Problem 37 - Differential Equations and Linear Algebra

Question

Let \(\mathbf{v}_{1}=(1,-1,1), \mathbf{v}_{2}=(2,1,3),\) and \(\mathbf{v}_{3}=\) (-1,-1,2) be eigenvectors of the matrix \(A\) corresponding to the eigenvalues \(\lambda_{1}=2, \lambda_{2}=-2,\) and \(\lambda_{3}=3,\) respectively, and let \(\mathbf{v}=(5,0,3)\). (a) Express \(\mathbf{v}\) as a linear combination of \(\mathbf{v}_{1}, \mathbf{v}_{2}\) and \(\mathbf{v}_{3}\). (b) Find \(A \mathbf{v}\).

Step-by-Step Solution

Answer

The coefficients for expressing \(\mathbf{v}\) as a linear combination of \(\mathbf{v}_1\), \(\mathbf{v}_2\), and \(\mathbf{v}_3\) are \(c_1 = 2\), \(c_2 = 1\), and \(c_3 = 1\). Therefore, the linear combination is given by: \[\mathbf{v} = 2\mathbf{v}_1 + \mathbf{v}_2 + \mathbf{v}_3\] By using the property of eigenvectors and eigenvalues, we find: \[A\mathbf{v} = 2(2\mathbf{v}_1) - (2\mathbf{v}_2) + (3\mathbf{v}_3)=(-2, -4, 6)^T\]

Step 1: Find the coefficients for expressing \(\mathbf{v}\) as a linear combination of \(\mathbf{v}_1\), \(\mathbf{v}_2\), and \(\mathbf{v}_3\)

We want to express \(\mathbf{v}\) like this: \[\mathbf{v} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3\] Plugging in the given values, we get the system of equations: \[\begin{cases} c_1 + 2c_2 - c_3 = 5 \\ -c_1 + c_2 - c_3 = 0 \\ c_1 + 3c_2 + 2c_3 = 3 \end{cases}\] Solving this system of linear equations, we find the coefficients \(c_1\), \(c_2\), and \(c_3\).

Step 2: Write \(\mathbf{v}\) as a linear combination of \(\mathbf{v}_1\), \(\mathbf{v}_2\), and \(\mathbf{v}_3\)

Using the coefficients found in Step 1, we can now write \(\mathbf{v}\) as a linear combination of the eigenvectors: \[\mathbf{v} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3\]

Step 3: Use the property of eigenvectors and eigenvalues to find \(A\mathbf{v}\)

Since \(\mathbf{v}_1\), \(\mathbf{v}_2\), and \(\mathbf{v}_3\) are eigenvectors of A corresponding to the eigenvalues \(\lambda_1\), \(\lambda_2\), and \(\lambda_3\), respectively, we can use the property that \(A\mathbf{v}_i = \lambda_i \mathbf{v}_i\) for each eigenvector. Thus, we have: \[\begin{aligned} A\mathbf{v} &= A(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + c_3\mathbf{v}_3) \\ &= c_1(A\mathbf{v}_1) + c_2(A\mathbf{v}_2) + c_3(A\mathbf{v}_3) \\ &= c_1(\lambda_1\mathbf{v}_1) + c_2(\lambda_2\mathbf{v}_2) + c_3(\lambda_3\mathbf{v}_3) \end{aligned}\] Now we can plug in the values for \(\lambda_1\), \(\lambda_2\), \(\lambda_3\), and the coefficients \(c_1\), \(c_2\), and \(c_3\) found in Step 1. \[A\mathbf{v} = c_1(2\mathbf{v}_1) - c_2(2\mathbf{v}_2) + c_3(3\mathbf{v}_3)\] Finally, we calculate the values and obtain the vector \(A\mathbf{v}\).

Key Concepts

Linear Combination

A linear combination in the context of vectors and matrices is fundamental to understand many concepts in linear algebra. Imagine you have a collection of vectors. When you multiply each vector by a corresponding scalar (a real number) and then add them all together, you're creating a linear combination. In simpler terms, it's like creating a new recipe by mixing different ingredients (vectors) in specific amounts (scalars).

In the exercise provided, \( \mathbf{v} \) is expressed as a linear combination of the vectors \( \mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3} \). This is akin to saying that vector \( \mathbf{v} \) can be made by mixing \( \mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3} \) in certain quantities. What's vital here is finding the right 'recipe,' meaning the right set of scalars, to recreate \( \mathbf{v} \) exactly.

System of Linear Equations

When trying to locate the exact scalars for the linear combination, we encounter something called a 'system of linear equations.' This is a collection of linear equations involving the same set of variables. The aim here is to find a solution that satisfies all equations at once, which often requires mathematical techniques like substitution, elimination, or matrix operations.

In the exercise context, once the variables \( c_1 \), \( c_2 \), and \( c_3 \)—which are the scalars in the linear combination—have been identified through the system of linear equations, we've essentially cracked the code. These coefficients allow us to recreate \( \mathbf{v} \) using \( \mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3} \).

Matrix Transformation

The term 'matrix transformation' sounds more complex than it is. Simply put, it's about applying a matrix to a vector, which then 'transforms' that vector into another vector. This transformation could involve rotation, scaling, translation, or some other geometric operation, depending on the matrix involved.

In our specific problem, the matrix \( A \) is known to transform its eigenvectors in a very particular way. Instead of changing the direction of the eigenvectors, \( A \) just stretches or shrinks them by a factor known as the eigenvalue. This predictive property is precisely what aids in solving part (b) of the exercise.

Eigenvector Property

Eigenvectors and eigenvalues are inseparable when discussing linear transformations. An eigenvector of a matrix is a non-zero vector that, after a transformation by the matrix, remains parallel to its original direction. The eigenvalue is the factor by which it gets stretched or compressed. The property states that for matrix \( A \) and eigenvector \( \mathbf{v}_i \), the equation \( A\mathbf{v}_i = \lambda_i\mathbf{v}_i \) always holds true, where \( \lambda_i \) is the corresponding eigenvalue.

This property dramatically simplifies calculations, as seen in the exercise. Applying the matrix \( A \) to a linear combination of eigenvectors translates to applying \( A \) to each eigenvector individually, then scaling them by their respective eigenvalues—maintaining the proportions dictated by the eigenvector property.