Linear Independence

Let V be a finite vector space over the set of rationals, and F be a linear transformation from V to V. Suppose x, y and z are vectors in V such that

(i) F(x) = y

(ii) F(y) = z

(iii) F(z) = x+y

Suppose x is nonzero. Show that x, y and z are linearly independent vectors.

Proof: We first want to show that x and y are linearly independent. Suppose not. Then there exists a nonzero rational a such that y = ax.

Applying F to this equation yields z = ay = a(ax) = a^2 x. Thus the three vectors can be written x,ax and a^2x .

Now, consider F(z) = x + y = x + ax = x(1+a). On the other hand, F(z) = F(a^2 x) = a^2 F(x) = a^2y = a^3x.

So that by transitivity, (a^3-a-1) x=0. Since x is nonzero, a^3-a-1=0. However, you can check that this polynomial does not have rational roots (e.g. by remainder theorem). Thus a does not exist and therefore x and y are linearly independent.

It is now left to show that z is linearly independent to x and y. Suppose there exists rational numbers a and b such that z = ax+by. Applying F twice gives us y+z = az + b(x+y). That is, (1-a)z = bx + (b-1)y.

a cannot be equal to 1, otherwise, x and y would be linearly dependent, a contradiction. Thus we can write z = \displaystyle\frac{b}{1-a}x + \frac{b-1}{1-a}y. But x and y are linearly independent so a representation of z in terms of them will be unique. Finally, we will now compare coefficients and come up with a = \displaystyle\frac{b}{1-a} and b=\displaystyle\frac{b-1}{1-a}, which simplify to a-a^2 = b and ab =1. Solving for a and b, we get a^3-a^2+1=0 and b^3-b+1. These polynomials have no rational roots. Thus, by the same argument as of that of the above claim, a and b do not exist and therefore we can conclude that x,y and z are linearly independent.

This entry was posted on Sunday, April 6th, 2008 at 3:40 pm and is filed under Linear Algebra. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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