Tag: Identities

Prove some vector identities using the “cab minus bac” formula

by RoRi

May 5, 2016

In the previous exercise (Section 13.14, Exercise #9) we proved the “cab minus bac” formula:

$A \times (B \times C) = (C \cdot A) B - (B \cdot A) C.$

Using this formula prove the following identities:

$(A \times B) \times (C \times D) = (A \times B \cdot D)C - (A \times B \cdot C) D$ .
$A \times (B \times C) + B \times (C \times A) + C \times (A \times B) = O$ .
$A \times (B \times C) = (A \times B) \times C$ if and only if $B \times (C \times A) = O$ .
$(A \times B) \cdot (C \times D) = (B \cdot D)(A \cdot C) - (B \cdot C)(A \cdot D)$ .

Proof. Using the cab minus bac formula with $A \times B$ in place of $A$ , $C$ in place of $B$ , and $D$ in place of $C$ we have
$\begin{align*} (A \times B) \times (C \times D) &= (D \cdot (A \times B))C - (C \cdot (A \times B))D \\ &= (A \times B \cdot D) C - (A \times B \cdot C) D. \qquad \blacksquare \end{align*}$
Proof. Applying the cab minus back formula to each of the three terms in the sum we have
$\begin{align*} A \times (B \times C) &= (C \cdot A) B - (B \cdot A)C \\ B \times (C \times A) &= (A \cdot B) C - (C \cdot B)A \\ C \times (A \times B) &= (B \cdot C) A - (A \cdot C)B. \end{align*}$

So, putting these together we have

$\begin{align*} A &\times (B \times C) + B \times (C \times A) + C \times (A \times B) \\ &= (C \cdot A)B - (B\cdot A)C + (A \cdot B)C - (C \cdot B)A + (B \cdot C)A - (A \cdot C)B \\ &= O. \qquad \blacksquare \end{align*}$
Proof. From cab minus bac we have
$A \times (B \times C) = (C \cdot A)B - (B \cdot A)C.$

Furthermore, since $(A \times B) \times C = -C \times (A \times B)$ , we can apply bac minus cab to get

$\begin{align*} (A \times B) \times C &= -C \times (A \times B) &(\text{Thm 13.12(a)}) \\ &= - ((B \cdot C)A - (A \cdot C) B) \\ &= (A \cdot C)B - (B \cdot C)A \\ &= (C \cdot A)B - (B \cdot C)A + (A \cdot B)C - (A \cdot B)C \\ &= (C \cdot A)B - (B \cdot A)C + B \times (C \times A) \\ &= A \times (B \times C) + B \times (C \times A). \end{align*}$

Therefore,

$(A \times B) \times C = A \times (B \times C) \iff B \times (C \times A) = O. \qquad \blacksquare$
Proof. From a previous exercise (Section 13.14, Exercise #7(d)) we know the identity $A \cdot B \times C = C \cdot A \times B$ . In this case we have $A \times B$ in place of $A$ , $C$ in place of $B$ and $D$ in place of $C$ . This gives us
$\begin{align*} (A \times B) \cdot (C \times D) &= D \cdot ((A \times B) \times C)\\ &= ((A \times B) \times C) \cdot D \\ &= (-C \times (A \times B)) \cdot D \\ &= (-(B \cdot C)A + (A \cdot C)B) \cdot D \\ &= (B \cdot D)(A \cdot C) - (B \cdot C)(A \cdot D). \qquad \blacksquare \end{align*}$

Prove the “cab minus bac” formula

by RoRi

May 5, 2016

The “cab minus bac” formula is the vector identity

$A \times (B \times C) = (C \cdot A)B - (B \cdot A)C.$

Let $B = (b_1, b_2, b_3)$ and $C = (c_1, c_2, c_3)$ . Prove that

$\mathbf{i} \times (B \times C) = c_1 B - b_1 C.$

This is the “cab minus bac” formula in the case $A = \mathbf{i}$ . Prove similar formulas for the special cases $A = \mathbf{j}$ and $A = \mathbf{k}$ . Put these three results together to prove the formula in general.

