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My teacher in course Mat-2.3139 here claims that all positive-definite -definitions will result in the same result or I am misunderstanding something. I am clearly misunderstanding something because the below cases do not result in the same result: I get that the matrix $\left[\begin{array}{cc} 1 & 2 \\ -1 & 0 \\\end{array}\right]$ is and is not positively-definite, varying between the different -methods to check the positively-definiteness. Where am I misunderstanding things?

[CONFLICT] Not the same conclusion, why?

1. [YES] Determinant-of-All-Squares-check: calculations here

2. [NO] Square-form -check: calculations here

3. [YES|NO] Eigen-values -positive: Eigen-values are $\lambda _2=\frac{1}{2} \left(1-i \sqrt{7}\right)$ and $\lambda _2=\frac{1}{2} \left(1+i \sqrt{7}\right)$ according to this. Now the real -part is positive but the complex part is not, is this positively definite or not?

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    Well, actually it doesn't really matter, since $Q(c \mathbf{x})= c^2 Q(\mathbf{x})$. (So that it's enough to look, for example, at $\mathbf{x}$ on the unit sphere.)2012-10-11

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As Per Manne and EuYu commented, the positive definiteness only applies to symmetric matrices. Suppose $A$ is a symmetric square-matrix below. You can see here a basic example where it works.

Definitions for positively-definitess

  1. determinant-of-all-squares -check: $det(A_{i,i})>0$ where $A$ is a symmetric quare-matrix and $i\leq n$ where $n$ is the degree of the matrix $A$

  2. square-form -check: $\bar x^t A \bar x>0$ where $\bar x$ is inside an unit ball, defined in Finnish here where the unit-ball -definition here (pages 866-867 in the book here)

  3. eigen-value -check: all eigen-values must be positive $\lambda_i>0 \forall i\in\mathbb N$

Generalization from Real numbers to Complex numbers

Suppose $z \in\mathbb C^n$. If $z^t M z \in\mathbb C^m$ with some $m$ and even with Hermitian $M$, then not positively definite, by this example here but please note that some authors use the notation $\mathbb R(z^tMz)>0$ where $z\in\mathbb C$ for positively definiteness. In other words, if the square -form is complex, it is not positively definite. You must get a scalar of the complex square form, otherwise the inequality undefined so not positively definite.

The latter Wikipedia -article mentions a relaxation of symmetrism with the matrices in complex case. This weaker form of positively-definiteness requires only that $z^t M z>0$ in which case the other definitions of positively-definiteness do not match, getting your ambiguous results. You can try this easily for example with your complex eigen-values: if you assume that $\mathbb R(z^tMz)>0$ only required for the definition, then you get Yes/No/Yes so not matching.

So your teacher is considering the strong form of positively-definiteness in which the definiteness have the same conclusion. In the weaker form, your "conflict" is correct: you both are correct depending on the definition of positively-definiteness, please, note that $a_{ij}=a^*_{ji}$ where $.^*$ is the complex conjuagate (needed in the Hermitian -definition, generalization of the term symmetric to complex case).

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    Well, when most people say positive-definite, I think they implicitly mean the "strong-form" (symmetric) that you're talking about. The reason is simple enough. It's precisely because the non-symmetric matrices do not have such a nice chain of properties that we can exploit. For example, before you brought it to my attention, I wasn't even aware that positive-definiteness was even defined for non-symmetric matrices.2012-10-11