Finding Eigenvalues of a Matrix: A Comprehensive Understanding

Step 1: Introduction to Eigenvectors and Eigenvalues

In linear algebra, eigenvectors and eigenvalues are essential concepts that reveal hidden properties of a matrix. An eigenvector of a square matrix A is a special vector that remains in the same direction when multiplied by A. The eigenvalue, denoted by λ, is the scaling factor by which the eigenvector is stretched or compressed.

Mathematically, this relationship is expressed as:

A⋅v = λ⋅v

This equation tells us that when matrix A multiplies vector v, the result is the same vector v scaled by a factor λ.

Step 2: Rearranging to Understand the Difference

To understand the nature of eigenvectors and eigenvalues, we can rearrange the equation to see how λ affects the vector:

A⋅v – λ⋅v = 0

This form highlights the difference between the transformation of v by A and the simple scaling of v by λ.

Step 3: Connecting to the Identity Matrix

We can further express the equation in terms of a matrix product by introducing the identity matrix I:

(A – λ⋅I)⋅v = 0

The identity matrix I serves as a bridge to connect the eigenvalue λ with the matrix A. It allows us to see how λ alters the matrix A itself, not just the vector v.

Step 4: The Determinant and Singular Matrices

If the above equation has a non-zero solution for v, then the matrix A – λ⋅I must be singular. A singular matrix is one that does not have an inverse, and its determinant is zero:

det(A – λ⋅I) = 0

This characteristic equation is the gateway to finding the eigenvalues of A. It connects the geometric concept of stretching/compressing vectors to the algebraic concept of determinants and polynomials.

Conclusion: The process of finding eigenvalues is rich in mathematical meaning. It starts with the geometric idea of vectors remaining in the same direction under a transformation and leads to an algebraic polynomial equation. The subtraction of λ in this process is a key step that connects the eigenvalue to the matrix itself, revealing the hidden structure of the matrix. Understanding these concepts is vital in various applications, from physics to machine learning.

Finding Eigenvalues of the Matrix

Given matrix:

\[ \begin{bmatrix} 2 & 3 \\ 3 & -6 \\ \end{bmatrix} \]

1. Write down the characteristic polynomial by subtracting \(\lambda\) times the identity matrix:

\[ \begin{bmatrix} 2 – \lambda & 3 \\ 3 & -6 – \lambda \\ \end{bmatrix} \]

2. Calculate the determinant of the matrix:

\[ \text{det}(A – \lambda I) = (2 – \lambda)(-6 – \lambda) – (3 \cdot 3) \]

3. Expand the expression:

\[ (2 – \lambda)(-6 – \lambda) – 9 = (-\lambda + 2)(-\lambda – 6) – 9 = \lambda^2 + 4\lambda – 12 – 9 = \lambda^2 + 4\lambda – 21 \]

4. Factorize the polynomial:

\[ \lambda^2 + 4\lambda – 21 = (\lambda + 7)(\lambda – 3) = 0 \]

5. Solve for \(\lambda\):

\[ \lambda_1 = -7, \lambda_2 = 3 \]

Conclusion: The eigenvalues of the given matrix are \(-7\) and \(3\). This process involves finding the characteristic polynomial and factoring it to find the eigenvalues.

Finding Eigenvalues of the given matrix: \(\begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}\)

Step 1: Write down the given matrix

The matrix we are analyzing is \(\begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}\). We want to find the eigenvalues, which are the values of λ that satisfy the characteristic polynomial.

Step 2: Form the characteristic polynomial

We need to find the determinant of the matrix minus λ times the identity matrix. The characteristic polynomial is formed by setting this determinant equal to zero:

det\(\left(\begin{bmatrix} 1-λ & 4 \\ 2 & 3-λ \end{bmatrix}\right)\) = 0

Step 3: Expand the determinant

Expanding the determinant, we multiply the elements of the main diagonal and subtract the product of the other diagonal:

(1-λ)(3-λ) – (4 ⋅ 2) = (1-λ)(3-λ) – 8 = 0

Step 4: Simplify the polynomial

Multiplying out the brackets, we get:

