Show by Example that $\mathbb{H}^n$ to $\mathbb{H}^n$ is not necessarily $\mathbb{H}$-linear Show by example that for$ A \in M_n \mathbb{H}, L_A : \mathbb{H}^n \rightarrow\mathbb{H}^n $ is not necessarily $\mathbb{H}$-linear
So I thought it would be linear by definition. Because if we have $ A \in M_n \mathbb{K}, L_A : \mathbb{K}^n \rightarrow \mathbb{K}^n $ such that for $X \in \mathbb{K}^n$ we would have $L_A (X) := (X . A^T)^T$  which i believe believe always give a linear value for example where $ X = [x,y]$ and $A = [[1,2],[3,4]]$ would give us $[x+2y,3x+4y]$
 A: The dimension isn't relevant, so we might as well consider $n=1$ for simplicity.
Suppose $R$ is a ring and $a\in R$ does not commute with everything in the ring, i.e. $a\not\in Z(R)$. Then the left-multiplication-by-$a$ map $\phi_a(x)=ax$ is not $R$-linear, i.e. it does not satisfy $\phi_a(rx)=r\phi_a(x)$ for all $r\in R$. Can you convince yourself of this? Try writing out what the equality means using the definition of $\phi_a$, and maybe setting $x=1_R$ (supposing $R$ is unital) for even greater effect.
The idea here is that "$R$-linear" means that applying the matrix transformation and applying a scalar taken from $R$ commute (can be done in either order), and since the scalars themselves form a subring of the matrices (scalar multiples of the identity matrix), a necessary condition for the matrices to be $R$-linear is for all scalars to commute with all other scalars, i.e. $R$ commutative.
A: Your question is missing some details about the context, which makes it difficult to interpret. You are right that matrices are introduced to encode linear maps, so that application of a matrix to a vector should always be a linear map. What is not clear is whether that application of a matrix$~A$ to a vector corresponds to left multiplication $L_A$.
When doing linear algebra over skew field like  $\def\H{\Bbb H}\H$ one must decide first whether scalar multiplication is written with the scalar on the left or or the right. This is because one requires associativity of scalar multiplication, which could be written either as $\lambda.(\mu.v)=(\lambda\mu).v$ for any vector$~v$ (the Greek letters being scalars) if scalars are written on the left, or $(v.\lambda).\mu=v.(\lambda\mu)$ if scalars are written on the right. The two equations do not mean the same thing, and are in general incompatible with each other, so one cannot allow scalars to be applied from either side at will. (The vector space $\H^n$ is special in that left and right multiplication by scalars can be defined for it, but when using it for coordinate vectors of another vector space, only one of these two structures should be used: the term "linear" can refer only to one.) With the first rule one is defining a left $\H$ vector space, with the second rule a right $\H$ vector space.
Unfortunately, when doing linear algebra over commutative fields the most common convention is to write scalars on the left and to have matrices act on vectors by multiplication on the left. When working over a skew field one cannot maintain both conventions. Since your citation says that $L_A$ is not linear, I'll suppose that it has defined left $\H$ vector spaces (and so views $\H^n$ as $\H$-vector space by left-multiplication with scalars from$~\H$) and follow that convention.
A central question is how to represent linear maps $l:\H^m\to\H^n$ by matrices. It remains true over a skew field that any map defined on a basis of the vector space can be uniquely extended to a linear map, so given that $l(e_i)=a_{i,1}f_1+\cdots+a_{i,n}f_n$ for $i=1,\ldots,m$ (where $[e_1,\ldots,e_m]$ is the standard basis of $\H^m$ and $[f_1,\ldots,f_n]$ is the standard basis of $\H^n$), the scalars $a_{i,j}$ determine $l$ uniquely as linear map. Linearity means compatibility with addition and scalar multiplication $l(\alpha v)=\alpha l(v)$, so the effect of $l$ on a vector with coordinates $(c_1,\ldots,c_m)$ is given by
$$
\begin{align}
  l(c_1e_1+\cdots+c_me_m)&=c_1l(e_1)+\cdots+c_ml(e_m)
\\&=\sum_{i=1}^nc_i(a_{i,1}f_1+\cdots+a_{i,n}f_n)
\\&=\biggl(\sum_{i=1}^nc_ia_{i,1}\biggr)f_1+\cdots+\biggl(\sum_{i=1}^nc_ia_{i,n}\biggr)f_n
\end{align}
$$
and one sees that the list of parenthesised coordinates is the one-line matrix product
$$
  (c_1~~c_2~~\cdots~~c_m)*
\begin{pmatrix}
a_{1,1}&a_{1,2}&\ldots&a_{1,n}\\
a_{2,1}&a_{2,2}&\ldots&a_{2,n}\\
\vdots&\vdots&\ddots&\vdots\\
a_{m,1}&a_{m,2}&\ldots&a_{m,n}\\
\end{pmatrix}.
$$
The important point here is that the matrix is multiplied on the right.
There was some foresight on my part in choosing the order of the indices of $a_{i,j}$, but the essential point is that the coefficients $a_{i,j}$ describing $l$ get multiplied on the right to the coordinates $c_i$, and there is no way to achieve this by left-multiplication by a matrix. Concretely in your text, it must be that right-multiplication $R_A:\H^m\to\H^n$ is a linear map, but $L_A$ is not.
Working with right-$\H$ vector spaces (which I personally am more comfortable with) one would get left-multiplication by a matrix to describe application of a linear map. (The reason I am more comfortable is that if as above one is writing linear maps on the left, so $l(v)$ rather than $(v)l$, then composition of linear maps gives a matrix product in the opposite order if writing matrices on the right, but in the same order if writing them on the left; however the side to multiply matrices is dictated by where one writes the scalars). Some points that are true independent of the convention:


*

*Scalar multiplication and the application of linear maps commute with each other (but not with another operation of the same kind).

*If scalars are written on one side, then linear maps are given, in terms of coordinates on chosen bases, by matrix multiplication on the other side (and the two commute).

*Scalar multiplication by a non-central scalar $\lambda$ is not linear, and cannot be represented by any matrix. 

*Matrices themselves form a vector space of the same kind (left or right) as the vector spaces they represent linear maps on. However, with $\lambda$ as above, if a matrix $A$ represents (with respect to some bases) a linear map $l$, then its multiple $\lambda A$ (assuming left vector spaces here) does not represent the map $v\mapsto l(\lambda v)=\lambda l(v)$, since that map, though well-defined, is not linear; indeed the linear map that $\lambda A$ does describe cannot be expressed in terms of $l$ and $\lambda$ at all, since it also depends on the bases used to represent $l$ by the matrix $A$.

*In particular the scalar multiple $\lambda I_n$ does not represent scalar multiplication by $\lambda$ (confirming the second point), even though multiplying coordinate vectors on the wrong (here left) side by that matrix does have the effect of scalar multiplication by$~\lambda$.

