3.5 Note

Let $v,u$ be elements of $V$. The line segment between $v$ and $v+u$ is the set of all points

$v+tu$, $t \in [0,1]$.

If $v,w$ are elements of $V$, the line segment between $v$ and $w$ is

$t_1v + t_2w$, subject to $t_1, t_2 \geq 0$ and $t_1 + t_2 = 1$.

The parallelogram spanned by $v,w$ is

$t_1 v + t_2 w$, $0 \leq t_i \leq 1$ for $i=1,2$]

The triangle spanned by $O,v,w$ is

$t_1 v + t_2 w$, $0 \leq t_i$ and $t_1 + t_2 \leq 1$.

or

$t_1v_1 + t_2v_2 + t_3v_3$, $t_i \geq 0$ for $i=1,2,3$ and $t_1 + t_2 +t_3 = 1$

where $v = v_1 - v_3$ and $w = v_2 - v_3$

Corollary from Exercise 6

Let $L: V \rightarrow W$ be a linear map. Let $S'$ be a convex set in $W$. Let $S$ be the set of all elements $P$ in $V$ s.t. $L(P)$ is in $S'$, then $S$ is convex.

A brief summary of some basic but important properties about linear map and vector space.

 A linear map is a map $F: V \rightarrow U$ that satisfies those following properties:

1. $F(cA) = cF(A)$

2. $F(A+B) = F(A) + F(B)$.

In the above equations, $A,B \in V$. $V$ is called the domain of the linear map $F$ and $U$ is called the image of $L$.

$F$ is said to be injective if for $u,v \in V$, we have $F(u) = F(v)$ iff $u = v$. If $F$ is injective, then $\ker F = \{O\}$ and vice versa.

$F$ is said to be surjective if, for any $u \in U$, we have at least one element $v \in V$ s.t. $F(v) = u$.

If $F$ is both injective and surjective, then $F$ is bijective and has an inverse map $F^{-1}$ which is a linear map too.

If a linear map $F$ maps a vector space $V$ to itself, i.e. $F: V \rightarrow V$, then $F$ is called an operator sometime.

If a linear map $F: V \rightarrow U$ has an inverse map $F^{-1}: U \rightarrow V$, then we say $F$ is an isomorphism.

$\text{Im} F \subseteq U$, $\ker F \subseteq V$. If $\ker F = V$, then $F$ is a zero mapping.

Specially, if $F: V \rightarrow V$ and $\ker F = \{O\}$, then $F$ is invertible, i.e. $F$ is isomorphism. Because $\dim V = \dim \text{im}F + \dim \ker F$, since $\ker F = 0$, $\dim V = \dim \text{im} F$. Therefore, $V = \text{im}F$ and $F$ is surjective. Because $\ker F = \{O\}$, $F$ is injective, thus $F$ has an inverse linear map.

Useful corollaries in exercises of Chapter 3 Section 5

10. Let $V$ be a vector space. Let $P: V \rightarrow V$ be a linear map s.t. $P^2 = P$. Then

$V = \ker P + \text{im} P$ and $\ker P \cap \text{im}P = \{O\}$.

Check Chapter 1 Section 4 Theorem 4.1 for more details about the direct sum.

Two vector spaces $V$ and $U$ are isomorphic if there exists a bijective linear map s.t. $F: V \rightarrow U$.

Some important facts about composition of linear maps.

Suppose we have three linear mapping $F,G,H$ s.t. $F: V \rightarrow U$ and $G: U \rightarrow W$ and $H: W \rightarrow Z$.

We have

$F \circ ( G \circ H) $ = $ (F \circ G) \circ H$, i.e. the associative law. This property is on the textbook (Serge Lang, Linear Algebra 3rd edition) page 47

and

$F \circ (G + H ) = F \circ G + F \circ H$, i.e. the distributive law. See page 67 for more details.

However, the commutative law does not apply in general. In some special cases, it is applicable, though. For example, $F \circ F = F \circ F = F^2$.

A specific solution to Exercise 1 in Serge Lang Linear Algebra, Chapter 3 Section 3.

The answer given by the solution manual is somehow vague, therefore I wrote down my answer there for future reviewing.

