91Ó°ÊÓ

Suppose we have two linear combinations: 1. \(a_1u + a_2v = 0\) 2. \(b_1(u + v) + b_2(u - v) = 0\)

Assume {u, v} is linearly independent. This means that the only solution to equation (1) is when \(a_1=0\) and \(a_2=0\). Now, we will work with equation (2): \(b_1(u + v) + b_2(u - v) = (b_1+b_2)u + (b_1-b_2)v = 0\) Comparing this with equation (1), we get: \(a_1 = b_1+b_2\) and \(a_2 = b_1-b_2\) Since {u, v} is linearly independent, the only solution is \(a_1=0\) and \(a_2=0\). Therefore, we conclude that \(b_1+b_2=0\) and \(b_1-b_2=0\). Solving for \(b_1\) and \(b_2\), we find that \(b_1=0\) and \(b_2=0\). Thus, {u + v, u - v} is also linearly independent.

Assume {u + v, u - v} is linearly independent. This means that the only solution to equation (2) is when \(b_1=0\) and \(b_2=0\). Now, we will work with equation (1): \(a_1u + a_2v = (a_1+a_2)(u + v) + (a_1-a_2)(u - v) = 0\) Comparing this with equation (2), we get: \(b_1 = a_1+a_2\) and \(b_2 = a_1-a_2\) Since {u + v, u - v} is linearly independent, the only solution is \(b_1=0\) and \(b_2=0\). Therefore, we conclude that \(a_1+a_2=0\) and \(a_1-a_2=0\). Solving for \(a_1\) and \(a_2\), we find that \(a_1=0\) and \(a_2=0\). Thus, {u, v} is also linearly independent. In conclusion, we have proved that {u, v} is linearly independent if and only if {u + v, u - v} is linearly independent. #Part (b): Proving the equivalence of linear independence between {u, v, w} and {u + v, u + w, u + w}# (Note: I'll assume the third vector in the second set is indeed "u + w".)

Suppose we have two linear combinations: 3. \(c_1u + c_2v + c_3w = 0\) 4. \(d_1(u + v) + d_2(u + w) + d_3(u + w) = 0\)

Assume {u, v, w} is linearly independent. This means that the only solution to equation (3) is when \(c_1=0\), \(c_2=0\), and \(c_3=0\). Now, we will work with equation (4): \(d_1(u + v) + d_2(u + w) + d_3(u + w) = (d_1 + d_2 + d_3)u + (d_1)v + (d_2 + d_3)w = 0\) Comparing this with equation (3), we get: \(c_1 = d_1 + d_2 + d_3\), \(c_2 = d_1\), and \(c_3 = d_2 + d_3\) Since {u, v, w} is linearly independent, the only solution is \(c_1=0\), \(c_2=0\), and \(c_3=0\). Therefore, we conclude that \(d_1 + d_2 + d_3=0\), \(d_1=0\), and \(d_2 + d_3=0\). Solving for \(d_1\), \(d_2\), and \(d_3\), we find that \(d_1=0\), \(d_2=0\), and \(d_3=0\). Thus, {u + v, u + w, u + w} is also linearly independent.

Assume {u + v, u + w, u + w} is linearly independent. This means that the only solution to equation (4) is when \(d_1=0\), \(d_2=0\), and \(d_3=0\). Now, we will work with equation (3): \(c_1u + c_2v + c_3w = (c_1 + c_2)(u + v) + (c_1 + c_3)(u + w) + (-c_2 - c_3)(u + w) = 0\) Comparing this with equation (4), we get: \(d_1 = c_1 + c_2\), \(d_2 = c_1 + c_3\), and \(d_3 = -c_2 - c_3\) Since {u + v, u + w, u + w} is linearly independent, the only solution is \(d_1=0\), \(d_2=0\), and \(d_3=0\). Therefore, we conclude that \(c_1 + c_2=0\), \(c_1 + c_3=0\), and \(-c_2 - c_3=0\). Solving for \(c_1\), \(c_2\), and \(c_3\), we find that \(c_1=0\), \(c_2=0\), and \(c_3=0\). Thus, {u, v, w} is also linearly independent. In conclusion, we have proved that {u, v, w} is linearly independent if and only if {u + v, u + w, u + w} is linearly independent.