I read this sentence in Linear Algebra and its Applications: "Linear transformations preserve the operations of vector addition and scalar multiplication," but I don't know preserve means in this context. Alas, googling it and looking in the book's index found nothing relevant.
So, what does it mean?
It means it doesn't matter whether you do the operation first and the transformation second, or the transformation first and the operation second. You get the same result either way.
So for example for vector addition, if $f$ preserves vector addition then for any vectors $u, v$ in the domain of $f$ we have $$f(u + v) = f(u) + f(v).$$ On the left we did addition first, on the right we did the transformation first, and the equation says you get the same thing either way. For scalar multiplication the equation is $$f(cv) = cf(v)$$ for all vectors $v$ and scalars $c$.
Linear transformations preserve the operations of vector addition: $$T(a+b) = T(a) + T(b)$$ Linear transformations preserve the operations of scalar multiplication: $$T(\lambda a) = \lambda T(a) $$
Consider a definite integral over some interval $I \subseteq \mathbb{R}$. Then this is a linear map from the space of all real-valued integrable functions on $I$ to $\mathbb{R}$.
Let $I=[a,b]$ and consider $f,g$ to be integrable functions on $I$.
$$\int_a^{b} f(x)+g(x) \ dx = \int_a^b f(x) \ dx + \int_a^b g(x) \ dx$$
$$ \int_a^b \beta \cdot f(x) \ dx = \beta \cdot \int_a^b f(x) \ dx, \forall \beta \in \mathbb{R}$$
In the above argument if you consider the definite integral as a function say $H(\textbf{x})$ where $\textbf{x}$ corresponds to integral functions on a specified interval then we've shown that, $$H(\textbf{x}+\textbf{y}) = H(\textbf{x})+H(\textbf{y})\ \ \text{and}\ \ H(c \textbf{x}) = c H(\textbf{x})$$
Our two vector spaces are $U= \{\text{integrable functions over $I$}\}$ and $V=\mathbb{R}$; hence $H:U \to V$. One should also note that both are vector spaces over $\mathbb{R}$.
Now by the above linearity we can see that adding two definite integrals is the same as adding the two corresponding real numbers associated to them and multiplying a definite integral by a scalar is the same as multiplying the real number associated to it by the same scalar.
