Question

Let \( u_1 = (1, 0, 0, -1) \), \( u_2 = (2, 0, 0, -1) \), \( u_3 = (0, 0, 1, -1) \), \( u_4 = (0, 0, 0, 1) \) be elements in the real vector space \( \mathbb{R}^4 \). Then, which of the following is/are TRUE?

• A. \( \{u_1, u_2, u_3, u_4\} \) is a linearly independent set in \( \mathbb{R}^4 \).
• B. \( \{u_1 - u_2, u_3 - u_4, u_4 - u_1 \} \) is NOT a linearly independent set in \( \mathbb{R}^4 \).
• C. \( \{u_1, -u_2, u_3, -u_4 \} \) is NOT a linearly independent set in \( \mathbb{R}^4 \).
• D. \( \{u_1 + u_2, u_2 + u_3, u_3 + u_4, u_4 + u_1 \} \) is a linearly independent set in \( \mathbb{R}^4 \).

Hint:

To check if a set of vectors is linearly independent, solve the equation \( c_1 u_1 + c_2 u_2 + c_3 u_3 + c_4 u_4 = 0 \) and check if the only solution is \( c_1 = c_2 = c_3 = c_4 = 0 \).

Answer 1

The Correct Option is A

Step 1: Check the linear independence of the set \( \{u_1, u_2, u_3, u_4\} \).
To check if a set of vectors is linearly independent, we set up the equation: \[ c_1 u_1 + c_2 u_2 + c_3 u_3 + c_4 u_4 = 0, \] where \( c_1, c_2, c_3, c_4 \) are scalars. The equation expands to: \[ c_1 (1, 0, 0, -1) + c_2 (2, 0, 0, -1) + c_3 (0, 0, 1, -1) + c_4 (0, 0, 0, 1) = (0, 0, 0, 0). \] This gives the following system of equations: \[ \begin{aligned} c_1 + 2c_2 &= 0,
c_3 &= 0,
-c_1 - c_2 - c_3 + c_4 &= 0,
-c_1 - c_2 - c_3 + c_4 &= 0. \end{aligned} \] Solving this system of equations, we find that the only solution is \( c_1 = c_2 = c_3 = c_4 = 0 \). Thus, the set \( \{u_1, u_2, u_3, u_4\} \) is linearly independent. Final Answer: \[ \boxed{\{u_1, u_2, u_3, u_4\} \text{ is linearly independent in } \mathbb{R}^4.} \]