An example of linear transformation not matrix transformation I realized that matrix transformation must be a linear transformation, but linear is not necessary matrix. 
Can someone give me an example of a linear transformation that is not matrix transformation?
 A: I'm going to assume that, by "matrix transformation", you mean a linear transformation of the form
$$T : \Bbb{R}^n \to \Bbb{R}^m : v \mapsto Av$$
where $A$ is an $m \times n$ real matrix. We could also replace the reals with complex numbers (or indeed any field, if you're feeling extra spicy).
Now, the answer depends on how far down the rabbit hole you are. Some courses simply deal with $\Bbb{R}^n$ only, and linear maps can only be between various finite powers of $\Bbb{R}$. If this is the case, then all linear maps (between powers of $\Bbb{R}$) are matrix transformations.
You'll no doubt come across this in your course at some point. In essence, we can build $A$ by computing $Te_1, \ldots, Te_n$, where $e_1, \ldots, e_n$ are the standard basis for $\Bbb{R}^n$, and forming $A$ by putting $Te_1, \ldots, Te_n$ as the columns of $A$ (in order).
Other courses expand beyond this, and talk about finite-dimensional vector spaces (in the abstract sense). One discovers that, for example, the set $P_3(\Bbb{R})$ of polynomials of degree $3$ or less, with real coefficients, is a $4$-dimensional vector space over $\Bbb{R}$, and hence it makes perfect sense to make linear transformations to and from this space. For example,
$$T : P_3(\Bbb{R}) \to P_3(\Bbb{R}) : p(x) \mapsto p(0)x^2 + 3xp'(x)$$
is a linear transformation.
Note that it can't be a matrix transformation in the above sense, as it does not map between the right spaces. The vectors here are polynomials, not column vectors which can be multiplied to matrices.
That said, there still is a way to "represent" $T$ by a matrix. It's not that $T$ is multiplication by a matrix, but there are still matrices that can help us evaluate $T$ in a meaningful way. In fact, this can be done for any linear map $T$ between any two finite-dimensional spaces over the same field.
I won't go into too much detail about this here, but essentially it involves choosing bases $B$ and $B'$ for the domain space and the codomain space respectively, evaluating the map on the vectors in $B$, and writing the results as coordinate column vectors with respect to $B'$. These coordinate vectors are the columns of $A$. We can then multiply coordinate vectors with respect to $B$, to get the transformed vector as a coordinate vector with respect to $B'$. If you haven't covered this stuff yet, you probably will soon.
It's important to know that choosing different bases will yield different matrices. That is, you can't really "identify" a linear map with a single matrix. There has to be some choices of bases first.
And then, there are infinite-dimensional spaces, such as the set $C[0, 1]$ of continuous real-valued functions on the domain $[0, 1]$. We can define linear maps to and/or from such spaces too. In this case, there's no way to even represent linear maps like this as matrices; they're too complicated for just a finite array of numbers!
A: Let $X=C[0,1]$ be the linear space of continuous complex functions on $[0,1]$, and let $(Tf)(x)=xf(x)$. $T$ is a linear transformation on $C[0,1]$. $C[0,1]$ does not a have a finite basis. So you cannot represent $T$ as a matrix.
