If \(V\) and \(W\) are subvector spaces of \(\RNrSpc{n}\), then the kernel of a linear map \(L\from V\to W\) is a subspace of \(V\). Moreover, if the \((m,n)\)-matrix \(\Mtrx{A}\) represents \(L\), then \(\KerOf{L} = \NullSpc{\Mtrx{A}}\)
If \(V\) and \(W\) are subvector spaces of \(\RNrSpc{n}\), then the kernel of a linear map \(L\from V\to W\) is a subspace of \(V\). Moreover, if the \((m,n)\)-matrix \(\Mtrx{A}\) represents \(L\), then \(\KerOf{L} = \NullSpc{\Mtrx{A}}\)
To see that \(\KerOf{L}\) is a subspace of \(V\) observe the following
If \(\Vect{x}\) and \(\Vect{y}\) are in \(\KerOf{L}\), then this means that
\[L(\Vect{x}) = \Vect{0} = L(\Vect{y})\]But then
| \(L(\Vect{x} + \Vect{y})\) | \(= \) | \(L(\Vect{x}) + L(\Vect{y})\) |
| \(\) | \(=\) | \(\Vect{0} + \Vect{0} = \Vect{0}\) |
This means that \(\KerOf{L}\) is closed under addition.
If \(\Vect{x}\) is in \(\KerOf{L}\), then this means that \(L(\Vect{x}) = \Vect{0}\). Now, if \(t\) in \(\RNr\) is arbitrary, then
| \(L(t\cdot \Vect{x})\) | \(=\) | \(t\cdot L(\Vect{x})\) |
| \(\) | \(=\) | \(t\cdot \Vect{0} = \Vect{0}\) |
This means that \(\KerOf{L}\) is closed under scalar multiplication.
Thus \(\KerOf{L}\) is a subspace of \(V\).
To see the second part of the proposition, recall that \(L(\Vect{x}) = \Mtrx{A}\cdot \Vect{x}\), for all \(\Vect{x}\) in \(\RNrSpc{n}\). But this means exactly that \(\Vect{x}\) belongs to \(\KerOf{L}\) if and only if \(\Vect{x}\) is in \(\NullSpc{\Mtrx{A}}\). – The proof of the proposition is complete.