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Prove that some properties are still valid with an alternate definition of the dot product

by RoRi

If we define the dot product by the formula

    \[ (a_1, a_2, a_3) \cdot (b_1, b_2, b_3) = 2a_1 b_1 + a_2 b_2 + a_3 b_3 + a_1 b_3 + a_3 b_1 \]

prove that the properties of the dot product in Theorem 12.2 (page 451 of Apostol) continue to be valid.

Proof. Let A = (a_1, a_2, a_3), B = (b_1, b_2, b_3) and C = (c_1, c_2, c_3).

  1. We compute,

        \begin{align*} A \cdot B &= 2a_1 b_1 + a_2 b_2 + a_3 b_3 + a_1 b_3 + a_3 b_1 \\ &= 2b_1 a_1 + b_2 a_2 + b_3 a_3 + b_3 a_1 + b_1 a_3 \\ &= B \cdot A. \end{align*}

  2. We compute

        \begin{align*} A \cdot (B+C) &= A \cdot (b_1+c_1, b_2 + c_2, b_3 + c_3) \\ &= 2a_1 (b_1 + c_1) + a_2(b_2 + c_2) + a_3 (b_3 + c_3) + a_1(b_3 + c_3) + a_3 (b_1 + c_1) \\ &= 2a_1 b_1 + 2a_1 c_1 + a_2 b_2 + a_2 c_2 + a_3 b_3 + a_3 c_3 + a_1 b_3 + a_1 c_3 + a_3 b_1 + a_3 c_1 \\ &= (2a_1 b_1 + a_2 b_2 + a_3 b_3 + a_1 b_3 + a_3 b_1) + (2a_1 c_1 + a_2 c_2 + a_3 c_3 + a_1 c_3 + a_3 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_3 b_3 + a_1 b_3 + a_3 b_1) \\ &= 2(ca_1)b_1 + (ca_2)b_2 + (ca_3)b_3 + (ca_1)b_3 + (ca_3) b_1 \\ &= (cA) \cdot B. \end{align*}

    On the other hand,

        \begin{align*} c (A \cdot B) &= 2a_1 (cb_1) + a_2 (cb_2) + a_3 (cb_3) + a_1(cb_3) + a_3 (cb_1) \\ &= A \cdot (cB). \end{align*}

  4. We compute,

        \begin{align*} A \cdot A &= 2a_1^2 + a_2^2 + a_3^2 + a_1 a_3 + a_3 a_1 \\ &= a_1^2 + a_2^2 + (a_1 + a_3)^2 \\ &> 0 \end{align*}

    if A \neq O.

  5. If A = O then

        \[ A \cdot A = 2a_1^2 + a_2^2 + a_3^2 + a_1 a_3 + a_3 a_1 = 0. \qquad \blacksquare \]

Again, Cauchy-Schwartz will hold since the proof of Cauchy-Schwarz relied only on properties (a)-(e), which are all still valid under this new definition of the dot product.

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