On the definition of ‘kernel’:
| \(\KerOf{L}\) | \(\DefEq\) | \(\Set{ \Vect{x} \in V \st L(\Vect{x}) = \Vect{0} }\) |
Read this as: ‘kernel of L is, by definition, equal to the set of all those x in V such that L of x equals 0.’
Thus the kernel of \(L\) consists of all those \(\Vect{x}\) in \(V\) with \(L(\Vect{x}) = \Vect{0}\); that is the kernel of \(L\) consists of those vectors of \(V\) which get transformed into the \(\Vect{0}\)-vector by \(L\).
If several vectors of \(V\) get transformed into one, then this means that the distinction that previously existed between these vectors has been destroyed. Accordingly, a large kernel corresponds to a highly destructive map. A small kernel corresponds to a map which preserves most of the distinctions between various vectors of \(V\). So such a map is less destructive or more faithful.
We will see soon that the kernel of \(L\) is a subvector space of \(V\). So it has a dimension, and this dimension provides a measure for the destructiveness of \(L\).
On the definition of ‘image’:
| \(\ImgOf{L}\) | \(:=\) | \(\Set{ \Vect{y}\in W \st \Vect{y}=L(\Vect{x})\quad \text{for some}\quad \Vect{x}\in V }\) |
Read this as: ‘image of L is, by definition, equal to the set of all those y in W such that L of x equals y, for some x in V.’
In other words, a vector \(\Vect{w}\) in \(W\) belongs to the image of \(L\) if the equation
| \(L(\Vect{x})\) | \(=\) | \(\Vect{y}\) |
has a solution or, in equivalently, if there is a vector in \(V\) which gets transformed in \(\Vect{y}\) by \(L\).
Thus \(\ImgOf{L}\) consists of all those \(\Vect{y}\) in \(W\) into which \(L\) transforms some \(\Vect{x}\) in \(V\). Therefore \(L\) transforms \(V\) into its image \(\ImgOf{L}\).
We will soon see that \(\ImgOf{L}\) is a subspace of \(W\).