Problem 32 - Differential Equations and Linear Algebra

Question

Let \(T: V \rightarrow V\) be a linear transformation, and suppose that $$ \begin{aligned} T\left(2 \mathbf{v}_{1}+3 \mathbf{v}_{2}\right) &=\mathbf{v}_{1}+\mathbf{v}_{2} \\\ T\left(\mathbf{v}_{1}+\mathbf{v}_{2}\right) &=3 \mathbf{v}_{1}-\mathbf{v}_{2}. \end{aligned} $$ Find \(T\left(\mathbf{v}_{1}\right)\) and \(T\left(\mathbf{v}_{2}\right)\).

Step-by-Step Solution

Answer

\( T(\mathbf{v}_{1}) = 7\mathbf{v}_{1} - 4\mathbf{v}_{2} \) and \( T(\mathbf{v}_{2}) = 3\mathbf{v}_{2} - 5\mathbf{v}_{1} \)

Step 1: Find B\

\( -B = 5v_1 -3v_2 \Rightarrow B = 3v_2 - 5v_1 \) Now, we can plug in this value of B into the equation 1 and solve for A. \

Step 2: Find A\

\( 2A + 3(3v_2 - 5v_1) = v_1 + v_2 \Rightarrow 2A = 14v_1 - 8v_2 \Rightarrow A = 7v_1 - 4v_2 \) Now we have found T(v₁) and T(v₂): \

Step 3: Final Answer\

\( T(\mathbf{v}_{1}) = 7\mathbf{v}_{1} - 4\mathbf{v}_{2} \) and \( T(\mathbf{v}_{2}) = 3\mathbf{v}_{2} - 5\mathbf{v}_{1} \)

Key Concepts

Vector Spaces

In the context of linear algebra, a vector space is a fundamental concept referring to a collection of vectors that can be added together and multiplied by scalars (numbers) to produce another vector within the same space. Key properties of vector spaces include having a zero vector, associativity and commutativity of vector addition, and distribution of scalar multiplication.

A linear transformation, like the one considered in the exercise, operates within the confines of a vector space, mapping vectors from one space to another, or within the same space, while preserving vector addition and scalar multiplication. When we talk about finding images of specific vectors under transformation, it's essential to understand that we are working within the rules and limits of a particular vector space. This contextual understanding is crucial for solving the exercise and grasping the broader implications of linear transformations in vector spaces.

Linear Algebra

Linear algebra is the branch of mathematics concerned with vectors, vector spaces, linear transformations, and systems of linear equations. It's instrumental in almost every area of mathematics, including differential equations, as it provides the theoretical framework necessary to solve linear problems efficiently.

In solving the given exercise, we use methods of linear algebra to determine the output of a linear transformation on a particular set of basis vectors. Linear algebra provides the techniques like the step-by-step solution to manipulate and solve for unknowns in the equations representing the transformation, allowing us to express the transformations of individual vectors.

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives, representing physical phenomena like motion, heat, and wave propagation. Linear transformations often come into play when solving systems of differential equations, as they can simplify complex problems into more manageable forms by changing the basis of vector spaces.

The connection with our exercise lies in the properties of linear transformations that are used to solve linear differential equations. Although the original problem doesn't involve differential equations directly, the linear algebra skills employed to find the transformation of vectors are similarly used in differential equations to transform and solve for unknown functions.

Basis Vectors

Basis vectors are a set of vectors in a vector space that, through linear combinations, can represent every vector in that space. The concept is vital in linear algebra because once you know how a linear transformation affects the basis vectors, you can determine its effect on any vector in the space.

This central idea was used in the step-by-step solutions to find the images of the basis vectors under the given linear transformation. By writing equations based on the transformation properties as given, the problem was reduced to finding the representation of the transformed basis vectors, which in turn allowed the solver to express the transformation of any vector in the space in terms of these basis vectors.