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Prove or disprove some statements about vector in Rn

by RoRi

Prove or disprove:

  1. If A is orthogonal to B, then

        \[ \lVert A + xB \rVert \geq \lVert A \rVert \qquad \text{for all } x \in \mathbb{R}. \]

  2. If

        \[ \lVert A + xB \rVert \geq \lVert A \rVert \qquad \text{for all } x \in \mathbb{R} \]

    then A is orthogonal to B.

  1. Proof. Let A = (a_1, \ldots, a_n) and B = (b_1, \ldots, b_n). Then we have

        \begin{align*} \lVert A + xB \rVert &= \left( \sum_{i=1}^n (a_i + xb_i)(a_i xb_i) \right)^{\frac{1}{2}} \\[9pt] &= \left( \sum_{i=1}^n (a_i^2 + x^2 b_i^2 + 2xa_i b_i) \right)^{\frac{1}{2}}. \end{align*}

    But, A \cdot B = 0 by hypothesis, so \sum 2 x a_i b_i =0. Therefore,

        \[ \lVert A + xB \rVert = \left( \lVert A \rVert^2 + x^2 \lVert B \rVert^2 \right)^{\frac{1}{2}} \geq \lVert A \rVert^2 \]

    since x^2 \lVert B \rVert^2 \geq 0 for all real x. \qquad \blacksquare

  2. Proof. Suppose otherwise, that A \cdot B \neq 0. Then, we know that for all x \in \mathbb{R} we have

        \begin{align*} && \lVert A + xB \rVert & \geq 0 \\[9pt] \implies && \left( \sum_{i=1}^n (a_i^2 + x^2 b_i^2 + 2xa_i b_i) \right)^{\frac{1}{2}} &\geq 0\\[9pt] \implies && \left( \lVert A \rVert^2 + x^2 \lVert B \rVert^2 + 2x (A \cdot B) \right)^{\frac{1}{2}} &\geq 0 \\[9pt] \implies && x^2 \lVert B \rVert^2 + 2x (A \cdot B) &\geq 0 \\[9pt] \implies && x \lVert B \rVert^2 + 2(A \cdot B) &\geq 0 \end{align*}

    for all x \neq 0. So, in particular is must hold for both

        \[ x_1 = -\frac{3 (A \cdot B)}{\lVert B \rVert^2} \qquad \text{and} \qquad x_2 = \frac{3 (A \cdot B)}{\lVert B \rVert^2}. \]

    Both of these are nonzero since A \cdot B \neq 0 by assumption. But then,

        \[ x_1 \lVert B \rVert^2 + 2(A \cdot B) = -(A \cdot B) \geq 0 \]

    and

        \[ x_2 \lVert B \rVert^2 + 2(A \cdot B) = (A \cdot B) \geq 0. \]

    But since A \cdot B \neq 0 we cannot have both (A \cdot B) > 0 and (A \cdot B) < 0. Hence, this is a contradiction, so we must have A \cdot B = 0. \qquad \blacksquare.

4 comments

  1. Anonymous says:

    b is false, take for example x=1, and A=kB. The inequality holds without A beeing orthogonal to B (for infinite values of k).

    • Mohammad Azad says:

      I don’t know what you mean, but here is a proof that b is true:
      If B=0, the statement is clearly true, hence suppose B is not zero then norm(B)>0 now norm(A+xB)=sqrt(norm(A)^2 + 2x(A.B) + x^2norm(B)^2) >= norm(A) implies that 2x(A.B) + x^2norm(B)^2 >=0 and this is true for any real x, hence for any x>0 we may “solve” for x to get x >= -2(A.B)/norm(B)^2 this means that (A.B)>=0 otherwise, the term on the right will be a positive number less than or equal to all positive numbers which is absurd (There is a proof for this in the introduction). Similarly, if x <= 0 then by "solving" for x we get the similar inequality x <= -2(A.b)/norm(B)^2 this one implies that (A.B)=0 for otherwise -(A.B)/norm(B)^2 will be a negative number less than or equal to all negative numbers, this completes the proof.

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