Problem 1

1. For which real values of t do the following vectors form a linearly independent set in R⁴:

(t, 2t, 3, 4), (1, 2, 3t, 4t), (t, 2, 3t, 4), (t, 2, 3, 4t).

If this is linearly dependent, then there must exist an a, b, c, and d such that the following holds true:

a(t, 2t, 3 4)+ b(1, 2, 3t, 4t)+ c(t, 2, 3t, 4) +d(t, 2, 3, 4t) = 0

The system is linearly independent for all values of t such that this system has no solutions.

Let us represent this system as a matrix of coefficients, so that Av (where v=(a,b,c,d) ) is the system.

> with(linalg):
A:=matrix([[t,1,t,t],[2*t,2,2,2],[3,3*t,3*t,3],[4,4*t,4,4*t]]);

A :=	[ t	1	t	t ]
	[ 2t	2	2	2 ]
	[ 3	3t	3t	3 ]
	[ 4	4t	4	4t ]

> B:=det(A);

B := - 96 t³ + 96 t + 48 t⁴ - 48

> solve(B=0);

-1, 1, 1, 1

We know by a previous theorem that in order for the system to have only the trivial solution, it's determinant must be nonzero. Therefore, if the determinant of A is non-zero, there is only the trivial solution and the set of vectors is linearly independent. If the determinant is zero, the vectors are linearly dependent. Thus for every t except 1 and -1 the vectors are linearly independent.



Problem 2

2. Show that the following set of functions from C[0,1] is linearly independent: x, xe^x, x²e^x, x³e^x. We have a theorem that states that if the Wronskian of a set of functions is not identically zero, then the set of functions is linearly independent. We use this theorem to show that the above set of functions is indeed linearly independent.

Since there are four equations, we use the derivatives up to the third derivative in our matrix. The matrix of which we must find the determinant (Wronskian) is:

> A:=matrix([[x,x*e^x,x^2*e^x,x^3*e^x],[1,e^x+x*e^x,2*x*e^x+x^2*e^x,3*x^2*e^x+x^3*e^x],[0,2*e^x+x*e^x,2*e^x+4*x*e^x+x^2*e^x,6*x*e^x+6*x^2*e^x+x^3*e^x],[0,3*e^x+x*e^x,6*e^x+6*x*e^x+x^2*e^x,6*e^x+18*x*e^x+9*x^2*e^x+x^3*e^x]]);

A :=	[ x	xex	x2ex	x3ex ]
	[ 1	ex + xex	2xex + x2ex	3x2ex + x3ex ]
	[ 0	2ex + xex	2ex + 4xex + x2ex	6xex + 6x2ex + x3ex ]
	[ 0	3ex + xex	6ex + 6xex + x2ex	6ex + 18xex + 9x2ex + x3ex ]

> det(A);

2 x⁴ (e^x )³

The set of functions which we are given must be linearly independent if 2x⁴*(e^x)³ is not equal to zero. Since no values of x exist such that e^x and (e^x)³ are zero, we know that the only way for the Wronskian to be equal to zero is in the trivial case of x=0. Therefore, we know that the given set of functions is linearly independent.