1

On this page of Mathworld Wolfram regarding Positive Definite Matrix, it is stated that

"A necessary and sufficient condition for a complex matrix $A$ to be positive definite is that the Hermitian part is positive definite."

which further implies that:

"Therefore, a general complex (respectively, real) matrix is positive definite iff its Hermitian (or symmetric) part has all positive eigenvalues."

It is also known that:

A matrix is positive definite if and only if it is Hermitian and has all positive eigenvalues and

A positive definite matrix must be Hermitian.

However, from Wolfram it seems that a general complex matrix $A$ can be positive definite even if it is not Hermitian (as long as its Hermitian part is positive definite). Are the statements on mathworld wolfram right or wrong?

Ka Wa Yip
  • 1,116
  • Further down on the page you link discusses that Mathworld is not restricting attention to symmetric or Hermitian matrices even though most treatments do. – Michael Burr Jan 30 '24 at 10:25
  • The general idea here is that if you have a bi/sesquilinear form $y,x\mapsto x^* A y$, then by some change of basis, you can make $A$ hermitian/symmetric. So that is why, in some contexts, positive definite matrices are taken to be hermitian/symmetric. – student91 Jan 30 '24 at 10:27

1 Answers1

1

According to its own definition of positive definiteness (a complex square matrix $A$ is called positive definite if $\operatorname{Re}(x^\ast Ax)>0$ for every nonzero complex vector $x$; a real square matrix $A$ is called positive definite if $x^TAx>0$ for every nonzero real vector $x$), the two statements you have mentioned are correct. In the complex case, \begin{align*} &\operatorname{Re}(x^\ast Ax)>0\text{ for all }x\in\mathbb C\setminus0\\ &\Leftrightarrow (x^\ast Ax)+(x^\ast Ax)^\ast>0\text{ for all }x\in\mathbb C\setminus0\\ &\Leftrightarrow x^\ast(A+A^\ast)x>0\text{ for all }\in\mathbb C\setminus0\\ &\Leftrightarrow A+A^\ast\text{ is positive definite in conventional sense}\\ &\Leftrightarrow \frac{A+A^\ast}{2}\text{ is positive definite in conventional sense}\\ &\Leftrightarrow \frac{A+A^\ast}{2}\text{ is Hermitian and has a positive spectrum}.\\ \end{align*} The real case is similar.

However, some assertions on the same page are wrong:

  • “The determinant of a positive definite matrix is always positive”.
  • “The definition of positive definiteness is equivalent to the requirement that the determinants associated with all upper-left submatrices are positive.”

These two assertions apply to positive definite matrices in conventional sense only. They are wrong according to MathWorld’s definition. For a simple counterexample, consider the $1\times1$ complex matrix $A=1+i$. If we require $A$ to be real, the first assertion above is true, but sufficiency part in the second one is false — consider $A=\pmatrix{1&-1\\ 1&0}$ for instance.

user1551
  • 149,263