## The Kernel of a Linear Transformation

The kernel of a linear transformation
$T$
operating on a set
$S$
is the set of elements
$x \in S$
for which
$T(x)=0$
so that
$T$
sends
$x$
to the zero element of the codomain.
Example. 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 kernel of this transformation is the set of constants, since the differential of a constant is zero.
The kernel of a transformation has certain properties. The kernel of a transformation is a subspace of the domain. 1.
$T(0)=0$

2. If
$x,y \in ker(T), T(x)=T(y)=0 \rightarrow T(\alpha x+ \beta y)= \alpha T(x) + \beta T(y) =\alpha \times 0 + \beta \times 0 =0$
and
$\alpha x+ \beta y \in ker(T)$

The zero element of the domain is always in the kernel. The dimension of the kernel as a subspace is always less than or equal to the dimension of the domain, considered as a subspace.