\[T\]
operating on a set \[S\]
is the set of elements \[T(x) \in T(S)S\]
i.e.e the set of elements of the codomain that are the image of some element of \[\]
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 image of this transformation is the set of polynomials of degree 2, since the differentiation reduces the degree of polynomials by 1..
The image of a transformation has certain properties. The image of a transformation is a subspace of the codomain.
1.
\[T(0)=0\]
2. If
\[T(x),T(y) \in im(T) \rightarrow T(\alpha x+ \beta y)= \alpha T(x) + \beta T(y) \in im(T)\]
The dimension of the image as a subspace is always less than or equal to the dimension of the domain, considered as a subspace.