Tag: Complex Numbers

Prove that the unit coordinate vectors form a basis for Cⁿ

by RoRi

April 29, 2016

The unit coordinate vectors in $\mathbb{R}^n$ are defined by

$E_i = (0,0, \ldots, 0,1,0,\ldots, 0)$

where the 1 is in the $i$ th position. We know that these form a basis for $\mathbb{R}^n$ . Prove that they also form a basis for $\mathbb{C}^n$ .

Proof. First, $E_1, \ldots, E_n$ span $\mathbb{C}^n$ since if $(z_1, \ldots, z_n) \in \mathbb{C}^n$ is any vector then

$(z_1, \ldots, z_n) = z_1 E_1 + \cdots + z_n E_n.$

Further, $E_1 ,\ldots, E_n$ are independent in $\mathbb{C}^n$ since

$\begin{align*} z_1 E_1 + \cdots + z_n E_n &= 0 & \implies && z_1 &= 0 \\ &&&& z_2 &= 0 \\ &&&& &\vdots \\ &&&& z_n &= 0. \end{align*}$

Hence, $z_1 = z_2 = \cdots = z_n =0$ . Therefore, the unit coordinate vectors form a basis for $\mathbb{C}^n.\qquad \blacksquare$

Prove that a given set of vectors forms a basis for C³

by RoRi

April 29, 2016

Let
$A = (1,0,0), \qquad B = (0,i,0), \qquad C = (1,1,i)$

be three vectors in $\mathbb{C}^3$ . Prove that these vectors form a basis for $\mathbb{C}^3$ .
Write the vector $(5,2-i,2i)$ as a linear combination of $A,B,C$ .

Proof. We know that any set of three linearly independent vectors in $\mathbb{C}^3$ will span $\mathbb{C}^3$ , and thus form a basis. (This is from Theorem 12.10, which is valid for $\mathbb{C}^n$ .) Thus, it is sufficient to show that $A,B,C$ are linearly independent. To that end, let $z_1, z_2, z_3$ be scalars in $\mathbb{C}$ , then
$\begin{align*} z_1 A + z_2 B + z_3 C &= O & \implies && z_1 + z_3 &= 0 \\ &&&& iz_2 + z_3 &= 0 \\ &&&& iz_3 &= 0. \end{align*}$

From the third equation we have $z_3 = 0$ , and so the second equation implies $z_2 = 0$ , and finally the third equation implies $z_1 = 0$ . Hence, $z_1 = z_2 = z_3 = 0$ , and $A,B,C$ are linearly independent $. \qquad \blacksquare$
To express $(5,2-i,2i)$ as a linear combination of $A,B,C$ , let $z_1, z_2, z_3 \in \mathbb{C}$ be scalars. Then,
$\begin{align*} z_1 A + z_2 B + z_3 C &= (5,2-i,2i) & \implies && z_1 + z_3 &= 5 \\ &&&& iz_2 + z_3 &= 2-i \\ &&&& iz_3 &= 2i. \end{align*}$

From the third equation we have $z_3 = 2$ . Plugging this into the first and second equations we get $z_1 = 3$ and $z_2 = -1$ . Therefore,

$(5,2-i,2i) = 3A - B + 2C.$

Compute the angle between (1,0,i,i,i) and (i,i,i,0,i) in C⁵

by RoRi

April 29, 2016

We define the angle $\theta$ between two vectors $A$ and $B$ in $\mathbb{C}^n$ by the formula

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

Compute the angle between $A = (1,0,i,i,i)$ and $B = (i,i,i,0,i)$ in $\mathbb{C}^5$ .

We compute as follows:

$\begin{align*} \theta &= \arccos \frac{\frac{1}{2} ( A \cdot B + \overline{A \cdot B})}{\lVert A \rVert \lVert B \rVert} \\[9pt] &= \arccos \left( \frac{\frac{1}{2}(2-i + (2 + i))}{4} \right) \\[9pt] &= \arccos \left( \frac{1}{2} \right) \\[9pt] &= \frac{\pi}{3}. \end{align*}$

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),

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

Find a nonzero vector in C³ orthogonal to two given vectors

by RoRi

April 28, 2016

Let

$A = (2,1,-i), \qquad B = (i,-1,2i).$

Find a non-zero vector $C$ in $\mathbb{C}^3$ orthogonal to both $A$ and $B$ .

For $C = (c_1, c_2, c_3)$ to be orthogonal to both $A$ and $B$ we must have $C \cdot A = 0$ and $C \cdot B = 0$ . This gives us

$\begin{align*} C \cdot A &= 0 & \implies && 2c_1 + c_2 + ic_3 &= 0 \\ C \cdot B &= 0 & \implies && -ic_1 - c_2 - 2ic_3 &= 0. \end{align*}$

From the first equation we get $c_2 = -2c_1 - ic_3$ . Plugging this into the second equation we then have $c_3 = (-2i-1)c_1$ . Let $c_1 = 1+i$ , then we have

$c_3 = 1-3i, \qquad c_2 = -5-3i.$

Therefore, $C = (1+i, -5-3i, 1-3i)$ is a vector in $\mathbb{C}^3$ orthogonal to both $A$ and $B$ .

Compute some dot products of vectors in C²

by RoRi

April 28, 2016

Let

$A = (1,i), \qquad B = (i,-i), \qquad C = (2i,1)$

be vectors in $\mathbb{C}^2$ . Compute the following dot products.

$A \cdot B$ ;
$B \cdot A$ ;
$(iA) \cdot B$ ;
$A \cdot (iB)$ ;
$(iA) \cdot (iB)$ ;
$B \cdot C$ ;
$A \cdot C$ ;
$(B+C) \cdot A$ ;
$(A-C) \cdot B$ ;
$(A - iB) \cdot (A + iB)$ .

We compute,
$A \cdot B = (1)(-i) + (i)(i) = -1 -i.$
We compute,
$B \cdot A = (i)(1) + (-i)(-i) = -1 + i.$
We compute,
$(iA) \cdot B = (i, -1) \cdot (i, -i) = (i)(-i) + (-1)(i) = 1 - i.$
We compute,
$A \cdot (iB) = (1,i) \cdot (-1,1) = (1)(-1) + (i)(1) = -1 + i.$
We compute,
$(iA) \cdot (iB) = (i,-1)\cdot (-1,1) = (i)(-1) + (-1)(1) = -1-i.$
We compute,
$B \cdot C = (i)(-2i) + (-i)(1) = 2 - i.$
We compute,
$A \cdot C = (1)(-2i) + (i)(1) = -i.$
We compute,
$(B+C)\cdot A = (3i,1-i) \cdot(1,i) = (3i)(1) + (1-i)(-i) = -1 + 2i.$
We compute,
$(A-C) \cdot B = (1-2i, i-1) \cdot (i,-i) = (1-2i)(-i) + (i-1)(i) = -i - 2 -1 - i = -3-2i.$
We compute,
$(A-iB)\cdot(A+iB) = A\cdot A + A \cdot i B - iB \cdot A - B\cdot B = 2 + (-1+i) - (-1-i) -2 = 2i.$