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Find a linearly independent subset of each of the given sets of vectors

Find a maximal linearly independent subset of each of the following sets of vectors in \mathbb{R}^4.

  1. \{ (1,0,1,0), (1,1,1,1), (0,1,0,1), (2,0,-1,0) \}.
  2. \{ (1,1,1,1), (1,-1,1,1), (1,-1,-1,1), (1,-1,-1,-1) \}.
  3. \{ (1,1,1,1), (0,1,1,1), (0,0,1,1), (0,0,0,1) \}.

  1. This set is a dependent set since (1,1,1,1) = (1,0,1,0) + (0,1,0,1); hence, we cannot find a subset of size 4. Then, consider the three element subset \{ (1,0,1,0), (0,1,0,1), (2,0,-1,0) \}. This set is independent since

        \begin{align*}  x(1,0,1,0) + y(0,1,0,1) + z(2,0,-1,0) = O && \implies && x+2z &= 0 \\  && && y &= 0 \\  && && x-z &= 0. \end{align*}

    The third equation gives us x = z; hence, by the first equation x = z = 0. Therefore, x = y = z = 0 and the set is independent.

  2. The set \{ (1,1,1,1), (1,-1,1,1), (1,-1,-1,1), (1,-1,-1,-1) \} is independent since the equation

        \[ c_1 (1,1,1,1) + c_2(1,-1,1,1) + c_3 (1,-1,-1,1) + c_4 (1,-1,-1,-1) = (0,0,0,0) \]

    gives us the equations

        \begin{align*}  c_1 + c_2 + c_3 + c_4 &= 0 \\  c_1 - c_2 - c_3 - c_4 &= 0 \\  c_1 + c_2 - c_3 - c_4 &= 0 \\  c_1 + c_2 + c_3 - c_4 &= 0. \end{align*}

    The only solution to this system is c_1 = c_2 = c_3 = c_4 = 0; hence, this is a linearly independent subset.

  3. The set \{ (1,1,1,1), (0,1,1,1), (0,0,1,1), (0,0,0,1) \} is linearly independent since

        \[ c_1 (1,1,1,1) + c_2(0,1,1,1) + c_3 (0,0,1,1) + c_4 (0,0,0,1) = (0,0,0,0) \]

    gives us the system of equations

        \begin{align*}  c_1 &= 0 \\  c_1 + c_2 &= 0 & \implies && c_2 &= 0 \\  c_1 + c_2 + c_3 &=0 & \implies && c_3 &=0 \\  c_1 + c_2 + c_3 + c_4 &=0 & \implies && c_4 &= 0. \end{align*}

    Thus, this set is linearly independent.

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