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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  • It depends on convention. Notice a **real** square matrix $A$ satisfies $x^T A x \ge 0$ for all $x$ if and only if $x^T (A+A^T) x \ge 0$ for all $x$. – user251257 Oct 04 '16 at 22:51
  • Duplicate: http://math.stackexchange.com/questions/1555039/what-is-the-agreed-upon-definition-of-a-positive-definite-matrix – Arash Oct 04 '16 at 22:56
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    Possible duplicate of [What is the agreed upon definition of a "positive definite matrix"?](https://math.stackexchange.com/questions/1555039/what-is-the-agreed-upon-definition-of-a-positive-definite-matrix) – Hans Lundmark Jul 03 '17 at 06:34

3 Answers3


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.

Daniel Robert-Nicoud
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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.

Martin Argerami
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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.