiniteMatrix.html
- https://reference.wolfram.com/language/ref/PositiveDefiniteMatrixQ.html
- Zwillinger D., et al., "CRC standard mathematical tables and formulas, 33rd edition", Section 2.4.14, page 99.
Admittedly, both the first two are Wolfram sources.
The discrepancy of the definitions is acknowledged in the Math World reference where they state "confusingly, the discussion of positive definite matrices is often restricted to only Hermitian matrices, or symmetric matrices in the case of real matrices...". The Wikipedia source also acknowledges the discrepancy when they say "some authors use more general definitions of definiteness, including some non-symmetric real matrices, or non-Hermitian complex ones. "
C. Dunn
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There is also the case of infinite-dimensional operators; see e.g. https://en.wikipedia.org/wiki/Positive_operator_(Hilbert_space) – KBS May 05 '22 at 16:19
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A definition with scalar products $x^\top M x$ only involves the symmetric part of a matrix (note that all that follows can be directly adapted to complex matrices).
First, we can decompose any matrix $M$ under the form $$ M = S + A$$ with $S$ symmetric and $A$ anti-symmetric (or skew-symmetric). Consider for that $S = (M+M^\top)/2$ and $A = (M-M^\top)/2$.
Second, $x^\top A x = 0$ for all vectors $x$. Indeed, $$ x^\top A x = x^\top(Ax) = (Ax)^\top x = x^\top A^\top x = -x^\top A x.$$
Therefore, $x^\top M x = x^\top S x$ for all vectors $x$. Whatever definition you consider in the answer post of C. Dunn, it necessary involves the symmetric part of $M$. The difference is whether you impose $A=0$ or not.
reded
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