Problem 4 - Differential Equations and Linear Algebra

Question

Verify directly from Definition 6.1 .3 that the given mapping is a linear transformation. \(T: C^{2}(I) \rightarrow C^{0}(I)\) defined by $$ T(y)=y^{\prime \prime}-16 y $$.

Step-by-Step Solution

Answer

To verify that the given mapping \(T: C^{2}(I) \rightarrow C^{0}(I)\) defined by \(T(y)=y^{\prime \prime}-16 y\) is a linear transformation, we show that it satisfies superposition and homogeneity: 1. Superposition: For any functions \(y_1, y_2\) in \(C^2(I)\) and scalars \(c_1, c_2\), we compute \(T(c_1y_1 + c_2y_2)\) and find \(T(c_1y_1 + c_2y_2) = c_1T(y_1) + c_2T(y_2)\). 2. Homogeneity: For any function \(y\) in \(C^2(I)\) and scalar \(c\), we compute \(T(cy)\) and find \(T(cy) = c^2T(y)\). Since both conditions are satisfied, the given mapping is a linear transformation.

Step 1: Show Superposition

Let \(y_1, y_2\) be any functions in \(C^2(I)\) and \(c_1, c_2\) be any scalars. First, we need to compute \(T(c_1y_1 + c_2y_2)\). Using the definition of the given mapping: $$ T(c_1y_1 + c_2y_2) = (c_1y_1 + c_2y_2)^{\prime \prime} - 16(c_1y_1 + c_2y_2) $$ Now, take the second derivative and distribute the constant: $$ T(c_1y_1 + c_2y_2) = (c_1y_1^{\prime \prime} + c_2y_2^{\prime \prime}) - 16(c_1y_1 + c_2y_2) $$ Using the linearity of derivatives, we can separate each term: $$ T(c_1y_1 + c_2y_2) = c_1(y_1^{\prime \prime} - 16y_1) + c_2(y_2^{\prime \prime} - 16y_2) $$ We recognize that \(y_1^{\prime \prime} - 16y_1 = T(y_1)\) and \(y_2^{\prime \prime} - 16y_2 = T(y_2)\). So, we have: $$ T(c_1y_1 + c_2y_2) = c_1T(y_1) + c_2T(y_2) $$ We have now shown that the superposition condition is satisfied.

Step 2: Show Homogeneity

Let \(y\) be any function in \(C^2(I)\) and \(c\) be any scalar. Now, we need to compute \(T(cy)\). Using the definition of the given mapping: $$ T(cy) = (cy)^{\prime \prime} - 16(cy) $$ Take the second derivative and distribute the constant: $$ T(cy) = c^2y^{\prime \prime} - 16c^2y $$ Factoring \(c^2\) out, we get: $$ T(cy) = c^2(y^{\prime \prime} - 16y) $$ We recognize that \((y^{\prime \prime} - 16y) = T(y)\), so we have: $$ T(cy) = c^2T(y) $$ We have now shown that the homogeneity condition is satisfied. Since both superposition and homogeneity conditions are satisfied, we can conclude that the given mapping is a linear transformation.$$

Key Concepts

Definition of Linear Transformation

Understanding the foundation of a linear transformation is crucial for students delving into linear algebra. So, what exactly is a linear transformation? It is a type of function that maps one vector space to another and preserves the operations of vector addition and scalar multiplication. In essence, a linear transformation makes sure the structure of the space is maintained.

Consider a transformation \( T \) from a space called \( C^{2}(I) \) to another space \( C^{0}(I) \). In mathematical terms, we can express a linear transformation \( T \) by two main properties: \( T(v + w) = T(v) + T(w) \) for any vectors \( v \) and \( w \), and \( T(cv) = cT(v) \) for any vector \( v \) and scalar \( c \). If a mapping adheres to these rules, it can rightfully be called a linear transformation.

Superposition Principle

The superposition principle is a perfect illustration of what makes a transformation linear. What does it essentially state? If you have two inputs, let's call them \( y_1 \) and \( y_2 \), and you apply the transformation \( T \) to their sum, \( T(y_1 + y_2) \), it should be the same as if you transformed each one separately and then added the results, \( T(y_1) + T(y_2) \).

In the exercise, we used this principle to show that the transformation \( T(y) = y'' - 16y \) maintains superposition. By demonstrating that \( T(c_1y_1 + c_2y_2) = c_1T(y_1) + c_2T(y_2) \) for scalar coefficients \( c_1 \) and \( c_2 \), we've shown that the transformation respects the superposition principle, making it a candidate for being linear.

Homogeneity Property

Another important aspect of linear transformations is the homogeneity property. In layman's terms, a transformation is considered homogeneous if, when you scale an input by a certain factor, the output is scaled by the same factor.

Using the example from the provided exercise solution, you saw that when a scalar \( c \) multiplies an input function \( y \), the output of the transformation scales by \( c^2 \), which originally seems to violate the property. However, it is important to correct an error in the solution that assumed homogeneity would result in \( c^2T(y) \) while it should correctly be \( cT(y) \). Thus, the corrected form, which you should expect in a legitimate linear transformation, is \( T(cy) = c(y'' - 16y) \) which simplifies to \( T(cy) = cT(y) \). This correction is crucial, as homogeneity is a defining feature of linear transformations.

Second Derivative

When discussing functions in the context of calculus, the second derivative is a concept that pops up frequently, especially in conversations regarding concavity, acceleration, and inflection points. In a linear transformation task like the one we're examining, the second derivative is central to how the transformation itself is computed.

For a function \( y \) in the space \( C^{2}(I) \), the second derivative is denoted by \( y'' \). It represents how the rate of change of \( y \)'s rate of change is behaving. In our mapping \( T \), the second derivative interacts with the function - \( T(y) = y'' - 16y \). It's clear from this that the transformation involves taking the second derivative of the input function and then manipulating it further, emphasizing the importance of understanding second derivatives when dealing with such transformations.