Understanding Linear Algebra: Dependent vs. Independent Vectors Explained

dervast
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Hi i am new to linear algebra and i am not sure how i can translate some terms so i need some help with that
when we have vectorus u1,u2,.. un
and is k1*u1+k2*u2+...un*kn=0 then we say that these vectors are linear dependant or something like that
and if
k1*u1+k2*u2+...un*kn=0 and k1=0 and k2=0 and so on these vectores are linear independantly.
Do u know if my translation is correct?
 
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Posted in the wrong place, but oh well.

We say u_1...u_n are linearly independent if \sum{k_iu_i}=0 implies that all the k's are zero. They're linearly dependent if there are scalar k's such that not all of them are zero, and the above holds
 
And what is the physical meaning of linearly dependence and independence? What do u understand when y hear someone saying that something is linearly independent or dependent?
 
A set of vectors are linearly independent if, for n linearly independent vectors, you can describe n dimensions of space. So basically, it means that each vector you add describes a new dimension of direction. So the first vector points in a line, the second one lies in a plane with the first, the third lies in a volume with the first two, etc. If the vectors were linearly independent, then perhaps the first lies in a line, the second lies in a plane, and then the third also lies in that plane. So you don't get one dimension/vector
 
Thx a lot for the answers.but have u ever heard of any scientist to be using linear independability in a different fashion that the one u have just described?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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