3 – λ – 3λ + λ² – 8 = λ² – 4λ – 5 = 0

Step 5: Factorize the polynomial

We can factorize the polynomial by finding the roots:

(λ – 5)(λ + 1) = 0

Step 6: Solve for λ

The solutions to this equation are the eigenvalues of the given matrix:

λ₁ = 5, λ₂ = -1

Conclusion: Finding Eigenvalues of the given matrix \(\begin{bmatrix} 1 & 4 \\ 2 & 3 \end{bmatrix}\) results in λ₁ = 5 and λ₂ = -1. This process involves forming the characteristic polynomial, expanding it, and solving it to find the eigenvalues.

Finding Eigenvalues of the given matrix: A Comprehensive Guide

In this guide, we are focusing on Finding Eigenvalues of the given matrix \(\begin{bmatrix} 5 & 2 \\ 1 & 4 \end{bmatrix}\). Understanding eigenvalues is essential in various mathematical and engineering applications.

1. Write down the given matrix: The matrix we are analyzing is \(\begin{bmatrix} 5 & 2 \\ 1 & 4 \end{bmatrix}\). This is the starting point for Finding Eigenvalues of the given matrix.

2. Set up the characteristic polynomial: We need to find the values of \(\lambda\) that satisfy \(\text{det}(A – \lambda I) = 0\), where \(I\) is the identity matrix. This equation will help us find the eigenvalues.

3. Subtract \(\lambda\) from the diagonal elements: By subtracting \(\lambda\) from the diagonal elements, we form the matrix \(\begin{bmatrix} 5-\lambda & 2 \\ 1 & 4-\lambda \end{bmatrix}\). This step transforms the original matrix into a form that allows us to find the characteristic polynomial.

4. Calculate the determinant: We calculate the determinant of the transformed matrix. The expression is \((5-\lambda)(4-\lambda) – (2 \cdot 1) = \lambda^2 – 9\lambda + 18\). This step involves multiplying the diagonal elements and subtracting the product of the off-diagonal elements.

5. Factorize the polynomial: We factorize the polynomial to find its roots. In this case, we get \((\lambda – 6)(\lambda – 3) = 0\). Factoring helps us find the values of \(\lambda\) that satisfy the equation.

6. Find the eigenvalues: The eigenvalues of the matrix are the roots of the characteristic polynomial. In this case, \(\lambda_1 = 6\) and \(\lambda_2 = 3\). These values are the eigenvalues of the given matrix.

Conclusion: Finding Eigenvalues of the given matrix \(\begin{bmatrix} 5 & 2 \\ 1 & 4 \end{bmatrix}\) results in \(\lambda_1 = 6\) and \(\lambda_2 = 3\). This guide provides a step-by-step breakdown of each component, ensuring a deep understanding of the process of Finding Eigenvalues of the given matrix.

Finding Eigenvalues of the Given Matrix

Starting Point: You have the matrix equation \( Av = \lambda v \), where \( A \) is the given matrix, \( v \) is the eigenvector, and \( \lambda \) is the eigenvalue.

Rearranging: You can rewrite this as \( (A – \lambda I)v = 0 \), where \( I \) is the identity matrix.

Changing the Diagonal: When you subtract \( \lambda I \) from \( A \), you are subtracting \( \lambda \) from the diagonal elements of \( A \). This changes the \( x \) component of the first column vector and the \( y \) component of the second column vector.

Finding Parallel Vectors: The goal is to find the values of \( \lambda \) that make the matrix \( A – \lambda I \) singular (i.e., its determinant is zero). When this happens, the column vectors of \( A – \lambda I \) become linearly dependent (parallel), and the matrix cannot be inverted.

Solving for \( \lambda \): By setting the determinant of \( A – \lambda I \) equal to zero and solving for \( \lambda \), you find the eigenvalues of \( A \).

The subtraction of \( \lambda \) from the diagonal elements is a mathematical way to express the geometric concept of finding the values that make the column vectors parallel. It’s a beautiful connection between algebra and geometry, and it’s central to understanding eigenvectors and eigenvalues.