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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,

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

Prove an identity for the angle between vectors in Cn

The angle \theta between two vectors non-zero A, B \in \mathbb{C}^n is defined by the equation

    \[ \theta = \arccos \frac{\frac{1}{2}(A \cdot B + \overline{A \cdot B})}{\lVert A \rVert \lVert B \rVert}. \]

The inequality

    \[ -2 \leq \frac{A \cdot B + \overline{A \cdot B}}{\lVert A \rVert \lVert B \rVert} \leq 2 \]

we established in the previous exercise (Section 12.17, Exercise #6) show that there is a unique \theta \in [0, \pi] satisfying this equation. Prove that we have

    \[ \lVert A - B \rVert^2 = \lVert A \rVert^2 + \lVert B \rVert^2 - 2 \lVert A \rVert \lVert B \rVert \cos \theta. \]


Proof. From the definition of \theta we have

    \begin{align*}   && \theta &= \arccos \frac{\frac{1}{2} (A \cdot B + \overline{A \cdot B})}{\lVert A \rVert \lVert B \rVert} \\[9pt]  \implies && \cos \theta &= \frac{A \cdot B + \overline{A \cdot B}}{2 \lVert A \rVert \lVert B \rVert} \\[9pt]  \implies && 2 \lVert A \rVert \lVert B \rVert \cos \theta &= A \cdot B + \overline{A \cdot B}. \end{align*}

But then we know from this exercise (Section 12.17, Exercise #3) that

    \[ A \cdot B + \overline{A \cdot B} = \lVert A+B \rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2. \]

And, we know from this exercise (Section 12.17, Exercise #5) that

    \[ \lVert A+ B\rVert^2 = 2\lVert A \rVert^2 + 2 \lVert B \rVert^2 - \lVert A - B \rVert^2. \]

Therefore,

    \begin{align*}  && 2 \lVert A \rVert \lVert B \rVert \cos \theta &= \lVert A + B\rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2 \\  \implies && 2 \lVert A \rVert \lVert B \rVert \cos \theta &= 2 \lVert A \rVert^2 + 2 \lVert V \rVert^2  - \lVert A - B \rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2 \\[9pt]  \implies && \lVert A - B \rVert^2 &= \lVert A \rVert^2 + \lVert B \rVert^2 - 2 \lVert A \rVert \lVert B \rVert \cos \theta. \qquad \blacksquare \end{align*}