Question

A transformation T is given. Determine whether or not T is linear; if not, state why not. $$ T\left(\left[\begin{array}{l} x_{1} \\ x_{2} \end{array}\right]\right)=\left[\begin{array}{l} 0 \\ 0 \end{array}\right] $$

Short answer

The transformation \( T \) is linear.

Step-by-step solution

Understand the Definition of a Linear Transformation

A transformation \( T: \mathbb{R}^n \to \mathbb{R}^m \) is linear if for any vectors \( \,\mathbf{u}, \,\mathbf{v} \in \mathbb{R}^n \, \) and any scalars \( c \in \mathbb{R} \), it satisfies the following properties: \( T(\mathbf{u}+\mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \) and \( T(c\mathbf{u}) = cT(\mathbf{u}) \). The transformation is linear if both properties hold.

Apply Transformation to a Single Vector

Given the transformation \( T\left(\left[\begin{array}{l} x_{1} \ x_{2} \end{array}\right]\right) = \left[\begin{array}{l} 0 \ 0 \end{array}\right] \), we see that any vector \( \mathbf{u} = \left[\begin{array}{l} u_{1} \ u_{2} \end{array}\right] \), when transformed by \( T \), results in the zero vector \( \left[\begin{array}{l} 0 \ 0 \end{array}\right] \).

Check Additivity Property

Suppose \( \mathbf{u} = \left[\begin{array}{l} u_{1} \ u_{2} \end{array}\right] \) and \( \mathbf{v} = \left[\begin{array}{l} v_{1} \ v_{2} \end{array}\right] \). The transformation of their sum is: \[ T(\mathbf{u} + \mathbf{v}) = T\left(\left[\begin{array}{l} u_{1} + v_{1} \ u_{2} + v_{2} \end{array}\right]\right) = \left[\begin{array}{l} 0 \ 0 \end{array}\right]. \] Similarly, \[ T(\mathbf{u}) + T(\mathbf{v}) = \left[\begin{array}{l} 0 \ 0 \end{array}\right] + \left[\begin{array}{l} 0 \ 0 \end{array}\right] = \left[\begin{array}{l} 0 \ 0 \end{array}\right]. \] Thus, the additivity property holds.

Check Homogeneity Property

For a scalar \( c \in \mathbb{R} \), consider \( T(c\mathbf{u}) \). We have: \[ T\left(c\left[\begin{array}{l} u_{1} \ u_{2} \end{array}\right]\right) = T\left(\left[\begin{array}{l} cu_{1} \ cu_{2} \end{array}\right]\right) = \left[\begin{array}{l} 0 \ 0 \end{array}\right]. \] Meanwhile, \[ cT(\mathbf{u}) = c\left(\left[\begin{array}{l} 0 \ 0 \end{array}\right]\right) = \left[\begin{array}{l} 0 \ 0 \end{array}\right]. \] The homogeneity property is also satisfied.

Conclusion About Linearity

Since both additivity \( T(\mathbf{u}+\mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \) and homogeneity \( T(c\mathbf{u}) = cT(\mathbf{u}) \) properties are satisfied by \( T \), the transformation is linear.

Key concepts

Additivity and Linear Transformations

Linear transformations are key components in vector spaces. For a transformation to be linear, it must preserve vector addition. This is what we call the additivity property.
Imagine you have two vectors in the space, let’s call them \( \mathbf{u} \) and \( \mathbf{v} \).

• The transformation should give the same result whether you combine the vectors first and then apply the transformation, or transform them individually and then combine the results.
• This means, mathematically, the transformation \( T \) should satisfy: \( T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \).

In the provided example, the transformation of any vector results in a zero vector. So, when we transform the sum of two vectors, we still get a zero vector — just like when we add two zero vectors together. This clearly shows that the additivity property holds for our transformation, supporting its linearity.

Homogeneity in Vector Spaces

Apart from adding vectors, linear transformations must also respect scaling of vectors. This characteristic is known as the homogeneity property. To grasp this, think of any vector \( \mathbf{u} \) being scaled by a number, which we can call a scalar \( c \).

• For a transformation to be linear, multiplying a vector by a scalar and then applying the transformation should be the same as applying the transformation to the vector first and then multiplying the outcome by the scalar.
• In formula terms, \( T(c\mathbf{u}) = cT(\mathbf{u}) \) must hold.

In the given transformation example, a scaled vector “disappears” into the zero vector. Mathematically speaking, both transforming first or multiplying the zero vector by any real scalar yields a zero vector. This confirms that our transformation upholds the homogeneity property.

Zero Transformation

In some transformations, every input vector gets mapped to the zero vector. This special kind of transformation is called a zero transformation.
Why is it called zero? Because no matter what vector you feed into the transformation, the output is always the zero vector.

• The transformation shown \( T(x_1, x_2) = (0, 0) \), is a perfect example of this. Here, no matter the components \( x_1 \) and \( x_2 \), the resulting vector is always zero.

Zero transformations are inherently linear. Both the additivity and homogeneity properties are naturally satisfied. Think of it as giving the same 'reply' to every vector! This means zero transformations not only meet all the criteria for linearity but do so in a neat, consistent way.