# Example for linear transformations as vector spaces?

I am currently studying a book on Functional Analysis by George Bachman and Lawrence Narici. In one of the first chapters, they describe how a (linear) mapping $$A$$ can map from a vector space $$X$$ to a different vector space $$Y$$:

$$A:X\rightarrow Y$$

which has the (linear) properties that for any scalars $$\alpha$$ and $$\beta$$ and any vectors $$x$$ and $$y$$, we have:

$$A(\alpha x + \beta y) = \alpha A(x)+\beta A(y)$$

Now the critical part: They state that "the collection of all linear transformations mapping $$X$$ into $$Y$$ can be viewed as a vector space by defining addition of the linear transformations $$A$$ and $$B$$ to be that transformation which takes $$x$$ inton $$Ax+Bx$$; symbollically, we have: $$(A+B)x=Ax+Bx$$ and for scalar multiplication:" $$(\alpha A)x=\alpha A x$$

where they omit the parantheses around $$(x)$$ for brevity. I see how these operations above imply linearity, but how can the collection of these transformations be interpreted as a vector space? As far as I have gathered, a vector space should have basis vectors. What would these basis vectors be in the case of $$A$$?

I have an unfinished example below, which you could use to explain your answer. Otherwise feel free to use your own examples, if this is easier.

Example: $$x \in X \in \mathbb{R}^2$$ and $$y \in Y \in \mathbb{R}^3$$. In this case $$A$$ in $$y=Ax$$ would be a $$3 \times 2$$ matrix. What would be the basis vectors of the vector space interpretation of $$A$$?

Regarding your example: Any $$3 \times 2$$ matrix can be written as a linear combination of $$\{E_{ij} \}$$, where $$E_{ij}$$ is $$0$$ everywhere except in position $$(i,j)$$.

A better way of looking at vector spaces is to concentrate on their properties from an algebraic point of view: the essential property is linearity, the fact that all elements in the vector space can be multiplied by a scalar and added together to form new elements in the vector space. A function between vector spaces is linear if it preserves this essential property: that the image of a linear combination is again a linear combination is very simple way.

All vector spaces have a basis, and that is why linear functions can be expressed in terms of matrix multiplication.

Instead of vectors as a basis, you would have matrices as a basis (which we can still think of as vectors by the way). So, in your example, A can be written as a linear combination of 6 other matrices, where each matrix has only one 1, and five 0's.

Note that having basis vectors is not a defining property of vector spaces. Moreover, while in standard set theory you can prove that every vector space has a basis, you can't always explicitly give one.

In your example, it is easy to give a basis. A standard choice would be the set of matrices $$E_{ij}$$ that have an $$1$$ in row $$i$$/column $$j$$, and $$0$$ everywhere else: $$\pmatrix{a&b\\c&d\\e&f} = a\pmatrix{1&0\\0&0\\0&0} + b\pmatrix{0&1\\0&0\\0&0} + \ldots + f\pmatrix{0&0\\0&0\\0&1}$$

An example of a vector space where you can't explicitly give a basis is $$\mathbb R$$ as a vector space over $$\mathbb Q$$ (clearly $$\mathbb R$$ is a group under addition, and you certainly can multiply a real number by a rational number, giving again a real number; the remaining vector space axioms all reduce to the corresponding number arithmetic laws).