Exercise 1. Let $A, B$ be two vectors in $\mathbb{R}^2$ forming a basis of $\mathbb{R}^2$. Let $F: \mathbb{R}^2 \rightarrow \mathbb{R}^n$ be a linear map. Show that either $F(A), F(B)$ are linearly independent, or the image of $F$ has dimension 1, or the image of $F$ is $\{O\}$.

Solution:

Because $\dim V = \dim \ker F + \dim \text{im} F$, $\dim V = 2$.

1. If $F(A)$ and $F(B)$ are linearly independent, then $F(A)$ and $F(B)$ forming a basis of $\text{im} F$. The $\text{im}F = 2$ and $\ker F = 0$.

2. If $F(A)$ and $F(B)$ are linearly dependent, i.e. $F(A) = k_1 F(B)$. then $t_1 F(A) + t_2 F(B) = t_1 k_1 F(B) + F(B) = (t_2 + t_1 k_1) F(B)$.

2-1. If $t_1F(A) + t_2F(B) = O$, then $(t_2 + t_1 k_1) F(B) = O$.  Because $t_1 F(A) + t_2 F(B) = F(t_1 A + t_2 B)$ and $A,B$ are a basis of $\mathbb{R}^2$, therefore for all $ v \in \mathbb{R}^2$, $F(v) = O$, that is, $\text{im}F = 0$, i.e. $\ker F = 2$.

2-2. If $t_1F(A) + t_2F(B) \neq O$, then $(t_2 + t_1 k_1)F(B) \neq O$. Because $t_1 F(A) + t_2 F(B) = F(t_1 A + t_2 B) = (t_2 + t_1 k_1)F(B) \neq O$, therefore $F(B)$ is a basis of $\text{im}F$. That is, $\dim \text{im}F = 1$ and $\dim \ker F = 1$.

An interesting corollary about a line under the linear mapping.

Suppose we have a linear mapping $F: V \rightarrow U$. By simple geometry knowledge, we know that for two points $x,y \in V$, the segment between points $x,y$ are

$x+(1-t)y$, where $t \in [0,1]$.

Because $F$ is a linear mapping, we have $F(x + (1-t)y) = F(x) + F( (1-t)y ) = F(x) + (1-t)F(y)$, where $t \in [0,1]$.

If $y \in V$ lying in the $\ker F$, i.e. $F(y) = O$, then $F(x + (1-t)y) = F(x)$. Therefore, the image of the segment $x + (1-t)y$ in $V$ under the linear mapping $F$ in $U$ is a point $F(x)$.

Otherwise, we have two different points $F(x), F(y) \in U$.

The following part needs verifying once more.

We'll prove that if $x,y \notin \ker F$, then $F(x) = F(y)$ iff $x=y$ as follows.

Suppose we have $x,y \in V$ and $x \neq y$, $x,y \notin \ker F$ and $F(x) = F(y)$, we have

$F(x) - F(y) = O$. Because $F$ is a linear mapping, we got $F(x)-F(y) = F(x) + -F(y) = F(x) + F(-y) = F(x-y)$. However, because $F(x) = F(y)$ in our assumption, we have $O = F(x-y)$. Because $x,y \notin \ker F$ and $x-y \in \ker F$, we have $x-y = O$, i.e. $x=y$. Hence, we reached an contradiction, showing that if $x,y \notin \ker F$ and $F(x) = F(y)$, then $x=y$. Q.E.D.

Intuitively, $x \in V$ is a linear combination of the basis of $V$ and $y \in V$ is a linear combination of the basis of $V$. If $F(x) = F(y)$, we got $F(t_1 v_1 + t_2 v_2 + ... + t_n v_n ) = F(k_1 v_1 + k_2 v_2 + k_3 v_3 + ... + k_n v_n)$ and $F( (t_1 - k_1)v_1 + (t_2 - k_2)v_2 + ... (t_n - k_n) v_n) = O$. Because $v_1, v_2, ..., v_n$ are the basis, and $x,y \notin \ker F$, the linear combination of $x,y$ are not in $\ker F$, and therefore we have $t_i = k_i$ for $i = 1,...,n$ thus $x=y$.

