Problem 38 - Differential Equations and Linear Algebra

Question

If \(\mathbf{v}_{1}, \mathbf{v}_{2},\) and \(\mathbf{v}_{3}\) are linearly independent eigenvectors of \(A\) corresponding to the eigenvalue \(\lambda,\) and \(c_{1}, c_{2},\) and \(c_{3}\) are scalars (not all zero), show that \(c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+c_{3} \mathbf{v}_{3}\) is also an eigenvector of \(A\) corresponding to the eigenvalue \(\lambda\).

Step-by-Step Solution

Answer

We showed that \(A(c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + c_{3}\mathbf{v}_{3}) = \lambda(c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + c_{3}\mathbf{v}_{3})\), concluding that the linear combination \(c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + c_{3}\mathbf{v}_{3}\) is an eigenvector of \(A\) corresponding to the eigenvalue \(\lambda\).

Step 1: Define Eigenvectors and Eigenvalues

Recall that a vector \(\mathbf{v}\) is an eigenvector of a matrix \(A\) with eigenvalue \(\lambda\) if \(A \mathbf{v} = \lambda \mathbf{v}\). In our case, we have eigenvectors \(\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\) and eigenvalue \(\lambda\) such that: - \(A \mathbf{v}_{1} = \lambda \mathbf{v}_{1}\) - \(A \mathbf{v}_{2} = \lambda \mathbf{v}_{2}\) - \(A \mathbf{v}_{3} = \lambda \mathbf{v}_{3}\) We are tasked with showing that the linear combination \(c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+c_{3} \mathbf{v}_{3}\) is also an eigenvector of \(A\) with the same eigenvalue \(\lambda\).

Step 2: Calculate \(A(c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + c_{3}\mathbf{v}_{3})\)

To begin, let's calculate \(A(c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + c_{3}\mathbf{v}_{3})\). By the matrix-vector product properties, we will obtain the following: \(A(c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + c_{3}\mathbf{v}_{3}) = A(c_{1}\mathbf{v}_{1}) + A(c_{2}\mathbf{v}_{2}) + A(c_{3}\mathbf{v}_{3})\)

Step 3: Use Given Information to Rewrite Expression

Now, let's substitute the given eigenvector and eigenvalue relationships for each of the vectors \(\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\) into our previous expression: \(A(c_{1}\mathbf{v}_{1}) + A(c_{2}\mathbf{v}_{2}) + A(c_{3}\mathbf{v}_{3}) = c_{1}(A\mathbf{v}_{1}) + c_{2}(A\mathbf{v}_{2}) + c_{3}(A\mathbf{v}_{3})\) \(= c_{1}(\lambda \mathbf{v}_{1}) + c_{2}(\lambda \mathbf{v}_{2}) + c_{3}(\lambda \mathbf{v}_{3})\)

Step 4: Factor Out the Eigenvalue \(\lambda\)

Finally, let's factor out the common eigenvalue \(\lambda\) from our expression: \(c_{1}(\lambda \mathbf{v}_{1}) + c_{2}(\lambda \mathbf{v}_{2}) + c_{3}(\lambda \mathbf{v}_{3}) = \lambda(c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + c_{3}\mathbf{v}_{3})\) So, we have shown that \(A(c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + c_{3}\mathbf{v}_{3}) = \lambda(c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + c_{3}\mathbf{v}_{3})\).

Step 5: Conclude the Proof

Since we have shown that multiplying \(A\) by the linear combination of eigenvectors (i.e., \(c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + c_{3}\mathbf{v}_{3}\)) yields the same linear combination scaled by the eigenvalue \(\lambda\), we conclude that the linear combination \(c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + c_{3}\mathbf{v}_{3}\) is indeed an eigenvector of \(A\) corresponding to the eigenvalue \(\lambda\).

Key Concepts

Eigenvalues

Eigenvalues are a fundamental concept in linear algebra. They provide insights into how a matrix acts on a vector. If we have a matrix, \( A \), and a non-zero vector, \( \mathbf{v} \), the vector is called an eigenvector if when we multiply it by \( A \), the result is simply a scaled version of \( \mathbf{v} \). The scaling factor is known as the eigenvalue, denoted as \( \lambda \). This relationship is defined by the equation:

• \( A \mathbf{v} = \lambda \mathbf{v} \)

In the given problem, the vectors \( \mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3} \) are eigenvectors of the matrix \( A \) associated with the eigenvalue \( \lambda \). Each vector satisfies the equation \( A \mathbf{v} = \lambda \mathbf{v} \). These vectors present unique solutions that tell us about the transformation properties of the matrix \( A \). As you learn more about matrices, knowing their eigenvalues helps us to understand rotations, scalings, and other transformations applied by the matrix.

Linear Independence

Linear independence is a crucial concept when dealing with vectors. A set of vectors is called linearly independent if no vector in the set can be written as a combination of the others. This means that each vector adds a new dimension of information. In simple terms, they "point" in different directions. If given vectors \( \mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3} \) are linearly independent, it ensures:

• The only solution to \( c_{1} \mathbf{v}_{1} + c_{2} \mathbf{v}_{2} + c_{3} \mathbf{v}_{3} = \mathbf{0} \) is when \( c_{1} = c_{2} = c_{3} = 0 \).

In this exercise, knowing that the eigenvectors are linearly independent helps us conclude that the combination \( c_{1} \mathbf{v}_{1} + c_{2} \mathbf{v}_{2} + c_{3} \mathbf{v}_{3} \) created from these vectors is unique unless all coefficients are zero. Thus, this combination also leads to an eigenvector associated with the matrix \( A \), provided the constants \( c_{1}, c_{2}, c_{3} \) are not all zero.

Matrix-Vector Multiplication

Matrix-vector multiplication is the process of applying a matrix to a vector, transforming it into another vector. This is a cornerstone of linear algebra, enabling you to explore how matrices operate on vectors. When you multiply a matrix \( A \) by a vector \( \mathbf{v} \), the result is a new vector formed from linear combinations of the rows of \( A \). Here's the basic form:

• If \( A \) is an \( m \times n \) matrix and \( \mathbf{v} \) is a vector with \( n \) elements, then \( A \mathbf{v} \) is a vector with \( m \) elements.

In this exercise, the matrix-vector product \( A(c_{1}\mathbf{v}_{1} + c_{2}\mathbf{v}_{2} + c_{3}\mathbf{v}_{3}) \) illustrates how a linear combination of eigenvectors transforms under \( A \). Thanks to the property \( A \mathbf{v} \approx \lambda \mathbf{v} \), this multiplication shows us that \( A \) effectively stretches or shrinks the vectors uniformly by \( \lambda \). Therefore, the entire linear combination is also stretched or shrunk by \( \lambda \), proving it remains an eigenvector of \( A \). This demonstrates the consistency and predictability of matrix operations on vectors.