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Prove or disprove some statements about vector in Rn

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.

3 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).

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