A common trick for proving $\text{Im} F = C$ where $C$ is a geometric shape's analytic equation.

From Linear Algebra, Serge Lang 3rd Ed. Chapter 3 Section 1.

Suppose we have a linear mapping $F = V \rightarrow W$, and suppose we have a coordinate $(u,v) \in V$. After mapping, it became $(x,y) \in W$ and satisfy an equation $E:F(x,y) = 0$.
We want to prove that: (1) $\text{Im}F \subseteq E$ and $E \subseteq \text{Im} F$ so that we can conclude $E = \text{Im}F$.
Here's the trick. We firstly pick up an arbitrary coordinate $(u,v)$ from $V$, and follow the mapping $F$ so that we can get an coordinate after mapping, say $(x,y)$, and that $(x,y)$ satisfies an equation $E:F(x,y) = 0$. This proved $\text{Im} F \subseteq E$, i.e. the image of $F$ is some points satisfies the equation $E$.

Secondly, we pick up an arbitrary coordinate $(x,y)$ from the equation $E$, and find out a coordinate $(u,v)$ in V so that $F(u,v)=(x,y)$. This proved that each point on that equation, $E$, is an element of the $\text{Im}F$, i.e. every point satisfies $E$ is an element of $\text{Im}F$. Eventually we got $E \subseteq \text{Im}F$.

Because $E \subseteq \text{Im}F$ and $\text{Im}F \subseteq E$, we conclude that $E = \text{Im}F$.

Example (Exercise 12 from the Chapter 3 Section 1):

Let $F$ be the mapping defined by $F(x,y) = (x/3,y/4)$. (Errata: in the printing version I have, there's a typo stating $F(x,y) = (x/3, x/4)$ rather than $(x/3,y/4$).) What is the image under $F$ of the ellipse

$\frac{x^2}{9} + \frac{y^2}{16} = 1$?

We'll solve this problem by the procedure mentioned above.

1. We pick up an arbitrary coordinate $(u,v)$ in $V$, where $V$ is the ellipse $\frac{x^2}{9} + \frac{y^2}{16} = 1$ and substitute it into the mapping $F$, we got $F(u,v) = (u/3, v/4)$. Let $(x,y) = (u/3,v/4)$ (Notice here, $(x,y) \in W$, where $W$ is the image of $F$. ) Then $3x=u$ and $4y=v$. Because $\frac{u^2}{9} + \frac{v^2}{16} = 1$, we have $\frac{(3x)^2}{9} + \frac{(4y)^2}{16} = 1$. After using some arithmetic tricks, we found $x^2 + y^2 = 1$, that is a unit circle and we proved $\text{Im}F \subseteq E$, where $E$ is the equation of the unit circle $x^2 + y^2 = 1$.

2. In the second step, our aim is proving $E \subseteq \text{Im}F$. We pick an arbitrary coordinate from $E$ satisfying $x^2 + y^2 = 1$, say, $(x,y)$. Then, we need to find out a coordinate $(u,v) \in V$ s.t. $F(u,v) = (x,y)$.  Take an arbitrary coordinate $(u,v)$ that may belong or not belong to $V$, we have $F(u,v) = (u/3, v/4) = (x,y)$. Therefore, $x = u/3$ and $y = v/4$. Substitute $(x,y) = (u/3,v/4)$ into the unit circle equation, we have

$x^2 + y^2 = \frac{u^2}{9} + \frac{v^2}{16} = 1$

This proved that $(u,v) \in V$ (recall that $V$ is the ellipse equation) and every point on the unit circle $x^2 + y^2 = 1$ is in $\text{Im}F$. Therefore we got $E \subseteq \text{Im} F$. Thus, $\text{Im} F = E$ Q.E.D.

Addtional Tips:

1. Because for any $v \in V$ we proved that there is at least a image of $F$ at $v$ in $V$, namely, $F(v)$, so we implicitly proved that $F$ is surjective.

2. However, we didn't implicitly prove that $F$ is injective, because we didn't show that there is only an unique $v \in V$ s.t. $F(v)=w$ .