91Ó°ÊÓ

To show these vectors are linearly independent, we need to find constants \(c_1, c_2,\) and \(c_3\) such that: \(c_1(u+v+w) + c_2(v+w) + c_3w = 0\) We need to prove that the only solution for \(c_1, c_2,\) and \(c_3\) is \(c_1 = c_2 = c_3 = 0\). We can first do some rearranging: \(c_1(u+v+w) + c_2(v+w) + c_3w = 0\) \(c_1u + c_1v + c_1w + c_2v + c_2w + c_3w = 0\) \((c_1)u + (c_1+c_2)v + (c_1+c_2+c_3)w = 0\) Since \(\{u, v, w\}\) is a basis for V, and we know that basis vectors are linearly independent, for the above equation to be true, we must have: \(c_1 = 0\) \(c_1+c_2 = 0\) \(c_1+c_2+c_3 = 0\) Solving these equations, we get: \(c_1 = 0\) \(c_2 = 0\) \(c_3 = 0\) Thus, the vectors \(\{u+v+w, v+w, w\}\) are linearly independent.

Now, we need to prove that the vectors \(\{u+v+w, v+w, w\}\) span V. That means we can express the original basis vectors as a linear combination of the new vectors. To prove this, let's express the original vectors \(u, v\), and \(w\) as a linear combination of \(\{u+v+w, v+w, w\}\): 1. For \(u\), we can express it as: \(u = 1 \cdot (u+v+w) -1 \cdot (v+w) + 0 \cdot w\) 2. For \(v\), we can express it as: \(v = -1 \cdot (u+v+w) + 2 \cdot (v+w) - 1 \cdot w\) 3. For \(w\), we can express it as: \(w = 0 \cdot (u+v+w) - 0 \cdot (v+w) + 1 \cdot w\) Since we can express the original basis vectors as linear combinations of the new vectors, the new vectors span the same space as the original vectors. As we have proved that the vectors \(\{u+v+w, v+w, w\}\) are linearly independent and span the vector space V, the set \(\{u+v+w, v+w, w\}\) is indeed a basis for the vector space V.