Proof. For the case $A = \mathbf{i}$ we have

$\begin{align*} \mathbf{i} \times (B \times C) &= \mathbf{i} \times (b_2 c_3 - b_3 c_2, b_3 c_1 - b_1 c_3, b_1 c_2 - b_2 c_1) \\ &= (0, -b_1 c_2 + b_2 c_1, b_3 c_1 - b_1 c_3) \\ &= c_1 B - b_1 C. \end{align*}$

Similarly, for $A = \mathbf{j}$ and $A = \mathbf{k}$ we have

$\begin{align*} \mathbf{j} \times (B \times C) &= \mathbf{j} \times (b_2 c_3 - b_3 c_2, b_3 c_1 - b_1 c_3, b_1 c_2 - b_2 c_1) \\ &= (b_1 c_2 - b_2 c_1, 0, -b_2 c_3 + b_3 c_2) \\ &= c_2 B - b_2 C \\ \mathbf{k} \times (B \times C) &= \mathbf{k} \times (b_2 c_3 - b_3 c_2, b_3 c_1 - b_1 c_3, b_1 c_2 - b_2 c_1) \\ &= (-b_3 c_1 + b_1 c_3, b_2 c_3 - b_3 c_2, 0) \\ &= c_3 B - b_3 C. \end{align*}$

So, if $A = a_i \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k}$ is any vector in $\mathbb{R}^3$ then we have

$\begin{align*} A \times (B \times C) &= a_1 \mathbf{i} \times (B \times C) + a_2 \mathbf{j} \times (B \times C) + a_3 \mathbf{k} \times (B \times C) \\ &= a_1 (c_1 B - b_1 C) + a_2 (c_2 B - b_2 C) + a_3 (c_3 B - b_3 C) \\ &= (a_1 c_1 + a_2 c_2 + a_3 c_3) B - (a_1 b_1 + a_2 b_2 + a_3 b_3) C \\ &= (A \cdot C)B - (A \cdot B) C. \qquad \blacksquare \end{align*}$

Prove an identity of cross products and the unit coordinate vectors

by RoRi

May 4, 2016

Prove that we have the identity:

$\mathbf{i} \times (A \times \mathbf{i}) + \mathbf{j} \times (A \times \mathbf{j}) + \mathbf{k} \times (A \times \mathbf{k}) = 2A.$

Proof. Let $A = (a_1, a_2, a_3)$ and compute,

$\begin{align*} \mathbf{i} &\times (A \times \mathbf{i}) + \mathbf{j} \times (A \times \mathbf{j}) + \mathbf{k} (A \times \mathbf{k}) \\ &= \mathbf{i} \times (0, a_3, -a_2) + \mathbf{j} \times (-a_3, 0, a_1) + \mathbf{k} \times (a_2, -a_1, 0) \\ &= (0,a_2, a_3) + (a_1, 0, a_3) + (a_1, a_2, 0) \\ &= 2(a_1, a_2, a_3) \\ &= 2A. \end{align*}$

Prove an identity for the angle between vectors in Cⁿ

by RoRi

April 28, 2016

The angle $\theta$ between two vectors non-zero $A, B \in \mathbb{C}^n$ is defined by the equation

$\theta = \arccos \frac{\frac{1}{2}(A \cdot B + \overline{A \cdot B})}{\lVert A \rVert \lVert B \rVert}.$

The inequality

$-2 \leq \frac{A \cdot B + \overline{A \cdot B}}{\lVert A \rVert \lVert B \rVert} \leq 2$

we established in the previous exercise (Section 12.17, Exercise #6) show that there is a unique $\theta \in [0, \pi]$ satisfying this equation. Prove that we have

$\lVert A - B \rVert^2 = \lVert A \rVert^2 + \lVert B \rVert^2 - 2 \lVert A \rVert \lVert B \rVert \cos \theta.$

Proof. From the definition of $\theta$ we have

$\begin{align*} && \theta &= \arccos \frac{\frac{1}{2} (A \cdot B + \overline{A \cdot B})}{\lVert A \rVert \lVert B \rVert} \\[9pt] \implies && \cos \theta &= \frac{A \cdot B + \overline{A \cdot B}}{2 \lVert A \rVert \lVert B \rVert} \\[9pt] \implies && 2 \lVert A \rVert \lVert B \rVert \cos \theta &= A \cdot B + \overline{A \cdot B}. \end{align*}$

But then we know from this exercise (Section 12.17, Exercise #3) that

$A \cdot B + \overline{A \cdot B} = \lVert A+B \rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2.$

And, we know from this exercise (Section 12.17, Exercise #5) that

$\lVert A+ B\rVert^2 = 2\lVert A \rVert^2 + 2 \lVert B \rVert^2 - \lVert A - B \rVert^2.$

