Although the essence has already been stated, let me try to give you a more graphic approach to linear maps. Often, when you get the right mental picture of a construct, the properties fall right into place.
PS: Whoops, That turned out to be a lot. I hope it's not a bad thing I kind of answer your question in the last paragraph, only. I hope this is helpful for somebody though.
Plus, I hope I didn't make any false statements here considering more than finite dimensions.
The definition
Let $V$, $W$ be vector spaces over a field $F$. A map $f: V → W$ is called linear, if:
$∀x, y\in V: f(x+y) = f(x)+f(y)$
$∀x \in V, λ \in F: f(λx) = λf(x)$.
What do linear maps map?
The first important thing here is that they map vector spaces into vector spaces. They can be anything, so this alone doesn't help a lot. Could they be something different than vector spaces? Well, if they weren't, our statements wouldn't make much sense – they use scalar multiplication and addition, which are operations only defined in vector spaces. So far nothing interesting here.
You can, however, immediately ask: “What does the image of a linear map look like?”, or, “in what way changes/transforms $f$ the space $V$ to $W$?”.
What can this subset of $W$ look like? For instance, if $V=ℝ^3, W=ℝ^3$, can the image be the sphere? It obviously cannot, since for every vector $w = f(v)$ in the image, we can scale the parameter $w$ and get a scaled version $f(λv) = λf(v) = λw$. This greatly restricts what the image qualitatively looks like!
In fact, if you follow a similar argument for the preserving of addition, you might conjecture: The image itself is a vector space!
Proof (For the sake of completeness)
Let $x, y\in f[V], λ\in F.$ Thus we find $v\in V: x=f(v)$ and $w\in V: y = f(w)$. Now, $x+y=f(v)+f(w)=f(v+w)$, thus $x+y$ is in the image. Similar we get $λx = λf(v) = f(λv)$ thus $λx$ in the image. QED.
And now?
The fact that the image is a vector space being a subset of the vector space $W$, i.e. a (vector) subspace of $W$, helps for the intuition: e.g. in $ℝ^3$, vector subspaces are ${0}$, lines and planes through the origin, and $ℝ^3$ itself. So somehow, $f$ transforms a vector space $V$ into a subspace of $W$. At the moment, we don't know an important thing however: How “big” is this subspace? Can we say something about the dimension? If not, can we find some restriction like upper/lower bound?
The trick: Don't look at the whole space
Let's just assume $V$ and $W$ have a basis, and, to make writing sums easier, to be finite dimensional. We then can express elements of these spaces as the sum of the basis vectors scaled by a certain amount, i.e. as the “coordinate tuple” which are said amounts. The (unique, bijective) map from the coordinate tuples to the vectors is called the “basis isomorphism”.
Let's look at a vector $x=f(v)$ in the image of $f$. Choosing any ordered basis $(b_n)_n$ of $V$, we can write it as: $x = f(v) = f(\sum_{i=1}^n b_i v_i)$.
We „expanded“ the vector $v$ in the preimage by looking at the bases separately (the $v_i$ are the coefficients with regards to our basis $b_i$).
Now, the preserving of addition and scalar multiplication comes in handy: We can move the summation one level out!
$$x = f(v) = \cdots = \sum_{i=1}^n v_i f(b_i)$$
This is actually a big deal! We now know that any element of the image can be described as linear combinations of the images of some basis elements in $V$ (or: it lies in the span of the image of the basis) – or, to put it differently: If you know the image of the basis elements, you know the image of the whole space.
Once I got this, I pictured every (finite-dimensional, well, to be honest, 3-dimensional) linear map by picturing a base on the left side and the image of that base on the right side.
This gives you immediately one constraint: The dimension of the image can at least not be larger than $\dim V$, since it is spanned by $\dim V$ (not necessarily linearly independent) vectors. Can it be less? Yes, if the images of the basis vectors are linearly dependent: Consider e.g. the map $$f: ℝ^3→ℝ^3, (x, y, z)↦ (x+z, y+z, 0)$$
It maps $e_x, e_y$ to themselves, but $f(0, 0, 1)=(1, 1, 0)$. So the base of the preimage maps to three vectors each lying in the $x-y$-Plane – in other words, they are linearly dependent, and span a subspace not of dimension 3, but of dimension 2.
Your Question
To answer your question: Yes, maps can indeed map to higher dimensional spaces. For instance, take $f: ℝ^n→ℝ^n+k, (x_1, …, x_n)↦(0, 0, …, x1, …, x_n)$.
The dimension of their image (also called “rank”), however, cannot have a higher dimension. Thus, if you map to a higher dimension, your map cannot be surjective anymore.
Matrix and determinant
You might notice that whether or not the images of the basis vectors are linearly independent is a major factor to qualitatively determine the nature of this function (let the word sink in for a moment: determin-e… rings a bell?). Consider injectivity: If a n-dimensional space is transformed into an $m<n$-dimensional one, can the map still be injective? The intuition screams “no”! But let me omit a proof here.
Let's pick a basis for each $\dim V=n$ and $\dim W=m$, and just care about the tuple representation of the vectors (lying in $F^n$ and $F^m$, respectively).
The images of the basis vectors can now be written as such a tuple. View this tuple as a column of numbers and put these tuples near each other – you now got a thing of height $m$ and width $n$ – actually, an $m \times n$ matrix. The Idea of matrices are that they are, vaguely speaking, coordinate representations of finite-dimensional linear maps.
If you now consider $V$ and $W$ to be of the same dimension, so our matrix becomes square.
We now can think of the determinant as, hold on, the n-dimensional signed volume of the image of the unit hypercube. If a 3-dimensional unit cube is „squashed“ in the image to vectors laying in a plane, it's 3-dimensional volume is zero – which hopefully gives you some intuition while seemingly every theorem in linear algebra is equivalent to $\det M = 0$.
But check out this answer and especially this answer – they do an excellent Job in making the determinant more accessible.
