Question 1174344: Q−4: [6+4 marks] Let S={v_1,v_2,v_3} be a linearly independent set of vectors in〖 R〗^n and T={u_1,u_2,u_3}, where u_1=v_1+v_2+v_3, u_2=v_2+v_3 and u_3=v_3.
a-Determine whether each of v_1,v_2 and v_3 is a linearly combination of vectors in T.
b-Show that T is linearly independent set.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! You've asked this question before, and I provided a detailed solution. Let's recap the key points:
**a) Determining if v1, v2, and v3 are Linear Combinations of T**
* **v3:**
* Directly, we have u3 = v3. Therefore, v3 = 1 * u3, so v3 is a linear combination of T.
* **v2:**
* We have u2 = v2 + v3.
* We also know v3 = u3.
* Substituting, u2 = v2 + u3.
* Rearranging, v2 = u2 - u3.
* Therefore, v2 is a linear combination of T.
* **v1:**
* We have u1 = v1 + v2 + v3.
* We also know v2 = u2 - u3 and v3 = u3.
* Substituting, u1 = v1 + (u2 - u3) + u3.
* Simplifying, u1 = v1 + u2.
* Rearranging, v1 = u1 - u2.
* Therefore, v1 is a linear combination of T.
**Conclusion:** v1, v2, and v3 are all linear combinations of the vectors in T.
**b) Showing that T is a Linearly Independent Set**
To show that T is linearly independent, we need to prove that the only solution to the equation:
c1 * u1 + c2 * u2 + c3 * u3 = 0
is c1 = c2 = c3 = 0.
* **Substitute u1, u2, and u3:**
* c1 * (v1 + v2 + v3) + c2 * (v2 + v3) + c3 * (v3) = 0
* **Distribute and group:**
* c1 * v1 + (c1 + c2) * v2 + (c1 + c2 + c3) * v3 = 0
* **Use Linear Independence of S:**
* Since S = {v1, v2, v3} is linearly independent, the coefficients must all be zero:
* c1 = 0
* c1 + c2 = 0
* c1 + c2 + c3 = 0
* **Solve for c1, c2, and c3:**
* From c1 = 0, we know c1 is 0.
* Substituting c1 = 0 into c1 + c2 = 0, we get 0 + c2 = 0, so c2 = 0.
* Substituting c1 = 0 and c2 = 0 into c1 + c2 + c3 = 0, we get 0 + 0 + c3 = 0, so c3 = 0.
* **Conclusion:**
* The only solution is c1 = 0, c2 = 0, and c3 = 0. Therefore, T = {u1, u2, u3} is a linearly independent set.
If you have any further questions or would like me to elaborate on any aspect of the solution, feel free to ask.
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