Tag: norms

Prove that the norm of the cross product is the product of the norms if and only if A and B are orthogonal

by RoRi

May 1, 2016

Given vectors $A,B$ prove that

$\lVert A \times B \rVert =\lVert A \rVert \lVert B \rVert$

if and only if $A$ and $B$ are orthogonal.

Proof.

From Theorem 13.12(f) (page 483 of Apostol) we know

$\lVert A \times B \rVert^2 = \lVert A \rVert^2 + \lVert B \rVert^2 - (A \cdot B)^2.$

But $(A \cdot B)^2 = 0$ if and only if $A$ and $B$ are orthogonal (from the definition of orthogonality). Thus, $\lVert A \times B \rVert = \lVert A \rVert \lVert B \rVert$ if and only if $A$ and $B$ are orthogonal $. \qquad \blacksquare$

Prove some properties of two alternative definitions of the norm

by RoRi

April 28, 2016

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

by RoRi

April 28, 2016

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

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

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

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

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

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

by RoRi

April 28, 2016

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

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

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

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

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

by RoRi

April 26, 2016

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 $A$ ,

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

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 $B = (-1,1)$ .
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 $B = (-1,-1)$ .
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 $B = (-3,-2)$ .
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 $B = (-b,a)$ .

Compute the norms of given vectors

by RoRi

April 26, 2016

Given vectors $A = (2,-1,5)$ , $B = (-1,-2,3)$ and $C = (1,-1,1)$ in $\mathbb{R}^3$ calculate the norms of the following vectors

$A+B$ ;
$A - B$ ;
$A + B- C$ ;
$A - B + C$ ;

We compute
$\lVert A+B \rVert = \sqrt{ (A+B)\cdot(A+B)} = \sqrt{(1,-3,8) \cdot (1,-3,8)} = \sqrt{74}.$
We compute
$\lVert A-B \rVert = \sqrt{ (A-B) \cdot (A-B)} = \sqrt{(3,1,2) \cdot (3,1,2)} = \sqrt{14}.$
We compute
$\lVert A+B-C \rVert = \sqrt{(0,-2,7) \cdot (0,-2,7)} = \sqrt{53}.$
We compute
$\lVert A-B+C \rVert = \sqrt{(4,0,3) \cdot (4,0,3)} = \sqrt{25} = 5.$