Therefore,

$\begin{align*} && 2 \lVert A \rVert \lVert B \rVert \cos \theta &= \lVert A + B\rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2 \\ \implies && 2 \lVert A \rVert \lVert B \rVert \cos \theta &= 2 \lVert A \rVert^2 + 2 \lVert V \rVert^2 - \lVert A - B \rVert^2 - \lVert A \rVert^2 - \lVert B \rVert^2 \\[9pt] \implies && \lVert A - B \rVert^2 &= \lVert A \rVert^2 + \lVert B \rVert^2 - 2 \lVert A \rVert \lVert B \rVert \cos \theta. \qquad \blacksquare \end{align*}$

Prove some facts about vectors in Cⁿ

by RoRi

April 28, 2016

Prove that for $A, B \in \mathbb{C}^n$ we have
$\overline{A \cdot B} + A \cdot B \in \mathbb{R}.$
For non-zero vectors $A,B \in \mathbb{C}^n$ prove that
$-2 \leq \frac{A \cdot B + \overline{A \cdot B}}{\lVert A \rVert \lVert B \rVert} \leq 2.$

Proof. Let $A \cdot B = x + iy$ for $x, y \in \mathbb{R}$ .
$A \cdot B + \overline{A \cdot B} = (x+iy) + (x-iy) = 2x \in \mathbb{R}. \qquad \blacksquare$
Proof. Let $A \cdot B = x + iy$ for $x,y \in \mathbb{R}$ . Then,
$A \cdot B + \overline{A \cdot B} = 2x$

and

$\lVert A \rVert \lVert B \rVert = \left( A \cdot A \right)^{\frac{1}{2}} \left( B \cdot B \right)^{\frac{1}{2}} = \left( (A \cdot B)(A \cdot B) \right)^{\frac{1}{2}} = \sqrt{x^2+y^2}.$

Therefore,

$\begin{align*} \frac{A \cdot B + \overline{A \cdot B}}{\lVert A \rVert \lVert B \rVert} &= \frac{2x}{\sqrt{x^2+y^2}} \\ &\leq \frac{2x}{\sqrt{x^2}} \\ &= \frac{2x}{|x|} \\ &= \pm 2. \end{align*}$

Therefore,

$-2 \leq \frac{A \cdot B + \overline{A \cdot B}}{\lVert A \rVert \lVert B \rVert} \leq 2. \qquad \blacksquare$

Prove yet another identity for vectors in Cⁿ

by RoRi

April 28, 2016

Let $A, B \in \mathbb{C}^n$ be any two vectors. Prove that we have the following identity,

$\lVert A + B \rVert^2 + \lVert A - B \rVert^2 = 2 \lVert A \rVert^2 + 2 \lVert B \rVert^2.$

Proof. Using our computations of $\lVert A + B \rVert^2$ and $\lVert A - B \rVert^2$ in the previous exercise (Section 12.17, Exercise #4) we have

$\begin{align*} \lVert A + B \rVert^2 + \lVert A - B \rVert^2 &= \lVert A \rVert^2 + \lVert B \rVert^2 + A \cdot B + \overline{A \cdot B} + \lVert A \rVert^2 + \lVert B \rVert^2 - A \cdot B - \overline{A \cdot B} \\ &= 2 \lVert A \rVert^2 + 2 \lVert B \rVert^2. \qquad \blacksquare \end{align*}$

Prove another identity for vectors in Cⁿ

by RoRi

April 28, 2016

Let $A, B \in \mathbb{C}^n$ be any two vectors. Prove that we have the following identity:

$\lVert A + B \rVert^2 - \lVert A - B \rVert^2 = 2 (A \cdot B + \overline{A \cdot B}).$

Proof. From the previous exercise (Section 12.17, #3) we have the identity

$\lVert A + B \rVert = \lVert A \rVert^2 + \lVert B \rVert^2 + A \cdot B + \overline{A \cdot B}.$

This also give us

$\begin{align*} \lVert A - B \rVert &= \lVert A \rVert^2 + \lVert (-B) \rVert^2 + A \cdot (-B) + \overline{A \cdot (-B)} \\ &= \lVert A \rVert^2 + \lVert B \rVert^2 - A \cdot B - \overline{A \cdot B}. \end{align*}$

