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Prove that given properties are still valid under an alternate definition of the dot product

by RoRi

Suppose that in \mathbb{R}^2 we use an alternate definition of the dot product,

    \[ (a_1, a_2) \cdot (b_1, b_2) = 2a_1 b_1 + a_2 b_2 + a_1 b_2 + a_2 b_1. \]

Prove that all of the properties of Theorem 12.2 (page 451 of Apostol) are still valid. Is the Cauchy-Schwarz inequality still valid?

Proof. Let A = (a_1, a_2), B = (b_1, b_2) and C = (c_1, c_2).

  1. We compute,

        \begin{align*} A \cdot B &= 2a_1 b_1 + a_2 b_2 + a_1 b_2 + a_2 b_1 \\ &= 2 b_1 a_1 + b_2 a_2 + b_2 a_1 + b_1 a_2 \\ &= B \cdot A. \end{align*}

  2. We compute,

        \begin{align*} A \cdot (B+C) &= A \cdot ((b_1 + c_1, b_2 + c_2)) \\ &= 2a_1 (b_1 + c_1) + a_2 (b_2 + c_2) + a_1 (b_2 + c_2) + a_2 (b_1 + c_1) \\ &= 2a_1 b_1 + 2a_1 c_1 + a_2 b_2 + a_2 c_2 + a_1 b_2 + a_1 c_2 + a_2 b_1 + a_2 c_1 \\ &= 2a_1 b_1 + a_2 b_2 + a_1 b_2 + a_2 b_1 + 2a_1 c_1 + a_2 c_2 + a_1 c_2 + a_2 c_1 \\ &= A \cdot B + A \cdot C. \end{align*}

  3. We compute,

        \begin{align*} c (A \cdot B) &= c (2a_1 b_1 + a_2 b_2 + a_1 b_2 + a_2 b_1) \\ &= 2(ca_1) b_1 + (ca_2)b_2 + (ca_1)b_2 + (ca_2) b_1 \\ &= (cA) \cdot B. \end{align*}

    On the other hand,

        \begin{align*} c (A \cdot B) &= c (2a_1 b_1 + a_2 b_2 + a_1 b_2 + a_2 b_1) \\ &= 2a_1 (c b_1) + a_2 (cb_2) + a_1 (cb_2) + a_2 (cb_1) \\ &= A \cdot (cB). \end{align*}

  4. We compute,

        \begin{align*} A \cdot A &= 2a_1^2 + a_2^2 + a_1 a_2 + a_2 a_1 \\ &= a_1^2 + (a_1 + a_2)^2 \\ &> 0 \end{align*}

    if A \neq O.

  5. If A = O then

        \[ A \cdot A = 2a_1^2 + a_2^2 + 2a_1 a_2 = 0. \qquad \blacksquare \]

Cauchy-Schwarz still holds since we proved the Cauchy-Schwarz inequality using properties (a)-(e). Since these properties still hold, the same proof is valid.

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