Let V be the linear space consisting of all realvalued funct

Let V be the linear space consisting of all real-valued functions defined on the real line. Determine whether each of the following subsets of V is dependent or independent. Compute the dimension of the subspace spanned by each set. (a) {1, e^3x, e^x} (b) {sin x, sin 2x} (c) {e^x, e^-x, cosh x} (d) {1, cos2x, sin^2 x}

Solution

(a) If the given functions are linearly independent , then the equation p*1 + q*e^3x + r e^x = z(x) , (where p,q, r are arbitrary scalars and where z (x) is the zero function, should have only a trivial solution for p, q,r .

If x = 0, then p + q + r = 0…(1)

If x = 1, then p + qe³ + re = 0…(2)

If x = 1/3, then p + qe + re^1/3 = 0…(3)

From the 1^st equation, we have p = - q – r. On substituting this value of p in the 2^nd and the 3^rd equations, we have – q – r + qe³ + re = 0   or, q (e³ -1) + r ( e -1) = 0 …(4) and – r + qe + re^{1/ 3} = 0   or, q ( e-1) + r( e^1/3 – 1) = 0…(5)

On multiplying the equations, 4 and 5, by (e-1) and ( e³ -1), we get q(e³ -1)( e -1) + r ( e-1)² = 0…(6)   and q ( e-1) ( e³ -1) + r ( e^1/3 – 1) ( e³ -1) = 0…(7)

Now, on subtracting equation 6 from equation 5, we get r [( e-1)² - ( e^1/3 – 1) ( e³ -1)] = 0 so that r = 0( as the coefficient of r is not 0) . Then, from equation 4, we get q (e³ -1) = 0, so that q = 0. Now, from equation 1, we get p = -q – r = 0. Thus, the equation p*1 + q*e^3x + r e^x = z(x) has only a trivial solution so that the given functions are linearly independent.

(b) If the given functions are linearly independent , then the equation p sinx + q sin2x = z (x) (where p,q are arbitrary scalars and where z (x) is the zero function, should have only a trivial solution for p, q . Now, when x = /2, sinx = 1 and sin 2x = 0, so that p + q*0 = 0. Then p = 0 which implies that q = 0. Thus, the equation p sinx + q sin2x = z (x) has only a trivial solution so that the given functions are linearly independent.

(c) If the given functions are linearly independent , then the equation pe^x + qe^-x + r cosh x   = z (x) (where p,q,r are arbitrary scalars and where z (x) is the zero function, should have only a trivial solution for p, q ,r. Now, cosh x = ½( e^x + e^-x ) so that the above equation becomes pe^x + qe^-x + r ½( e^x + e^-x ) = z(x) or, ( p –r/2)e^x + ( q- r/2)e^-x = z (x) . Apparently, p = 1, q = 1 and r = -2 is a non-trivial solution of this equation. Therefore, the given functions are not linearly independent.

(d) If the given functions are linearly independent , then the equation p + q cos 2x + r sin² x   = z (x) (where p,q,r are arbitrary scalars and where z (x) is the zero function, should have only a trivial solution for p, q ,r. When x = /2, we have p – q + r = 0. Apparently, p = 1, q= 1 and r = 0 is is a non-trivial solution of this equation. Therefore, the given functions are not linearly independent.