Home » norms

Tag: norms

Prove some properties of two alternative definitions of the norm

Consider the following two definitions of the norm of a vector A = (a_1, \ldots, a_n) in \mathbb{R}^n.

    \[ \lVert A \rVert_1 = \sum_{k=1}^n |a_k|, \qquad \text{and} \qquad \lVert A \rVert_2 = \max_{1 \leq k \leq n} |a_k|. \]

Prove that we have the following inequalities for any vector A,

    \[ \lVert A \rVert_2 \leq \lVert A \rVert \leq \lVert A \rVert_1. \]

Give a geometric interpretation of this in the case that n = 2.


Proof. First, for the inequality on the left we have

    \begin{align*}  \lVert A \rVert_2 &= \max_{1 \leq k \leq n} |a_k| \\  &= \max_{1 \leq k \leq n} \left( a_k^2 \right)^2 \\  &\leq \sum_{k=1}^n (a_k^2)^{\frac{1}{2}} \\  &= \lVert A \rVert. \end{align*}

For the inequality on the right, consider

    \begin{align*}  \left(\sqrt{a_1^2 + a_2^2} \right)^2 &= a_1^2 + a_2^2 \\  &\leq a_1^2 + a_2^2 + 2 \sqrt{a_1^2}\sqrt{a_2^2} \\  &= \left( \sqrt{a_1^2} +\sqrt{a_2^2} \right)^2 \end{align*}

Taking square roots of both sides this gives us the inequality

    \[ \sqrt{a_1^2 + a_2^2} \leq \sqrt{a_1^2} + \sqrt{a_2^2}. \]

By induction we can then establish

    \[ \sqrt{a_1^2+\cdots+a_n^2} \leq \sqrt{a_1^2} + \cdots + \sqrt{a_n^2}. \]

Therefore,

    \begin{align*}  \lVert A \rVert &= \left( \sum_{k=1}^n a_k^2 \right)^{\frac{1}{2}} \\[9pt]  &\leq \sum_{k=1}^n \sqrt{a_k^2} \\[9pt]  &= \sum_{k=1}^n |a_k| \\[9pt]  &= \lVert A \rVert_1. \qquad \blacksquare \end{align*}

Prove some properties of a new definition of the norm

Consider an alternative definition for the norm of a vector A = (a_1, \ldots, a_n) given by

    \[ \lVert A \rVert = \max_{1 \leq k \leq n} | a_k |. \]

  1. Which properties of Theorems 12.4 and 12.5 (pages 453-454 of Apostol) are still valid with this new definition?
  2. Draw a figure which shows the set of points (x,y) \in \mathbb{R}^2 of norm 1 with this new definition.

  1. Claim. All of the properties of Theorems 12.4 and 12.5 hold with this new definition.
    Proof. For Theorem 12.4(a), if A \neq O then there is some a_i \neq 0 so \max_{1 \leq k \leq n} |a_k| > 0. Therefore \lVert A \rVert > 0.

    For Theorem 12.4(b), if A = O then a_i = 0 for all 1 \leq i \leq n. Therefore \max_{1 \leq k \leq n} |a_k| = 0 and so \lVert A \rVert = 0.

    For Theorem 12.4(c), we compute

        \begin{align*}  \lVert cA \rVert &= \max_{1 \leq k \leq n} |ca_k| \\  &= \max_{1 \leq k \leq n} |c| |a_k| \\  &= |c| \max_{1 \leq k \leq n} |a_k| \\  &= |c| \lVert A \rVert. \end{align*}

    For Theorem 12.5 we have

        \begin{align*}  \lVert A + B \rVert &= \max_{1 \leq k \leq n} | a_k + b_k | \\  &\leq \max_{1 \leq k \leq n} \left( |a_k| + |b_k| \right) \\  &\leq \max_{1 \leq k \leq n} |a_k| + \max_{1 \leq k \leq n} |b_k| \\  &= \lVert A \rVert + \lVert B \rVert. \qquad \blacksquare \end{align*}

  2. We have the following figure,

    Rendered by QuickLaTeX.com

Prove some properties of an alternative definition of the norm of a vector

Consider an alternative definition for the norm of a vector A = (a_1, \ldots, a_n) given by

    \[ \lVert A \rVert = \sum_{k=1}^n |a_k|. \]

  1. Prove that this definition satisfies all of the properties of Theorems 12.4 and 12.5 (pages 453-454 of Apostol).
  2. Consider this definition in \mathbb{R}^2 and draw the set of all points (x,y) which have norm 1.
  3. If we defined

        \[ \lVert A \rVert = \left| \sum_{k=1}^n a_k \right| \]

    which of the properties of Theorems 12.4 and 12.5 would hold?


  1. Proof. For Theorem 12.4(a) we have

        \[ \lVert A \rVert > 0 \]

    if A \neq O since all of the terms in the sum are greater than or equal to 0, with at least one non-zero since A \neq O.

    For Theorem 12.4(b) if A = O then \lVert A \rVert = \sum_{k=1}^n |0| = 0.

