Question

Find all eigenvalues and eigenvectors of the given matrix. $$ \left(\begin{array}{cc}{1} & {\sqrt{3}} \\ {\sqrt{3}} & {-1}\end{array}\right) $$

Short answer

A: The eigenvalues can be found by solving the characteristic equation: λ₁ = 0 and λ₂ = 2. The corresponding eigenvectors can be found by solving the system of linear equations: For λ₁ = 0, the eigenvector x₁ = (1/2)(1, -√3)ᵀ For λ₂ = 2, the eigenvector x₂ = (1/2)(1, √3)ᵀ

Step-by-step solution

Write down the given matrix

First, let's write down the given matrix A: $$ A=\left(\begin{array}{cc}{1} & {\sqrt{3}} \\\ {\sqrt{3}} & {-1}\end{array}\right) $$

Find the characteristic equation for the eigenvalues

The characteristic equation can be found by evaluating the determinant of \((A-\lambda I)\), denoted by \(|A-\lambda I|\), where \(\lambda\) is an eigenvalue, and \(I\) is the identity matrix. For given matrix A: $$ |A-\lambda I|=| \begin{pmatrix} 1-\lambda & \sqrt{3} \\ \sqrt{3} & -1 - \lambda \end{pmatrix} | = (1-\lambda)(-1-\lambda)-(\sqrt{3})^2 $$

Solve the characteristic equation

Next, let's solve the characteristic equation: $$ (1-\lambda)(-1-\lambda)-3=-(\lambda^2-2\lambda) $$ Solving the above quadratic equation, we get two eigenvalues: $$ \lambda_1=0, \: \lambda_2=2 $$

Find the eigenvectors for each eigenvalue

For each eigenvalue, let's find the eigenvectors by solving the following system of linear equations \((A-\lambda_i I)x=0\). For \(\lambda_1=0\), \((A-\lambda_1 I)x=0\) becomes: $$ A x = \left(\begin{array}{cc}{1} & {\sqrt{3}} \\\ {\sqrt{3}} & {-1}\end{array}\right) \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} =0 $$ Solving the above system, we get the eigenvector for \(\lambda_1\): $$ x_{1} = \frac{1}{2} \begin{pmatrix} 1 \\ -\sqrt{3} \end{pmatrix} $$ For \(\lambda_2=2\), \((A-\lambda_2 I)x=0\) becomes: $$ (A-\lambda_2 I)x = \left(\begin{array}{cc}{-1} & {\sqrt{3}} \\\ {\sqrt{3}} & {-3}\end{array}\right) \begin{pmatrix} x_3 \\ x_4 \end{pmatrix} =0 $$ Solving this system of equations, we get the eigenvector for \(\lambda_2\): $$ x_{2} = \frac{1}{2} \begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix} $$

State the final solution

The eigenvalues and corresponding eigenvectors for the given matrix are: $$ \lambda_{1} = 0, \: x_{1} = \frac{1}{2} \begin{pmatrix} 1 \\ -\sqrt{3} \end{pmatrix} $$ and $$ \lambda_{2} = 2, \: x_{2} = \frac{1}{2} \begin{pmatrix} 1 \\ \sqrt{3} \end{pmatrix} $$

Key concepts

Characteristic Equation

In the realm of linear algebra, the characteristic equation is a crucial tool used to find eigenvalues of a matrix. It involves calculating the determinant of the matrix \((A - \lambda I)\), where \(A\) is the given matrix, \(\lambda\) represents an eigenvalue, and \(I\) is the identity matrix of the same size. For the exercise at hand, the matrix \(A\) was:

• \(\begin{pmatrix} 1 & \sqrt{3} \ \sqrt{3} & -1 \end{pmatrix}\)

To find the characteristic equation, we compute the determinant:\[|A - \lambda I| = \begin{vmatrix} 1-\lambda & \sqrt{3} \ \sqrt{3} & -1-\lambda \end{vmatrix}\]This results in the equation:\[(1-\lambda)(-1-\lambda) - 3\]Solving this quadratic equation gives the eigenvalues \(\lambda_1 = 0\) and \(\lambda_2 = 2\). The roots of this equation are key in understanding the behavior of the matrix in various transformations.

Matrix Algebra

Matrix algebra is a branch of mathematics dealing with matrices and operations like addition, multiplication, and finding the determinant. In our case, matrix algebra helps in deriving the characteristic equation and solving for eigenvectors. When the characteristic equation gives us eigenvalues, we use them in the expression \((A - \lambda I)x = 0\) to find eigenvectors. During the solution:

• For \(\lambda_1 = 0\), solving \(Ax = 0\) led to the eigenvector \(\begin{pmatrix} 1/2 \ -\sqrt{3}/2 \end{pmatrix}\).
• For \(\lambda_2 = 2\), it resulted in the eigenvector \(\begin{pmatrix} 1/2 \ \sqrt{3}/2 \end{pmatrix}\).

These eigenvectors provide insight into the directions along which the matrix transformation acts as simple scaling.

Linear Algebra

Linear algebra is the foundation of vector spaces and linear transformations. It allows us to study systems of linear equations, which is essential in finding eigenvalues and eigenvectors. In this context, understanding linear spaces, basis, and dimension plays a significant role. The aim is to find vectors \(x\) such that transforming them by the matrix \(A\) merely stretches or compresses them by a factor \(\lambda\), without changing their direction.
Eigenvalues express this factor, while eigenvectors represent the direction. This includes understanding:

• The significance of linear transformations and how they can be decomposed into simpler actions using eigenvectors.
• The connection between eigenvalues and matrix stability, important in numerous applications, including systems dynamics and machine learning.

Mastering the concepts of linear algebra opens doors to analyzing more complex matrices and transformations across various fields.