Home » Blog » Prove that a given set of vectors forms a basis for C3

Prove that a given set of vectors forms a basis for C3

  1. Let

        \[ A = (1,0,0), \qquad B = (0,i,0), \qquad C = (1,1,i) \]

    be three vectors in \mathbb{C}^3. Prove that these vectors form a basis for \mathbb{C}^3.

  2. Write the vector (5,2-i,2i) as a linear combination of A,B,C.

  1. Proof. We know that any set of three linearly independent vectors in \mathbb{C}^3 will span \mathbb{C}^3, and thus form a basis. (This is from Theorem 12.10, which is valid for \mathbb{C}^n.) Thus, it is sufficient to show that A,B,C are linearly independent. To that end, let z_1, z_2, z_3 be scalars in \mathbb{C}, then

        \begin{align*}  z_1 A + z_2 B + z_3 C &= O & \implies && z_1 + z_3 &= 0 \\  &&&& iz_2 + z_3 &= 0 \\  &&&& iz_3 &= 0. \end{align*}

    From the third equation we have z_3 = 0, and so the second equation implies z_2 = 0, and finally the third equation implies z_1 = 0. Hence, z_1 = z_2 = z_3 = 0, and A,B,C are linearly independent. \qquad \blacksquare

  2. To express (5,2-i,2i) as a linear combination of A,B,C, let z_1, z_2, z_3 \in \mathbb{C} be scalars. Then,

        \begin{align*}  z_1 A + z_2 B + z_3 C &= (5,2-i,2i) & \implies && z_1 + z_3 &= 5 \\  &&&& iz_2 + z_3 &= 2-i \\  &&&& iz_3 &= 2i. \end{align*}

    From the third equation we have z_3 = 2. Plugging this into the first and second equations we get z_1 = 3 and z_2 = -1. Therefore,

        \[ (5,2-i,2i) = 3A - B + 2C. \]

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):