How do linear transformation properties preserve vector space structure?

Question

Currently I'm reading the Mathematics for Machine Learning book, and in the beginning of section 2.7 of the book, the following is given:

2.7 Linear Mappings
In the following, we will study mappings on vector spaces that preserve their structure, which will allow us to define the concept of a coordinate. In the beginning of the chapter, we said that vectors are objects that can be added together and multiplied by a scalar, and the resulting object is still a vector. We wish to preserve this property when applying the mapping: Consider two real vector spaces V,W. A mapping Φ : V → W preserves the structure of the vector space if

Φ(x + y) = Φ(x) + Φ(y) (2.85)

Φ(λx) = λΦ(x) (2.86)

for all x, y ∈ V and λ ∈ R.

Generally, when vectors are added together, or multiplied by a scalar, the result is also a vector (the same type of object).

How are the above properties of a linear transformation preserving this?

As long as the LHS and RHS of both given properties belong to the same vector space, can I not consider the vector properties preserved? Why do they have to be equal?

What does it mean by they preserve the structure of a vector space?

Is there a particular reason Φ(λx + ψy) = λΦ(x) + ψΦ(y) has a form similar to a linear combination (y = a1.x1 + a2.x2 +...)?

Answer 1

The reason the author says this is that it's a kind of motivation for the definition of a linear mapping. Just saying a linear map is any $f : V \rightarrow W$ for vector spaces $V$ and $W$ which satisfies:

$$f(u+v) = f(u) + f(v)$$

and

$$f(\lambda u) = \lambda f(u)$$

is indeed the definition, but it is not enlightening in the slightest, so the author is attempting to give some insight into why we have this definition.

Generally, when vectors are added together, or multiplied by a scalar, the result is also a vector (the same type of object). How are the above properties of a linear transformation preserving this?

Let's go over what vector properties mean. First, we have that adding vectors produces a vector, so, for vectors in boldface, we have:

$$\mathbf{u + v} = \mathbf{u} + \mathbf{v}$$

For example:

$$\begin{bmatrix} 1 + 2 \\ 4+5 \\ -1 + 15 \end{bmatrix} = \begin{bmatrix} 1 \\ 4 \\ -1 \end{bmatrix} + \begin{bmatrix}2 \\ 5 \\ 15 \end{bmatrix}$$

We also have the ability to pull out scalars:

$$\mathbf{au} = a\mathbf{u}$$

For example:

$$\begin{bmatrix} 10 \\ 40 \\ -10 \end{bmatrix} = 10\begin{bmatrix} 1 \\ 4 \\ -1 \end{bmatrix}$$

What this tells us is that the linear map is compatible with the rules of vector arithmetic. This is one of the author's intentions behind saying that it maintains structure.

Indeed, if $\mathbf{w} = \lambda \mathbf{u} + \delta \mathbf{v}$, if $f$ is a linear mapping, then $f(\mathbf{w}) = \lambda f(\mathbf{u}) + \delta f(\mathbf{v})$ (can you prove this?). So a linear map uniquely preserves the relationship between different vectors. Can you find three vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ such that $\mathbf{a} = \mathbf{b} + \mathbf{c}$, but for $g(x) = x^2$, we have $g(\mathbf{a}) \neq g(\mathbf{b}) + g(\mathbf{c})$? Why does this tell us that $g$ is not a linear map? Can you find a counterexample to prove that $g$ is not a linear map?

This idea that maps "preserve the decomposition of a vector into a sum of scaled vectors" will be vital to defining things like a basis of a vector space, and other useful properties.

As long as the LHS and RHS of both given properties belong to the same vector space, can I not consider the vector properties preserved? Why do they have to be equal?

In this case, once again consider $g(x) = x^2$. This function runs from $\mathbb{R}$ to $\mathbb{R}$, both of which are vector spaces, but it does not satisfy our properties. Hence while the output is a vector, the way that the output relates to its component parts is not the same as the way that the input relates to its component parts.

Once again, this is related to the idea of a basis, in which we find we may write a vector as a unique linear combination of a finite set of vectors (if the vector space is finite dimensional), and linear maps preserve that combination, but I do not know if you have been exposed to a basis yet.

Answer 2

A mapping preserves some structure if the image of the mapping has the same structure as the domain. Examples include group and ring homomorphism, linear transformations and graph isomorphisms.

For linear algebra in particular we have a geometric notion that we can work that can give us a meaningful interpretation of this. Say we have a projection onto a plane in $\mathbb{R}^3$. This is linear transformation which means if we add the vectors, or scale the vectors before the projection it would be the same as scaling and adding them after the projection. It allows us to calculate in either space with the assurance that the image will be the same.

Answer 3

Once upon a time there was a Halloween party to which all vectors of $\mathbb R^n$ were invited. Of course it was a costume party and the vectors did their best to choose the most creative costumes.

Besides drinking, eating and dancing, the vectors also did what they love to do most, which is to be multiplied by scalars and added together. People had a lot of fun when a prize was announced for whoever could prove or disprove that there is a scalar $\lambda $ such that $$ \lambda \times \text{Robin} = \text{Batman}, $$ but the eventual revelation that $$ \text{Frankenstein} + \text{Dracula} = \text{Frankenstein} $$ was a big embarrassment for Dracula who thought his expensive costume would not give away his true identity!

In any case, the function $$ T:\mathbb R^n \to \text{Party}, $$ assigning to each vector $v$, the party guest played by the vector $v$ is a linear function and so it preserves the vector operations in the sense that, e.g. the sum of two vectors doesn't care about which costume each vector is wearing.

The expression of this fact is nothing but $$ T(v+u)=T(v)+T(u), $$ and I encourage you to read the above formula aloud, replacing "$T$" for "the character represented by".