Proving Vector Addition with Linear Independence

  • Thread starter Thread starter b2386
  • Start date Start date
  • Tags Tags
    Proof
b2386
Messages
34
Reaction score
0
Hi all,

While working on my differential equations homework, I encountered a proof dealing with linear independence and vector addition. I sort of know how to proceed but, not having dealt with formal proofs much, I am afraid that I may not be addressing all necessary apects of the proof. Anyway, here is the question: Prove that if the vectors x = (x_1)i + (x_2)j and y = (y_1)i + (y_2)j
are linearly independent, then any vector z = (z_1)i + (z_2)j can be expressed as a linear combination of x and y.

The linear combination of x and y gives us (x_1)i + (x_2)j + (y_1)i + (y_2)j. Rearranging terms, [(x_1)+(y_1)]i + [(x_2)+(y_2)]j = x+y. We can now define x+y = z. Therefore, z = (z_1)i + (z_2)j

Where do I bring in the necessity of linear independence?
 
Physics news on Phys.org
b2386 said:
Prove that if the vectors x = (x_1)i + (x_2)j and y = (y_1)i + (y_2)j are linearly independent, then any vector z = (z_1)i + (z_2)j can be expressed as a linear combination of x and y.

Actually, this question is kind of 'definition-like'. You have two vectors in V^2(O). Any set of two vectors in V^2(O) which are linearly independent (i.e. non collinear) form a basis for V^2(O), and hence every vector from V^2(O) can be represented uniquely as a linear combination of these two independent vectors.
 
What exactly is V^2(O)? I haven't had linear algebra so I am probably unfamiliar with some terminology.

EDIT: Is that just a 2-D vector space?
 
Last edited:
b2386 said:
What exactly is V^2(O)? I haven't had linear algebra so I am probably unfamiliar with some terminology.

My apologies for not pointing it out - V^2(O) (or call it whatever you like) is the set of all radius vectors in the Euclidean plane, where your story is, of course, set up in a Cartesian coordinate system.

EDIT: it could be, but be careful when using that terminology; formally, a 2-D vector space is any 2-dimensional vector space - its elements don't need to be radius vectors!
 
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...
Back
Top