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 be a linear transformation and let be a basis for such that is a basis for . Then is a basis for .

**Proof**:

- are linearly independent.
Consider . By linearity, .

Thus .

As is a basis for ,

for some .

Rearranging, .

As is a basis (so is linearly independent), .

i.e. is the possible coefficients such that .

- span .
Let . Then for some .

As is a basis for , .

Thus

As , .

This holds for all , thus .

Besides, today we proved the following result.

Let be a linear transformation, and let . Suppose . Then

[Remark. We write , the preimage of .]

**Proof**: Suppose and . Then , i.e. so for some .

Conversely, for any , . 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.

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