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Do positive semidefinite matrices have to be symmetric? Can you have a non-symmetric matrix that is positive definite? I can't seem to figure out why you wouldn't be able to have such a matrix, but all my notes specify positive definite matrices as "symmetric $n \times n$ matrices."

Can anyone help me with an example of a non-symmetric positive definite matrix, or some insight into a proof for why it would need to be symmetric should that be the case? Thanks!

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3 Answers 3

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No, they don't, but symmetric positive definite matrices have very nice properties, so that's why they appear often.

An example of a non-symmetric positive definite matrix is $$M=\pmatrix{2&0\\2&2}.$$ Indeed, $$\pmatrix{x\\y}^T\pmatrix{2&0\\2&2}\pmatrix{x\\y} = (x+y)^2 + x^2 + y^2$$ which is strictly greater than $0$ whenever the vector is non-zero.

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    $\begingroup$ but in both wikipedia and wolfram mathworld, the definition of PSD matrix involves being symmetric. $\endgroup$
    – Sam
    Nov 11, 2022 at 11:31
  • $\begingroup$ @Sam Note that while PD doesn't imply symmetric when restricted to reals, the stronger PD over complex vectors does indeed imply Hermiticity. Proof: $x^* M x>0$ is real, so equals its conjugate, thus $x^* M x=x^* M^* x$ for all $x$, so $M=M^*$; $\endgroup$ Mar 21, 2023 at 2:38
  • $\begingroup$ @DavidRaveh I checked again. In wikipedia it does require the matrix PSD to be symmetric ('An $n\times n$ symmetric real matrix M M is said to be positive-semidefinite or non-negative-definite if $x^T M x ≥ 0$ for all x in $\mathbb {R} ^{n}$'. Is that a misconception? $\endgroup$
    – Sam
    Mar 21, 2023 at 5:36
  • $\begingroup$ This is discussed in a subsection of the wikipedia page. They repeat the requirement for complex matrices, but they use the letter $\textbf{z}$ to imply it holds for complex vectors, but don't say it explicitly. They also discuss the restriction to real matrices. en.wikipedia.org/wiki/… $\endgroup$ Mar 21, 2023 at 15:14
  • $\begingroup$ A factor of 2 is missing in the second equation. By the way, from the second equation, we can see that in the context of quadratic form, we can find an equivalent symmetric matrix ${1\over2}(M+M^T)$. $\endgroup$
    – syockit
    Apr 7, 2023 at 10:34
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As Daniel mentions in his answer, there are examples, over the reals, of matrices that are positive definite but not symmetric. The usefulness of the notion of positive definite, though, arises when the matrix is also symmetric, as then one can get very explicit information about eigenvalues, spectral decomposition, etc.

In the complex case, any positive-definite matrix is selfadjoint.

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Let me just add that there exist a branch of optimization and variational analysis in which the notion of a nonnegative definite matrix which is not symmetric is fundamental. Consider a smooth convex function; by convexity its Hessian is nonnegative positive definite while by Schwarz Theorem, it is also symmetric. However, there exist "non-gradient operators" i.e. which do not integrate into a function but still possess a notion of convexity. This is the notion of a monotone operator, which in the smooth case, its Jacobian is a nonnegative definite matrix which may not be symmetric.

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