Problem 13 - Differential Equations and Linear Algebra

Question

Find \(\operatorname{Ker}(T)\) and \(\operatorname{Rng}(T),\) and hence, determine whether the given transformation is one-to-one, onto, both, or neither. If \(T^{-1}\) exists, find it. $$T(\mathbf{x})=A \mathbf{x}, \text { where } A=\left[\begin{array}{rrr} 1 & 2 & -1 \\ 2 & 5 & 1 \end{array}\right]$$

Step-by-Step Solution

Answer

The kernel of the transformation is \(\operatorname{Ker}(T) = \operatorname{Span}\left\{\begin{bmatrix} -7 \\ -3 \\ 1 \end{bmatrix}\right\}\) and the range is \(\operatorname{Rng}(T) = \operatorname{Span}\left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right\}\), which is the entire target space R^2. Since the kernel is not only the zero vector, the transformation is not one-to-one, and as the range is the entire target space R^2, the transformation is onto. Therefore, the transformation is neither one-to-one nor bijective, and an inverse does not exist.

Step 1: Find Ker(T)

To find the kernel of the transformation, we want to solve the homogeneous system of equations Ax = 0: \[\left[\begin{array}{rrr} 1 & 2 & -1 \\\ 2 & 5 & 1 \end{array}\right] \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.\] First, we write down the augmented matrix, and then perform Gaussian elimination to find the solutions: \[\left[\begin{array}{rrr|r} 1 & 2 & -1 & 0 \\\ 2 & 5 & 1 & 0 \end{array}\right] \overset{R_2-2R_1}\longrightarrow \left[\begin{array}{rrr|r} 1 & 2 & -1 & 0 \\\ 0 & 1 & 3 & 0 \end{array}\right] \overset{R_1-2R_2}\longrightarrow \left[\begin{array}{rrr|r} 1 & 0 & 7 & 0 \\\ 0 & 1 & 3 & 0 \end{array}\right].\] So we have the system of equations: \( x_1 + 7x_3 = 0 \), \( x_2 + 3x_3 = 0 \). The general solution is given by \(x_3t\), where \(t\) is a parameter: \[\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -7t \\ -3t \\ t \end{bmatrix} = t\begin{bmatrix} -7 \\ -3 \\ 1 \end{bmatrix},\] which means \(\operatorname{Ker}(T) = \operatorname{Span}\left\{\begin{bmatrix} -7 \\ -3 \\ 1 \end{bmatrix}\right\}\).

Step 2: Find Rng(T)

Since the given transformation is a 3x2 matrix (from R^3 to R^2), its columns span the range. The range of the transformation is the column space of the matrix \(A\), which we can find by looking at the pivots in the reduced row echelon form computed above: \[\left[\begin{array}{rrr} 1 & 0 & 7 \\\ 0 & 1 & 3 \end{array}\right].\] The first and second columns are the pivot columns, so the range is \(\operatorname{Rng}(T) = \operatorname{Span}\left\{\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}\right\}\), which is the entire target space R^2.

Step 3: Determine the Type of Transformation

Since the kernel is not only the zero vector, the transformation is not one-to-one. Since the range is the entire target space R^2, the transformation is onto. So the transformation is neither one-to-one nor bijective, and an inverse does not exist.

Key Concepts

Kernel of a Transformation

The kernel of a transformation, often denoted as \( \operatorname{Ker}(T) \), is a key concept in linear algebra. It is the set of all vectors \( \mathbf{x} \) in the domain of \( T \) such that \( T(\mathbf{x}) = \mathbf{0} \). In simpler terms, it includes all the vectors that get mapped to the zero vector when the transformation is applied.

To find the kernel of the transformation \( T(\mathbf{x}) = A \mathbf{x} \) given a matrix \( A \), solve the equation \( A \mathbf{x} = \mathbf{0} \). This involves setting up a system of linear equations that you can solve using methods such as Gaussian elimination. This refers to strategically applying row operations to simplify the matrix and discover the solutions.

In our example, the kernel forms a one-dimensional subspace, spanned by \( \begin{bmatrix} -7 \ -3 \ 1 \end{bmatrix} \). This means that the transformation \( T \) maps all linear combinations of this vector to zero. It's important to note that if the kernel contains only the zero vector, the transformation is called "one-to-one" or injective.

Range of a Transformation

The range of a transformation, denoted as \( \operatorname{Rng}(T) \), refers to the set of all possible output vectors, or images, that can be produced by applying \( T \) to any vector in the domain. Think of it as all the directions we can reach by transforming the input vectors through \( T \).

For a matrix \( A \) representing the transformation \( T(\mathbf{x}) = A \mathbf{x} \), the range is the column space of \( A \). It consists of all linear combinations of the columns of \( A \). To find it, we focus on the pivot columns in the reduced row echelon form (RREF) of \( A \). These pivot columns highlight the essential directions that span the range.

In our given example, the transformation \( T \) maps \( \mathbb{R}^3 \) to \( \mathbb{R}^2 \). We've identified the pivot columns corresponding to the full \( \mathbb{R}^2 \) space, meaning \( \operatorname{Rng}(T) = \mathbb{R}^2 \). When the range is the entire target space, the transformation is called "onto" or surjective.

Gaussian Elimination

Gaussian elimination is a fundamental technique in linear algebra used to solve systems of linear equations. It's also instrumental in finding the rank, nullity, and inverse of a matrix. The process involves performing operations on the rows of the matrix to bring it into a simpler form, namely the row echelon form or, ideally, the reduced row echelon form.

These row operations include swapping two rows, multiplying a row by a non-zero scalar, and adding or subtracting the multiple of one row from another. The goal is to create zeroes below the pivots (leading 1s) of each matrix column.

In the example problem, Gaussian elimination helped simplify the augmented matrix:

• It identified the kernel by creating equations from the simplified matrix form.
• It also highlighted the columns that contribute to the range of the transformation (pivots).

This efficient and systematic procedure allows us to dissect and understand the intricacies of linear transformations effectively.