## Is the Set of Linear Transformations a Group?

For the set of all linear transformations
$\left\{ T \right\}$
to be a group, thye group axioms must be satisfied.
1. The identity element must be in the set. This condition is not satisfied. Differentiation is linear. We can define a linear transformation on the set of polynomials of degree 2.
$\frac{d}{dx}(1)=0,\frac{d}{dx}(x)=1, \frac{d}{dx}(x^2)=2x$
.
We represent
$1.x.x^2$
by the vectors
$\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\0\end{pmatrix}, \begin{pmatrix}0\\0\\1\end{pmatrix}$
Hence
$\frac{d}{dx}(a+bx+cx^2)=b+2cx$
.
The columns of the matrix representing
$T$
can be found by differentiating
$1.x.x^2$
in turn and representing the results as vectors.
The matrix representing th linear transformation is
$\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 &2 \end{array} \right)$
.
The polynomial
$2+3x+5x^2$
is represented by the vector
$\begin{pmatrix}2\\3\\5\end{pmatrix}$

$T(2+3x+5x^2)= \left( \begin{array}{cc} 0 & 1 & 0 \\ 0 & 0 & 2 \end{array} \right) \begin{pmatrix}2\\3\\5\end{pmatrix}=\begin{pmatrix}3\\10\\0\end{pmatrix}$
which returns the polynomial
$3+10x$
.
The image of this transformation is the set of polynomials of degree 2, since the differentiation reduces the degree of polynomials by 1. There is thus no identity element. 2. Inverse.
$T_1 \in \left\{ T \right\} \rightarrow T_1^{-1} \in \left\{ T \right\}$
. This condition is also not satisfied since the matrix above is not square, so has no inverse. 3. Closure
$T_1 , T_2 \in \left\{ T \right\} \rightarrow T_1 T_2 \in \left\{ T \right\}$
. This condition is not satisfied either.
Take
$T_1 = \left( \begin{array}{ccc} 1 & 2 & 0 \\ 6 & 4 & 0 \\ 1 & 0 & 2 \end{array} \right) , T_2 = \left( \begin{array}{cc} 1 & 0 \\ 6 & 4 \end{array} \right)$
.
These matrices cannot be multiplied.
4. Associativity is satified for matrix multiplication
Three of the four group axioms are not satisfied so the set of linear transformations is not a group. Howver, the set of linear transformations associted with invertible square matrices do form groups for each order.