Problem 12 - 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. $$T(\mathbf{x})=A \mathbf{x}, \text { where } A=\left[\begin{array}{rr} 1 & 2 \\ -2 & -4 \end{array}\right]$$

Step-by-Step Solution

Answer

The kernel of the given linear transformation \(T(\mathbf{x})=A\mathbf{x}\) with \(A=\left[\begin{array}{rr} 1 & 2 \\ -2 & -4 \end{array}\right]\) is \(\operatorname{Ker}(T) = \operatorname{span}\{(-2, 1)\}\). The range is \(\operatorname{Rng}(T)=\operatorname{span}\left\{\begin{bmatrix}1 \\ -2\end{bmatrix}\right\}\). The transformation is neither one-to-one nor onto, and it doesn't have an inverse transformation \(T^{-1}\).

Step 1: Define the Linear Transformation

We are given the matrix \(A\) for the linear transformation \(T(\mathbf{x})=A\mathbf{x}\) as: \[ A=\left[\begin{array}{rr} 1 & 2 \\ -2 & -4 \end{array}\right] \]

Step 2: Find the Kernel

To find the kernel of the transformation \(T\), \(\operatorname{Ker}(T)\), we need to find the vectors \(\mathbf{x}\) that satisfy \(T(\mathbf{x})=\mathbf{0}\), where \(\mathbf{0}\) is the zero vector. This means solving the following equation for \(\mathbf{x}\): \[A\mathbf{x}=\mathbf{0}\] \[ \left[\begin{array}{rr} 1 & 2 \\ -2 & -4 \end{array}\right] \left[\begin{array}{c} x_{1} \\ x_{2} \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \end{array}\right] \] Now we can rewrite the system of equations: \[\text {Case 1: } x_{1}+2 x_{2}=0\] \[\text {Case 2: } -2x_{1}-4 x_{2}=0\] Notice that Case 2 is actually a multiple of Case 1, we only need to solve Case 1: \[x_{1} = -2x_{2}\] Any vector with these coordinates will be in the Kernel, so \(\operatorname{Ker}(T) = \operatorname{span}\{(-2, 1)\}\).

Step 3: Find the Range

To find the range, \(\operatorname{Rng}(T)\), we need to find the span of the columns of the matrix \(A\). The columns of \(A\) are linearly dependent since the second column is a multiple of the first. Thus, the range of the transformation is given by: \[\operatorname{Rng}(T)=\operatorname{span}\left\{\begin{bmatrix}1 \\ -2\end{bmatrix}\right\}\]

Step 4: Determine if T is One-to-One, Onto, Both or Neither

1. \(T\) is one-to-one if and only if \(\operatorname{Ker}(T)\) contains only the zero vector. In our case, \(\operatorname{Ker}(T)\) contains more than the zero vector. Therefore, \(T\) is not one-to-one. 2. \(T\) is onto if and only if \(\operatorname{Rng}(T)\) spans the output vector space (\(\mathbb{R}^2\) in this case). In our case, \(\operatorname{Rng}(T)\) doesn't span the entire output vector space, since the columns of \(A\) are linearly dependent and form a line in \(\mathbb{R}^2\). Therefore, \(T\) is not onto. Thus, the linear transformation \(T\) is neither one-to-one nor onto.

Step 5: Determine if There Exists an Inverse Transformation

A linear transformation \(T\) has an inverse \(T^{-1}\) if and only if it is both one-to-one and onto. In our case, since the transformation is neither one-to-one nor onto, it does not have an inverse transformation \(T^{-1}\). #Conclusion# In conclusion, for the given linear transformation \(T(\mathbf{x})=A\mathbf{x}\) with \(A=\left[\begin{array}{rr} 1 & 2 \\ -2 & -4 \end{array}\right]\), the kernel \(\operatorname{Ker}(T) = \operatorname{span}\{(-2, 1)\}\) and the range \(\operatorname{Rng}(T)=\operatorname{span}\left\{\begin{bmatrix}1 \\ -2\end{bmatrix}\right\}\). The linear transformation is neither one-to-one nor onto, and it doesn't have an inverse transformation \(T^{-1}\).

Key Concepts

Kernel of a Linear Transformation

The kernel of a linear transformation, often denoted as \( \operatorname{Ker}(T) \), is a fundamental concept in linear algebra. It consists of all the vectors \( \mathbf{x} \) that, when transformed by \( T \), become the zero vector. Simply put, it’s the set of vectors that "get lost" under the transformation.

For our given transformation, which is defined by the matrix \( A = \begin{bmatrix} 1 & 2 \ -2 & -4 \end{bmatrix} \), the kernel can be found by solving the equation \( A\mathbf{x} = \mathbf{0} \). When we perform this calculation, we end up with the equation \( x_1 + 2x_2 = 0 \). This means that every vector of the form \( \mathbf{x} = \begin{bmatrix} -2c \ c \end{bmatrix} \) is in the kernel, where \( c \) is any real number. The kernel is spanned by \( \operatorname{span}\{ \begin{bmatrix} -2 \ 1 \end{bmatrix} \} \).

The size of the kernel tells us a lot about the transformation's properties. If the kernel only includes the zero vector, the transformation is injective or one-to-one. In our case, because the kernel contains more than just the zero vector, \( T \) is not one-to-one.

Range of a Linear Transformation

The range of a linear transformation, \( \operatorname{Rng}(T) \), refers to the set of all possible outputs when vectors from the domain are transformed by \( T \). In other words, it's the subspace of the codomain that \( T \) maps onto.

To determine the range in our example, we must find the span of the columns of the matrix \( A = \begin{bmatrix} 1 & 2 \ -2 & -4 \end{bmatrix} \). Observing these columns, we notice that the second column \( \begin{bmatrix} 2 \ -4 \end{bmatrix} \) is merely a multiple of the first column \( \begin{bmatrix} 1 \ -2 \end{bmatrix} \). Therefore, they don't form a linearly independent set; rather, they form a span that is one-dimensional.

Thus, the range of \( T \) is \( \operatorname{Rng}(T) = \operatorname{span}\{ \begin{bmatrix} 1 \ -2 \end{bmatrix} \} \). This means \( T \) maps its inputs to a line along \( \begin{bmatrix} 1 \ -2 \end{bmatrix} \) in \( \mathbb{R}^2 \). Because the range is not the entirety of \( \mathbb{R}^2 \), \( T \) is not onto (surjective).

One-to-One and Onto Transformations

Understanding whether a transformation is one-to-one or onto involves exploring its injective and surjective properties.

A transformation is **one-to-one** (injective) if it maps distinct inputs to distinct outputs. In practical terms, this means the kernel must include only the zero vector. For our example, since the kernel contains non-zero vectors, \( T \) is not one-to-one.

• Kernel implies no loss of distinctness, only zero vector = one-to-one

When a transformation is **onto** (surjective), it means that every possible output (in the codomain) can be traced back to some original input. In our scenario, the range of \( T \), which is just a single line in \( \mathbb{R}^2 \), matches only part of the codomain. Hence, \( T \) is not onto.

• Range covers entire codomain = onto

For a linear transformation to be invertible, it must be both one-to-one and onto. Given that \( T \) together lacks both these properties, an inverse transformation does not exist. Ensure you understand these concepts by seeing them in the matrix's specific context.