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The "positive semi-definiteness" definition of a Matrix $K$ may be formulated as follows:

$$\sum_{i,j=1}^n c_i c_j K_{i,j} \ge 0 \equiv c^T K c \ge 0$$

for any $c_1, ... , c_n \in \Bbb{R}$

But I can't figure out what do the coefficients $c_1, ..., c_n$ add to the definition. One told me they are used for the $=$ par of the $\ge$ when $c_1, ..., c_n$ are all equals to zero but without certainty.

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    It's much more clear to say that $K$ is positive semidefinite if and only if $c^T K c \geq 0$ for all vectors $c \in \mathbb R^n$. – 2017-02-15
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    I agree. But I've not seen this notation so far. Nevertheless, I still don't understand what it is the point of this vector c. – 2017-02-15
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    The matrix $K$ is used to define a function $\Bbb R^n\longrightarrow\Bbb R$. – 2017-02-15
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    I don't understand what you mean. – 2017-02-15

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For any matrix $K$, we define the quadratic form associated with $K$ to be the function $q:\Bbb R^n \to \Bbb R$ given by $$ q(c) = q(c_1,c_2,\dots,c_n) = \sum_{ij}K_{ij}c_ic_j $$ The vector $c$ here is an input to a function; it's just like the $x$ in $f(x) = x^2$. We say that $K$ is positive semidefinite if $q(c) \geq 0$ for every possible $c$.

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    Ok, thank you. I understand that but if $(1)~\sum_{ij} K_{ij} c_i cj \ge 0$ then $(2)~\sum_{ij} K_{ij} \ge 0$, right? (just take the vector $c$ composed of 1s). So why do we need to say it works with every vector $c$? Is there an exemple of symetric Matrix which verify $(2)$ but not $(1)$ ?? – 2017-02-15
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    Yes there is! For example: take $$ K = \pmatrix{1&2\\2&1}, \qquad c = (1,-1) $$ – 2017-02-15