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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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Positive definite is normally only defined for symmetric matrices. All the criterion you mentioned assumes the matrix is symmetric. –  EuYu Oct 11 '12 at 2:07
    
The theorem is true for all symmetric matrices $A$, and your example is not symmetric. –  Per Manne Oct 11 '12 at 2:07
    
By the way: none of those are definitions; they're criteria for positive-definiteness. –  Hans Lundmark Oct 11 '12 at 7:01
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Definition: A quadratic form $Q$ is positive definite if $Q(\mathbf{x}) > 0$ for every vector $\mathbf{x} \neq \mathbf{0}$. A symmetric matrix $A$ is positive definitite if the quadratic form $Q(\mathbf{x}) = \mathbf{x}^t A \mathbf{x}$ is positive definite. –  Hans Lundmark Oct 11 '12 at 8:56
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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.) –  Hans Lundmark Oct 11 '12 at 20:17

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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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I'm not sure what exactly you mean. If you're considering positive-definite matrices, then they are symmetric and hence have all real eigenvalues. What do you mean by "imply positive definiteness in the complex sense"? Do you mean something to do with Hermitian matrices? P.S. You need an @ in front of my name too, you can't chain the notifications it seems. –  EuYu Oct 11 '12 at 3:04
    
@EuYu clarified it. I think I am starting to understand my teacher: he meant the strong form of positively-definite. I am uncertain whether the weaker form of positively-definiteness actually result in different conclusions, investigating. –  hhh Oct 11 '12 at 3:23
    
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. –  EuYu Oct 11 '12 at 3:35

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