# Linear Algebra II

## September 30, 2014

### A brief summary

Filed under: 2014 Fall — Y.K. Lau @ 1:08 PM

We have finished Sections 7.1-7.2, please note that most of the examples are for your reading so we did not spare time to discuss them in lecture. This post is to give a brief summary and some remarks.

Firstly recall that a linear transformation is a function that satisfies the linearity conditions (T1) and (T2), see Definition 7.1. Verbally the linearity conditions preserve addition and scalar multiplication. Thm 1 shows basic properties of linear transformations. Thm 3 looks a bit complicated but is indeed important — it tells how to construct linear transformations. Next, we introduce the conept of kernel and image of a linear transformation. They can be viewed as a generalization of nullspace and column space of a matrix. In fact, kernel and image are subspaces and more importantly, satisfy the dimension theorem, see Thm 4. The proof of Thm 4 is a little technical but interesting The method of proof can be applied to get another result, that is, Thm 5. Below are the details.

Theorem 5. Let ${T:V\rightarrow W}$ be a linear transformation and let ${\{e_1,\cdots, e_r, e_{r+1}, \cdots, e_n\}}$ be a basis for ${V}$ such that ${\{e_{r+1},\cdots, e_n\}}$ is a basis for ${\ker T}$. Then ${\{T(e_1),\cdots, T(e_r)\}}$ is a basis for ${T(V)}$.

Proof:

• ${T(e_1),\cdots, T(e_r)}$ are linearly independent.

Consider ${c_1T(e_1)+\cdots+c_r T(e_r)=0}$. By linearity, ${T(c_1e_1+\cdots+c_r e_r)=0}$.

Thus ${c_1e_1+\cdots+c_r e_r\in \ker T}$.

As ${\{e_{r+1},\cdots, e_n\}}$ is a basis for ${\ker T}$,

$\displaystyle c_1e_1+\cdots+c_r e_r = d_{r+1}e_{r+1}+\cdots + d_ne_n$

for some ${d_{r+1},\cdots, d_n\in {\mathbb R}}$.

Rearranging, ${c_1e_1+\cdots+c_r e_r +(- d_{r+1})e_{r+1}+\cdots + (-d_n)e_n=0}$.

As ${\{e_1,\cdots, e_n\}}$ is a basis (so is linearly independent), ${c_1=\cdots=c_r=-d_{r+1}=\cdots = -d_n=0}$.

i.e. ${c_1=\cdots = c_r=0}$ is the possible coefficients such that ${c_1T(e_1)+\cdots+c_r T(e_r)=0}$.

• ${T(e_1),\cdots, T(e_r)}$ span ${T(V)}$.

Let ${w\in T(V)}$. Then ${w=T(v)}$ for some ${v\in V}$.

As ${\{e_1,\cdots, e_n\}}$ is a basis for ${V}$, ${v= a_1e_1+\cdots +a_re_r+a_{r+1}e_{r+1}+\cdots a_ne_n}$.

Thus ${T(v)= a_1T(e_1)+\cdots +a_rT(e_r)+a_{r+1}T(e_{r+1})+\cdots a_nT(e_n).}$

As ${T(e_{r+1})=\cdots = T(e_n)=0}$, ${w=a_1T(e_1)+\cdots +a_rT(e_r)}$.

This holds for all ${w\in T(V)}$, thus ${T(V)\subset {\rm Span}(T(e_1),\cdots, T(e_r))}$.

Besides, today we proved the following result.

Let ${T:V\rightarrow W}$ be a linear transformation, and let ${w\in W}$. Suppose ${T(x_0)=w}$. Then

$\displaystyle \{v\in V: \ T(v)=w\} = \{ x_0+u: \ u \in \ker T\}.$

[Remark. We write ${T^{-1}\{w\}= \{v\in V: \ T(v)=w\}}$, the preimage of ${w}$.]

Proof: Suppose ${T(v)=w}$ and ${T(x_0)=w}$. Then ${T(v-x_0)=T(v)-T(x_0)=0}$, i.e. ${v-x_0\in \ker T}$ so ${v=x_0+u}$ for some ${u\in \ker T}$.

Conversely, for any ${u\in \ker T}$, ${T(x_0+u)= T(x_0) +T(u)=w+0=w}$. This completes the proof.

Remark. This result is the theoretical basis for the solving method of some differential equations — a good example to demonstrate why we need to learn abstract vector spaces.