Let \(T:V\rightarrow W\) be a linear transformation where \(V\) and \(W\) be vector spaces with scalars coming from the same field \(\mathbb{F}\). The kernel of \(T\), denoted by \(\ker(T)\), is the set of vectors from \(V\) that gets mapped to the zero vector in \(W\); that is, \[\ker(T) = \{ v \in V : T(v) = 0_W\}.\]
Let \(T\) be given by \(T(x) = Ax\) for some \(A \in \mathbb{R}^{m \times n}\). By definition, the kernel of \(T\) is given by the set of \(x\) such that \(T(x) = 0\). But \(T(x) = 0\) precisely when \(Ax = 0\). Therefore, \(\ker(T) = N(A)\), the nullspace of \(A\).
Let \(T\) be a linear transformation from \(P_2\) to \(\mathbb{R}^2\) given by \(T(ax^2 + bx + c) = \begin{bmatrix} a+3c \\ a-c\end{bmatrix}\). The kernel of \(T\) is the set of polynomials \(ax^2+bx+c\) such that \(\begin{bmatrix} a+3c \\a - c\end{bmatrix}= \begin{bmatrix} 0\\0\end{bmatrix}\). Solving for \(a\) and \(c\), we get \(a = c = 0\). So \(\ker(T) = \{ bx : b \in \mathbb{R}\}\).
Let \(T:V\rightarrow W\) be a linear transformation. Suppose that you are asked to find all solutions to \(T(x) = b\) for some \(b \in W\). If you have found one solution, say \(\tilde{x}\), then the set of all solutions is given by \(\{\tilde{x} + \phi : \phi \in \ker(T)\}\). In other words, knowing a single solution and a description of the kernel of \(T\) tells you all the solutions to \(T(x) = b\).
When \(T(x)\) is given by \(Ax\), this amounts to saying that to specify all solutions to \(Ax = b\), we just need one particular solution and a description of the nullspace of \(A\).
Give the kernel for each of the following linear transformations.
\(T(ax+b) = a-b\) where \(a,b\in \mathbb{R}\).
\(T(x) = Ax\) where \(x = \begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix}\) and \(A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\).