91Ó°ÊÓ

We can express the linear transformation T as a matrix product. Given that the transformation is defined as \(T(x, y, z, t) = (x-y + z + t, x + 2z - t, x + y + 3z - 3t)\), we represent this transformation as: \[T \begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix} = \begin{pmatrix} 1 & -1 & 1 & 1 \\ 1 & 0 & 2 & -1 \\ 1 & 1 & 3 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix} \] The matrix representation of T is: \[A = \begin{pmatrix} 1 & -1 & 1 & 1 \\ 1 & 0 & 2 & -1 \\ 1 & 1 & 3 & -3 \end{pmatrix}\]

The column space of A is the subspace of \(\mathrm{R}^{3}\) spanned by the columns of A, which is also the image of T. We can compute the column space by performing Gaussian elimination on the matrix A: \[\begin{pmatrix} 1 & -1 & 1 & 1 \\ 1 & 0 & 2 & -1 \\ 1 & 1 & 3 & -3 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & -1 & 1 & 1 \\ 0 & 1 & 1 & -2 \\ 0 & 0 & 2 & -4 \end{pmatrix}\] After Gaussian elimination, the resulting matrix is in row-echelon form. We can see that the first three columns are linearly independent (due to non-zero leading entries), and the last column is linearly dependent. The column space (the image of T) is spanned by the first three columns of A: \[C(A) = \operatorname{span}\left\{ \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \right\}\] Thus, the basis of the image of T is: \[\operatorname{B}_{\operatorname{im}(T)} = \left\{ \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \right\}\] And the dimension of the image of T is: \[\operatorname{dim}(\operatorname{im}(T)) = 3\]

The nullspace (kernel) of T (or matrix A) is the set of all vectors that satisfy \(A\vec{x} = 0\). Since the fourth column is linearly dependent, it can be ignored in this calculation. From the row-echelon form of A, we can write the following system of equations: 1. \(x - y + z = -t\) 2. \(y + z = 2t\) 3. \(2z = 4t\) Dividing equation (3) by 2, we get: \[z = 2t\] Using equation (2), we can eliminate z and obtain: \[y = 0\] From equation (1): \[x-t = y+z = z = 2t\] So, \[x = 3t\] Therefore, the kernel of T can be written as the span of vector \(t\begin{pmatrix} 3 \\ 0 \\ 2 \\ 1 \end{pmatrix}\) for arbitrary constant t. Thus, the basis for the kernel of T is: \[\operatorname{B}_{\operatorname{ker}(T)} = \left\{ \begin{pmatrix} 3 \\ 0 \\ 2 \\ 1 \end{pmatrix} \right\}\] And the dimension of the kernel of T is: \[\operatorname{dim}(\operatorname{ker}(T)) = 1\] #Summary# For the linear transformation T, we have found: 1. A basis for the image of T: \(\operatorname{B}_{\operatorname{im}(T)} = \left\{ \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \right\}\) 2. The dimension of the image of T: \(\operatorname{dim}(\operatorname{im}(T)) = 3\) 3. A basis for the kernel of T: \(\operatorname{B}_{\operatorname{ker}(T)} = \left\{ \begin{pmatrix} 3 \\ 0 \\ 2 \\ 1 \end{pmatrix} \right\}\) 4. The dimension of the kernel of T: \(\operatorname{dim}(\operatorname{ker}(T)) = 1\)