Lecture 2

The image of $T$ is defined by $RanT=\{Tv;v\in V\}\subseteq W$. This is also called the range of $T$.

The rank of $T$, is the dimension of its image: $rankT=dimRanT$.

The kernel of $T$ is defined by $KerT:=\{v\in V;Tv=0\}\subseteq V$. What this means is that $KerT$ consists of all the vectors $v$ that are mapped to zero by $T$.

A mapping $T$ is called injective if its left-invertible, and $T{v}_1≠T{v}_2⟹v_1≠v_2.$

A mapping $T$ is called surjective if its right-invertible, and $RanT=W$.

Example:

Let

The matrix $A$ defines a linear map

This is represented by

In general, a matrix $A\in {𝕂}^{m\times n}$ is a linear map

Scalars: Let $\alpha \in ℝ,\text{then }\alpha :{ℝ}^2\to {ℝ}^2,x⟼\alpha x.$ The matrix equivalence of this is

Even rotation with an angle $\phi$ is a linear map on ${ℝ}^2$: ${ℝ}_{\phi }:x⟼{ℝ}_{\phi }x$.

Exercise: Fix an orthonormal (and oriented) base = $({𝐞}_1,{𝐞}_2)$.

Let ${ℝ}_{\phi }$ be the counter-clockwise rotation by an angle $\phi$. The matrix for ${ℝ}_{\phi }$ is then the $2\times 2$-matrix

Sketch the basis vectors ${𝐞}_1,{𝐞}_2$, and their images $R_{\phi }({𝐞}_1),R_{\phi }({𝐞}_2)$ in the plane.

Projection: Let $L\subseteq {ℝ}^2$ be a linear subspace. Choose an orthonormal basis $E=({𝐞}_1,{𝐞}_2)$ such that

Reflection across $L$ leaves the part of the vector along $L$ unchanged and flips the part perpendicular to $L$:

The matrix equivalent to mirroring a vector is given by

and the matrix equivalent to projecting is given by

Define the following:

Now we ask ourselves, is integration: $\int dt:f(t)⟼\int f(t)dt$ a map?

Solution: We have to fix the integration points. $f(t)⟼{\displaystyle \underset{t_0}{\overset{t}{\int }}f(s)ds}$ is a map.

Proposition: If $T:V\to W$ is a linear map, then:

Definition: Let $T:V\to W$ be a linear mapping, then:

Example and reminder

Let

viewed as a linear map $A:{ℝ}^3\to {ℝ}^2$. Its kernel is

The equation $Ax=0$ is equivalent to

Since the second equation is twice the first, we obtain

Setting $x_2=s$ and $x_3=t$, we find

Thus, $\dim \ker A=2$ and $\mathrm{rank}A=1$.

In this example,

Thus, $\mathrm{rank}A=1$. $A$ is not injective because $\ker A\ne \{0\}$: the equation $Ax=0$ has at least one free variable. $\square$

Theorem (Theorem 2.1.5, Proposition 2.1.7, and Theorem 2.2.12)

Let $T:V\to W$ be a linear map between finite-dimensional vector spaces.

1. The following are equivalent:

   $$\begin{array}{cc}T\text{ is injective}& ⟺\ker T=\{0_V\}\\ & ⟺\text{there exists a linear map }S:W\to V\text{ such that }S\circ T=I_V.\end{array}$$

   The map $S$ is called a left inverse of $T$:

   $$V\stackrel{\stackrel{}{T}}{\to }W\stackrel{\stackrel{}{S}}{\to }V.$$

2. The following are equivalent:

   $$\begin{array}{cc}T\text{ is surjective}& {⟺}^{\mathrm{def}}\mathrm{Ran}T=W\\ & ⟺\text{there exists a linear map }R:W\to V\text{ such that }T\circ R=I_W.\end{array}$$

   The map $R$ is called a right inverse of $T$:

   $$W\stackrel{\stackrel{}{R}}{\to }V\stackrel{\stackrel{}{T}}{\to }W.$$

Let $𝒜$ and $ℬ$ be bases of $V$ and $W$. The matrix of a linear map $T:V\to W$ with respect to these bases is defined by

Its columns are the coordinates of $ran(\cdot )$ of the basis vectors:

Example: Mirroring with respect to $x_1=x_2$.} Let $S:{ℝ}^2\to {ℝ}^2$ be the mirror across the line $x_1=x_2$. Choose the basis

Since $b_1$ lies along the reflection line and $b_2$ is perpendicular to it,

. You can see this since $b_1$ rotates around the line it sits on. Thus,

Note: The standard bases $({𝐞}_{\mathrm{𝟏}},{𝐞}_{\mathrm{𝟐}})$ when mapped by this mirroring will change places across the axles.

Example: The identity map and change of coordinates.

Let $ℰ$ be the standard basis of ${ℝ}^2=({𝐞}_{\mathrm{𝟏}},{𝐞}_{\mathrm{𝟐}})$. The identity map $I:{ℝ}^2\to {ℝ}^2$ leaves every vector unchanged, but its matrix converts $ℬ$-coordinates into $ℰ$-coordinates:

Since $I(b_1)=b_1$ and $I(b_2)=b_2$, we get

The reflection matrix in the standard basis is

Thus, in standard coordinates, $S(x_1,x_2)=(x_2,x_1)$.