Therefore,

$\begin{align*} \lVert A + B \rVert^2 - \lVert A - B \rVert^2 &= \lVert A \rVert^2 + A \cdot B + \overline{A \cdot B} - \lVert A \rVert^2 - \lVert B \rVert^2 + A \cdot B + \overline{A \cdot B} \\ &= 2 \left( A \cdot B + \overline{A \cdot B} \right). \qquad \blacksquare \end{align*}$

Prove an identity for vectors in Cⁿ

by RoRi

April 28, 2016

Let $A,B \in \mathbb{C}^n$ be any two vectors. Prove the following identity:

$\lVert A + B \rVert^2 = \lVert A \rVert^2 + \lVert B \rVert^2 + A \cdot B + \overline{A \cdot B}.$

Proof. We can compute, noting that $B \cdot A = \overline{A \cdot B}$ by Theorem 12.11(a),

$\begin{align*} \lVert A + B \rVert^2 &= (A+B) \cdot (A+B) \\ &= A \cdot A + A \cdot B + B \cdot A + B \cdot B \\ &= \lVert A \rVert^2 + \lVert B \rVert^2 + A \cdot B + \overline{A \cdot B}. \qquad \blacksquare \end{align*}$

Use vector methods to prove the trig identity cos (a-b) = cos a cos b + sin a sin b

by RoRi

April 28, 2016

Prove the trig identity

$\cos (a-b) = \cos a \cos b + \sin a \sin b$

by taking the dot product of the vectors $(\cos a, \sin a)$ and $(\cos b, \sin b)$ .

Proof.
The angle between the vectors $(\cos a, \sin a)$ and $(\cos b, \sin b)$ is given by

$\cos (a-b) = \frac{(\cos a, \sin a) \cdot(\cos b, \sin b)}{1} = \cos a \cos b + \sin a \sin b. \qquad \blacksquare$

Prove an identity of given finite sums

by RoRi

December 19, 2015

Prove the identity:

$\sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{k+m+1} = \sum_{k=0}^m (-1)^k \binom{m}{k} \frac{1}{k+n+1}.$

Proof. Using the hint (that $\frac{1}{k+m+1} = \int_0^1 t^{k+m} \, dt$ ) we start with the expression on the left,

$\begin{align*} \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{k+m+1} &= \sum_{k=0}^n (-1)^k \binom{n}{k} \int_0^1 t^{m+k} \, dt \\[10pt] &= \sum_{k=0}^n \int_0^1 (-1)^k \binom{n}{k} t^m t^k \, dt \\[10pt] &= \int_0^1 \sum_{k=0}^n (-1)^k \binom{n}{k} t^m t^k \, dt &(\text{finite sum}) \\[10pt] &= \int_0^1 t^m \sum_{k=0}^n \binom{n}{k} (-t)^k \, dt \\[10pt] &= \int_0^1 t^m \sum_{k=0}^n \binom{n}{k} (-t)^k (1)^{n-k} \, dt \\[10pt] &= \int_0^1 t^m (1-t)^n \, dt &(\text{Binomial theorem}). \end{align*}$

(The interchange of the sum and integral is fine since it is a finite sum. Those planning to take analysis should note that this cannot always be done in the case of infinite sums.) Now, we have a reasonable integral, but we still want to get everything back into the form of the sum on the right so we make the substitution $s = 1-t$ , $ds = -dt$ . This gives us new limits of integration from 1 to 0. Therefore, we have

$\begin{align*} \sum_{k=0}^n (-1)^k \binom{n}{k} \frac{1}{k+m+1} &= \int_0^1 t^m (1-t)^n \, dt \\[10pt] &= -\int_1^0 (1-s)^m s^n \, ds \\[10pt] &= \int_0^1 (1-s)^m s^n \, ds \\[10pt] &= \int_0^1 s^n \sum_{k=0}^m \binom{m}{k} (-s)^k 1^{m-k} \, ds &(\text{Binomial theorem})\\[10pt] &= \int_0^1 \sum_{k=0}^m \binom{m}{k} (-1)^k s^{k+n} \, ds \\[10pt] &= \sum_{k=0}^m (-1)^k \binom{m}{k} \int_0^1 s^{k+n} \, ds \\[10pt] &= \sum_{k=0}^m (-1)^k \binom{m}{k} \frac{1}{k+n+1}. \qquad \blacksquare \end{align*}$