# Find a matrix which is positive definite but its square is not

Is there a $2\times2$ matrix which is positive definite (not necessarily symmetric) but its square is a not positive definite matrix?

## 1 Answer

A positive definite matrix is symmetric by definition (https://en.wikipedia.org/wiki/Positive-definite_matrix). What definition are you using?

If we agree on the matrix being symmetric, then what you ask for does not exist even if you allow not just 2x2 matrices but square matrices of any dimension.

A PD matrix is diagonalizable and has positive eigenvalues. Therefore the square of it is obtained by just squaring the eigenvalues. It follows that the squared matrix also has positive eigenvalues and therefore PD.

If we extend the definition to non symmetric matrices through the definition: $M$ is PD $\iff$ for all $x\in\mathbb{R}^n$ either $x=0$ or $x^\intercal M x>0$, then the answer is a yes.

Hint: For two dimensions, think of a rotation matrix that rotates any vector by slightly less than $\pi/2$ radians. See here for the rotation matrix.

• An example of a matrix which which is positive definite but not symmetric is A = [1 1; -1 1] which is not symmetric. Commented Sep 22, 2016 at 23:55
• The usual definition of PD is for symmetric matrices, in which case the eigenvalues are real, therefore you can define PD as having all eigenvalues being positive. If you want to allow non-symmetric matrices, please let me know how it is defined. Commented Sep 22, 2016 at 23:58
• Are you defining it through the quadratic form? Commented Sep 22, 2016 at 23:58
• If w is a vector in R², of the form (w₁, w₂) w' A w = w₁(w₁ + w₂) + w₂(-w₁ + w₂) = w₁² + w₂² > 0 Doesn't that make the matrix A positive definite? Or, does it have to have positive eigenvalues too? Commented Sep 23, 2016 at 0:02
• Okay so you define a square matrix $M$ as real PD if for all $x\in\mathbb{R}^n$ not equaling zero, $x^\intercal M x>0$. Commented Sep 23, 2016 at 0:05