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Column vectors are written as rows, such as \(\mathbf{x}=\left(x_{1}, x_{2}\right),\) and \(T(\mathbf{x})\) is written as \(T\left(x_{1}, x_{2}\right)\). Show that the transformation \(T\) defined by \(T\left(x_{1}, x_{2}\right)=\) \(\left(4 x_{1}-2 x_{2}, 3\left|x_{2}\right|\right)\) is not linear.

Short Answer

Expert verified
The transformation \(T\) is not linear because it fails the homogeneity condition.

Step by step solution

01

Understanding linearity

A transformation \(T\) is linear if for any vectors \(\mathbf{u}\) and \(\mathbf{v}\), and any scalar \(c\), it satisfies both the additivity condition \(T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})\) and the homogeneity condition \(T(c\mathbf{u}) = cT(\mathbf{u})\).
02

Evaluate homogeneity condition

Test the homogeneity condition of linearity on a simple vector \(\mathbf{x} = (x_1, x_2)\) and a scalar \(c\). Specifically, check whether \(T(c x_1, c x_2) = c T(x_1, x_2)\). Compute both sides using the definition of \(T\).
03

Compute T(c*x)

Compute \(T(c x_1, c x_2) = (4c x_1 - 2c x_2, 3|c x_2|)\).
04

Compute c*T(x)

Compute \(c T(x_1, x_2) = c(4 x_1 - 2 x_2, 3|x_2|) = (c(4 x_1 - 2 x_2), c\cdot 3|x_2|)\).
05

Comparison of results

Compare \(T(c x_1, c x_2) = (4c x_1 - 2c x_2, 3|c x_2|)\) and \(c T(x_1, x_2) = (c(4 x_1 - 2 x_2), 3c|x_2|)\). Notice that the first component matches, but the second component does not match because \(3|c x_2|\) does not equal \(3c|x_2|\) unless \(c\) is positive.
06

Conclude non-linearity due to homogeneity failure

Since \(T(c x_1, c x_2)\) does not equal \(c T(x_1, x_2)\) for any scalar \(c\leq 0\) (due to the absolute value operation inserting a non-linear aspect), the transformation \(T\) is not homogeneous. Thus, \(T\) is not linear.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Homogeneity Condition
The homogeneity condition is a critical aspect of linear transformations. This property states that a transformation, say \(T\), is homogeneous if it scales a scaled input vector correctly.
Mathematically, for a vector \( \mathbf{u} = (u_1, u_2) \) and a scalar \( c \), homogeneity requires that:
  • \( T(c \mathbf{u}) = c T(\mathbf{u}) \)
In simpler terms, this means that if you scale a vector by a factor \(c\), and then apply the transformation \(T\), it should be the same as first applying \(T\) to the vector and then scaling by \(c\).
For the transformation given by \(T(x_1, x_2) = (4x_1 - 2x_2, 3|x_2|)\), the issue arises with the second component, \(3|c x_2|\) does not equal \(3c|x_2|\) when \(c\leq 0\). The absolute value introduces a non-linearity here.
Hence, the homogeneity condition is violated, demonstrating non-linearity.
Additivity Condition
The additivity condition is another important criterion for a transformation to be linear. It states that the transformation of a sum must equal the sum of the transformations.
In formula form, for vectors \( \mathbf{u} \) and \( \mathbf{v} \):
  • \( T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \)
This means that applying a transformation \(T\) to the sum of two vectors should yield the same result as transforming each vector separately and then adding the results.
For the transformation \(T(x_1, x_2) = (4x_1 - 2x_2, 3|x_2|)\), the structure suggests issues might arise particularly in the second component due to the absolute value.
While the step by step analysis focuses on homogeneity, a similar detailed exploration can be done for additivity to confirm non-linearity.
Linear Transformation
Understanding linear transformations is essential when analyzing any function or transformation. Linear transformations abide by both the homogeneity and additivity conditions, which ensure constancy in vector spaces.
In essence, for any vectors and scalars, the transformation behaves consistently and predictably.
  • Two key properties of a linear transformation include the ability to scale and add vectors while maintaining structure.
  • Mathematically, they can be represented as matrix multiplications.
In contrast, the transformation \(T(x_1, x_2) = (4x_1 - 2x_2, 3|x_2|)\) is not linear.
This is primarily due to its violation of the homogeneity condition, a manifestation of the absolute value in the second component which does not conform to linear algebra norms.
Absolute Value Function
The absolute value function, |x|, uniquely impacts the linearity of transformations. It measures the magnitude of a number or vector component, disregarding its sign.
While valuable in many contexts, this property disrupts linearity.
Here's why:
  • When scalars are negative, absolute values alter scaling, affecting outcomes predicted by both homogeneity and additivity.
  • This non-linear behavior inhibits uniform scaling and consistent additive properties.
For the transformation \(T(x_1, x_2) = (4x_1 - 2x_2, 3|x_2|)\), its presence in the second component is a root cause of its non-linear nature.
The absolute value function introduces complexity not present in purely linear scenarios, making these transformations particularly unique and sometimes challenging to work with.

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Most popular questions from this chapter

If a linear transformation \(T : \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}\) maps \(\mathbb{R}^{n}\) onto \(\mathbb{R}^{m}\) , can you give a relation between \(m\) and \(n ?\) If \(T\) is one-to-one what can you say about \(m\) and \(n ?\)

Suppose \(A\) is a \(3 \times 3\) matrix and \(y\) is a vector in \(\mathbb{R}^{3}\) such that the equation \(A \mathbf{x}=\mathbf{y}\) does not have a solution. Does there exist a vector \(\mathbf{z}\) in \(\mathbb{R}^{3}\) such that the equation \(A \mathbf{x}=\mathbf{z}\) has a unique solution? Discuss.

Balance the chemical equations in Exercises \(5-10\) using the vector equation approach discussed in this section. The following reaction between potassium permanganate \(\left(\mathrm{KMnO}_{4}\right)\) and manganese sulfate in water produces manganese dioxide, potassium sulfate, and sulfuric acid: \(\mathrm{KMnO}_{4}+\mathrm{MnSO}_{4}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{MnO}_{2}+\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4}\) [For each compound, construct a vector that lists the numbers of atoms of potassium (K), manganese, oxygen, sulfur, and hydrogen.]

Let \(A\) be an \(m \times n\) matrix and let \(\mathbf{u}\) be a vector in \(\mathbb{R}^{n}\) that satisfies the equation \(A \mathbf{x}=\mathbf{0} .\) Show that for any scalar \(c,\) the vector \(c \mathbf{u}\) also satisfies \(A \mathbf{x}=\mathbf{0} .[\text { That is, show that } A(c \mathbf{u})=\mathbf{0} .]\)

Should be solved without performing row operations. [Hint: Write \(A \mathbf{x}=\mathbf{0}\) as a vector equation. Given \(A=\left[\begin{array}{rrr}{4} & {1} & {6} \\ {-7} & {5} & {3} \\ {9} & {-3} & {3}\end{array}\right],\) observe that the first column. plus twice the second column equals the third column. Find a nontrivial solution of \(A \mathbf{x}=\mathbf{0} .\)

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