# What is the agreed upon definition of a "positive definite matrix"?

A positive definite matrix is a symmetric matrix A for which all eigenvalues are positive. - Gilbert Strang

I have heard of positive definite quadratic forms, but never heard of definiteness for a matrix.

• Because definiteness is higher dimensional analogy for whether if something is convex (opening up) or concave (opening down). It does not make sense to me to say a matrix is opening up, or matrix is opening down.

Therefore it does not make sense to say that a matrix has definiteness.

• In addition, when we say $M \in \mathbb{R}^{n \times n}$ positive definite, what is the first thing we do? We plug $M$ into a function(al) $x^T (\cdot) x$ and check whether the function is positive for all $x \in \mathbb{R}^n$. Clearly, that means we are defining this definiteness with respect to $x^T (\cdot) x$ and NOT $M$ itself.

Furthermore, when matrix have complex eigenvalues, then we ditch the notion of definiteness property all together. Clearly, definiteness is a flimsy property for matrices if we can just throw it away when it becomes inconvenient.

I will grant you that if we were to define positive definite matrices, we should only define with respect to symmetric matrices. This is the definition on Wikipedia, the definition used by numerous linear algebra books and many applied math books.

But then when confronted with a matrix of the form

$$\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}$$

I still firmly believe that this matrix is not positive definite because it is not symmetric. Because to me positive definiteness implies symmetry.

To what degree is it widely agreed upon in the math community that a positive definite matrix is defined strictly with respect to symmetric matrices and why only with respect to symmetric matrices?

• A matrix $A$ is said to be positive definite if $x^TAx>0$ Dec 1, 2015 at 16:48
• @Quintic Consider matrix $A$ = $\begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix}$ it has eigenvalues $1 \pm i$, none of which is real, which by implication positive (or negative for that matters). But the quadratic form is $x_1^2 + x_2^2$ > 0 for all $x_1, x_2 \in \mathbb{R}$ not equal to zero. none of the eigenvalues are positive. So are you saying that a matrix can be positive definite even if none of the eigenvalues are real and positive? Dec 1, 2015 at 17:58
• @Fraïssé Since it's used in quadratic form, it can be converted into a symmetric matrix. In this case, your matrix becomes the identity matrix which is positive definite. The complex eigenvalues in this case matter in linear form, not quadratic. Apr 7 at 10:52

The positive definiteness (as you already pointed out) is a property of quadratic forms. However, there is a "natural" one-to-one correspondence between symmetric matrices and quadratic forms, so I really cannot see any reason why not to "decorate" symmetric matrices with positive definiteness (and other similar adjectives) just because it is in "reality" the form they define which actually has this property. I can see this one-to-one correspondence as one of the reasons why the symmetry should be implicitly assumed when talking about positive definite matrices.

One can of course devise a different name for this property, but why? In addition, positive definite matrix is a pretty standard term so if you continue reading on matrices I'm sure you will find it more and more often.

Some authors (not only on Math.SE) allow positive definite matrices to be nonsymmetric by saying that $M$ is such that $x^TMx>0$ for all nonzero $x$. In my opinion this adds more confusion than good (not only on Math.SE). Also note that (with a properly "fixed" inner product) such a definition would not even make sense in the complex case if the matrix was allowed to be non-Hermitian ($x^*Mx$ is real for all $x$ if and only if...).

Anyway, for real matrices, it of course makes sense to study nonsymmetric matrices giving a positive definite quadratic form through $x^TMx$ (which effectively means that the symmetric part is positive definite). However, I find denoting them as positive definite quite unlucky.

If the square matrix A is positive definite then the hypersurface defined by

$$Y=X^T.A.X$$

is convex.

Symmetric real-valued matrices are diagonizable, i.e., even when viewed as complex matrices their eigenvalues are all real.

For general complex square matrices replace 'symmetric' with 'hermitean'.

Unfortunately, there is an ambiguity over the definition of positive definiteness, and probably there is no agreement among the authors. I also stumbled upon the Gilbert Strang's definition and asked to myself "wait, does positive definiteness imply symmetry?".

The ambiguity is that:

definition 1: positive definiteness does imply symmetry. Therefore, positive definiteness is a subclass of the symmetric matrices, in which $$\mathbf{x}^T \mathbf{Ax} > 0 \;\forall\; \mathbf{x} \in \mathbb{R}^n\backslash\{\mathbf{0}\}$$. Some authors, such as Gilbert Strang, use this definition

definition 2: positive definiteness is simply defined as $$\mathbf{x}^T \mathbf{Ax} > 0 \;\forall\; \mathbf{x} \in \mathbb{R}^n\backslash\{\mathbf{0}\},$$ Such a definition is adopted Golub. Therefore, we can have a matrix that is not symmetric but it is positive define (according to the definition 2). For instance, the matrix $$\mathbf{A} = \begin{bmatrix} 2 & 0 \\ 2 & 2 \end{bmatrix}$$ is not definite positive by the definition 1 (because $$\mathbf{A}$$ is not symmetric), but is definite positive by the definition 2 (because $$\mathbf{x}^T \mathbf{Ax} > 0 \;\forall\; \mathbf{x} \in \mathbb{R}^n\backslash\{\mathbf{0}\}$$). In fact, Golub calls such matrices "Unsymmetric Positive Definite" matrices (vide section 4.2.2).

As far as I could notice, many books about statistical signal processing follows the definition 2, while books about fundamentals of linear algebra follow either the definition 1 or 2.

Personally, I acknowledge that both definitions exist, but I follow the def. 2. And, as the authors that follow the def. 2, when $$\mathbf{A} = \mathbf{A}^T$$ also satisfies $$\mathbf{x}^T \mathbf{Ax} > 0 \;\forall\; \mathbf{x} \in \mathbb{R}^n\backslash\{\mathbf{0}\}$$, I say that $$\mathbf{A}$$ is symmetric and positive definite.