Question

A transformation T is given. Determine whether or not T is linear; if not, state why not. $$ T\left(\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right]\right)=\left[\begin{array}{l} 1 \\ 1 \end{array}\right] $$

Short answer

T is not linear as it fails both additivity and homogeneity tests.

Step-by-step solution

Understand the Transformation

The given transformation \( T \) maps any vector \( \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \) to a constant vector \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \). This means \( T \) outputs the same vector regardless of the input vector.

Check Additivity

To check if \( T \) is linear, first verify additivity: for vectors \( \mathbf{u} \) and \( \mathbf{v} \), \( T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \). Here, \( T(\mathbf{u} + \mathbf{v}) = \begin{bmatrix} 1 \ 1 \end{bmatrix} \) while \( T(\mathbf{u}) = \begin{bmatrix} 1 \ 1 \end{bmatrix} \) and \( T(\mathbf{v}) = \begin{bmatrix} 1 \ 1 \end{bmatrix} \). So, \( T(\mathbf{u}) + T(\mathbf{v}) = \begin{bmatrix} 2 \ 2 \end{bmatrix} \), which does not equal \( T(\mathbf{u} + \mathbf{v}) \). Hence, \( T \) is not additive.

Check Homogeneity

Next, verify homogeneity: for a scalar \( c \) and vector \( \mathbf{u} \), \( T(c\mathbf{u}) = cT(\mathbf{u}) \). Here, \( T(c\mathbf{u}) = \begin{bmatrix} 1 \ 1 \end{bmatrix} \) while \( cT(\mathbf{u}) = c \begin{bmatrix} 1 \ 1 \end{bmatrix} \). Unless \( c = 1 \), these two expressions are not equal. Hence, \( T \) is not homogeneous.

Conclusion on Linearity

Since \( T \) fails the additivity test and the homogeneity test, it is not linear. Linear transformations must satisfy both properties, which \( T \) does not.

Key concepts

Additivity Property

The additivity property is a critical feature of linear transformations. It states that if a transformation is linear, then the transformation of the sum of two vectors will equal the sum of the transformations of the individual vectors.
Mathematically, for vectors \( \mathbf{u} \) and \( \mathbf{v} \), the transformation \( T \) should satisfy:

• \( T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \)

In the given problem, the transformation \( T \) outputs the fixed vector \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) for any input.
So when tested for additivity with vectors \( \mathbf{u} \) and \( \mathbf{v} \), it results in \( T(\mathbf{u} + \mathbf{v}) \) still being \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \).
But the transformation's output given in the problem means \( T(\mathbf{u}) + T(\mathbf{v}) = \begin{bmatrix} 2 \ 2 \end{bmatrix} \), failing the additivity test. Thus, \( T \) is not additive.

Homogeneity Property

For a transformation to be linear, it should also satisfy the homogeneity property. This property states that scaling a vector should result in a scaled transformation output.
In mathematical terms, if \( c \) is a scalar and \( \mathbf{u} \) is a vector, a linear transformation \( T \) must fulfill:

• \( T(c\mathbf{u}) = cT(\mathbf{u}) \)

In the given exercise, the transformation \( T \) transforms any scaled vector \( c\mathbf{u} \) to the same vector \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \).
On the other side, scaling the transformation output would produce \( c \begin{bmatrix} 1 \ 1 \end{bmatrix} \), making it dependent on the value of \( c \).
Unless \( c = 1 \), these two results will differ, showing that \( T \) fails the homogeneity check.

Vector Mapping

Vector mapping is a key idea in understanding transformations. It refers to how vectors from one space (input space) are mapped or transformed into another space (output space).
In a linear transformation, the mapping should revolve around the origin and maintain linearity characteristics, such as additivity and homogeneity.
However, the exercise transformation \( T \) maps every input vector \( \begin{bmatrix} x_1 \ x_2 \end{bmatrix} \) to the fixed output vector \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \), regardless of \( x_1 \) and \( x_2 \).
This implies a constant transformation, which does not consider the direction or magnitude of the input vector's components.
Linear transformations map origin to origin, but this transformation ignores input values completely, indicating that it is not following a linear vector mapping.

Non-linear Transformation

Non-linear transformations differ significantly from linear ones in that they do not adhere to the rigid rules of additivity and homogeneity.
A transformation is non-linear if it fails to map vectors in a way that maintains these two principles, as seen in the exercise problem.
The transformation \( T \) assigns the same output vector regardless of input variations, breaking the rules of linearity. This lack of adherence to the linearity rules clearly categorizes it as a non-linear transformation.
Non-linear transformations are useful in contexts where simplicity is not adequate, and they can capture more complex relationships between inputs and outputs.
In this exercise, however, the transformation's non-linearity mainly signifies a deviation from expected linear behavior due to its constant mapping.