    For Theorem 12.4(c) we have

        \[ \lVert cA \rVert = \sum_{k=1}^n |ca_k| = \sum_{k=1}^n |c| |a_k| = |c| \sum_{k=1}^n |a_k| = |c| \lVert A \rVert. \]

    For Theorem 12.5 we have

        \begin{align*}  \lVert A + B \rVert &= \sum_{k=1}^n |a_k + b_k| \\  &\leq \sum_{k=1}^n (|a_k| + |b_k|) \\  &= \sum_{k=1}^n |a_k| + \sum_{k=1}^n |b_k| \\  &= \lVert A \rVert + \lVert B \rVert. \qquad \blacksquare \end{align*}

  2. We have the following diagram

    Rendered by QuickLaTeX.com

  3. Property 12.4(a) fails since if we take A = (-1,1) then \lVert A \rVert = 0, but A \neq O.
    Property 12.4(b) holds since if A = O then \lVert A \rVert =0.
    Property 12.4(c) holds since

        \begin{align*}  \lVert cA \rVert &= \left| \sum_{k= 1}^n ca_k \right| \\[9pt]  &= \left| c \sum_{k=1}^n a_k \right| \\[9pt]  &= |c| \left| \sum_{k=1}^n a_k \right| \\[9pt]  &= |c| \lVert A \rVert. \end{align*}

    Property 12.5 holds since

        \begin{align*}  \lVert A +B \rVert &= \left| \sum_{k=1}^n (a_k + b_k) \right| \\[9pt]  &= \left| \sum_{k=1}^n a_k + \sum_{k=1}^n b_k \right| \\[9pt]  &\leq \left| \sum_{k=1}^n a_k \right| + \left| \sum_{k=1}^n b_k \right| \\[9pt]  &= \lVert A \rVert + \lVert B \rVert.  \end{align*}

Find a vector satisfying given relations with a given vector

Find a vector B \in \mathbb{R}^2 such that B \cdot A = 0 and \lVert B \rVert = \lVert A \rVert for the following vectors A,

  1. A = (1,1);
  2. A = (1,-1);
  3. A = (2,-3);
  4. A = (a,b).

  1. If A = (1,1) then letting B = (b_1, b_2) we have

        \[ B \cdot A = 0 \quad \implies \quad b_1 + b_2 = 0 \quad \implies \quad b_2 = -b_1. \]

    Since \lVert A \rVert = \sqrt{2} we have \lVert B \rVert = \sqrt{2} therefore,

        \[ \sqrt{b_1^2 + b_2^2} =\sqrt{2} \quad \implies \quad 2b_1^2 = 2 \quad \implies \quad b_1 = \pm 1, \ b_2 = \mp 1. \]

    Therefore, B = (1,-1) or B = (-1,1).

  2. If A = (1,-1) then letting B = (b_1, b_2) we have

        \[ B \cdot A = 0 \quad \implies \quad b_1 - b_2 = 0 \quad \implies \quad b_2 = b_1. \]

    Since \lVert A \rVert = \sqrt{2} we have \lVert B \rVert = \sqrt{2} therefore,

        \[ \sqrt{b_1^2 + b_2^2} =\sqrt{2} \quad \implies \quad 2b_1^2 = 2 \quad \implies \quad b_1 = \pm 1, \ b_2 = \pm 1. \]

    Therefore, B = (1,1) or B = (-1,-1).

  3. If A = (2,-3) then letting B = (b_1, b_2) we have

        \[ B \cdot A = 0 \quad \implies \quad 2b_1 - 3b_2 = 0 \quad \implies \quad b_1 = \frac{3}{2} b_2. \]

    Since \lVert A \rVert = \sqrt{13} we have \lVert B \rVert = \sqrt{13} therefore,

        \[ \sqrt{b_1^2 + b_2^2} =\sqrt{13} \quad \implies \quad \sqrt{ \left( \frac{9}{4} \right) b_2^2 + b_2^2 } = \sqrt{13} \quad \implies \quad b_2 = \pm 2, \ b_2 = \pm 3. \]

    Therefore, B = (3,2) or B = (-3,-2).

  4. If A = (a,b) then letting B = (b_1, b_2) we have

        \[ B \cdot A = 0 \quad \implies \quad ab_1 + bb_2 = 0 \quad \implies \quad b_1 = \frac{b}{a} b_2. \]

    Since \lVert A \rVert = \sqrt{a^2 + b^2} we have \lVert B \rVert = \sqrt{a^2 + b^2} therefore,

        \[ \sqrt{b_1^2 + b_2^2} =\sqrt{2} \quad \implies \quad \sqrt{ \left( \frac{b}{a} \right) b_2^2 + b_2^2 } = \sqrt{a^2+b^2} \quad \implies \quad b_1= \pm b \quad b_2 = \mp a. \]

    (The final equalities follow from b_2^2 = \pm (a^2+b^2) \left( \left(\frac{b}{a}\right)^2 + 1 \right)^{-1}.) Therefore we have B = (b,-a) or B = (-b,a).