Question

Find all eigenvalues and eigenvectors of the given matrix. $$ \left(\begin{array}{rr}{-2} & {1} \\ {1} & {-2}\end{array}\right) $$

Short answer

Matrix A: $$ \left(\begin{array}{rr}{-2} & {1} \\\ {1} & {-2}\end{array}\right) $$ Answer: The eigenvalues are λ₁ = -1 and λ₂ = -3. The corresponding eigenvectors are: For λ₁ = -1: $$ v₁ = \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$ For λ₂ = -3: $$ v₂ = \begin{pmatrix} 1 \\ -1 \end{pmatrix} $$

Step-by-step solution

Calculate the Characteristic Equation

First, we need to find the determinant of the matrix subtracted by a scalar (associate with the eigenvalue) multiplied by the identity matrix. Given Matrix A: $$ \left(\begin{array}{rr}{-2} & {1} \\\ {1} & {-2}\end{array}\right) $$ Let λ be the scalar (representing the eigenvalue), subtract λ times the identity matrix (I) from A: $$ A - λI = \left(\begin{array}{rr}{-2-λ} & {1} \\\ {1} & {-2-λ}\end{array}\right) $$ Now, find the determinant of (A - λI) and equate it to 0 to get the characteristic equation: $$ |A - λI| = \begin{vmatrix} {-2-λ} & {1} \\\ {1} & {-2-λ} \end{vmatrix} = (-2-λ)(-2-λ) - (1)(1) = 0 $$

Solve the Characteristic Equation

Now, we need to solve the quadratic equation derived from the determinant in the previous step to find the eigenvalues: $$ λ^2 + 4λ + 3 =λ^2 + 3λ + λ + 3 = (λ+1)(λ+3) = 0 $$ The solutions for λ are: $$ λ_1 = -1 \\ λ_2 = -3 $$ We have found the eigenvalues as -1 and -3.

Find the Eigenvectors

Now, for each eigenvalue, we will find the corresponding eigenvectors by solving the linear system: For λ = -1, substitute λ in the matrix (A - λI): $$ \left(\begin{array}{rr}{-1} & {1} \\\ {1} & {-1}\end{array}\right) \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 $$ Notice that the first and second rows are linearly dependent. That means, we only need to solve one of them: $ - x_1+ x_2 = 0 $ Let \(x_2\) = t, then \(x_1 = \)t. So, we have the eigenvector for λ = -1 as: $$ v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} $$ For λ = -3, substitute λ in the matrix A - λI: $$ \left(\begin{array}{rr}{1} & {1} \\\ {1} & {1}\end{array}\right) \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = 0 $$ Again, the first and second rows are linearly dependent. Thus, we only need to solve one of them: $ x_1 + x_2 = 0 $ Let \(x_1\) = t, then \(x_2 = -t\). So, we have the eigenvector for λ = -3 as: $$ v_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} $$ Now, we have found all the eigenvalues and their corresponding eigenvectors: Eigenvalue λ₁ = -1: Eigenvector: $$v₁ = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$$ Eigenvalue λ₂ = -3: Eigenvector: $$v₂ = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$$

Key concepts

Characteristic Equation

In the realm of linear algebra, the characteristic equation is a crucial concept when dealing with eigenvalues. To determine the eigenvalues of a matrix, we begin by formulating this equation. The key idea is that the characteristic equation emerges from the determinant of a modified version of the original matrix.
The modified matrix is created by subtracting \( \lambda I \), where \( \lambda \) represents the eigenvalue and \( I \) is the identity matrix. This adjustment yields the matrix \( A - \lambda I \).
Once this modified matrix is established, its determinant can be calculated.

• The determinant of \( A - \lambda I \) provides us with a polynomial expression in terms of \( \lambda \).
• Setting this polynomial's determinant equal to zero gives us the characteristic equation.

Solving this equation reveals the eigenvalues of the matrix.
In our specific exercise, the characteristic equation derived was\( (\lambda + 1)(\lambda + 3) = 0 \), leading to the eigenvalues \( \lambda_1 = -1 \) and \( \lambda_2 = -3 \). These values are critical because they pave the way for identifying the corresponding eigenvectors.

Determinant Calculation

Calculating the determinant is a fundamental step in finding the characteristic equation.
Determinants provide a scalar value that indicates various properties of matrices, such as whether they are invertible or not.
When we perform determinant calculation on \( A - \lambda I \), it involves a simple algebraic process for a 2x2 matrix:

• Take the product of the main diagonal elements (\( a_{11} \cdot a_{22} \)).
• Subtract the product of the off-diagonal elements (\( a_{12} \cdot a_{21} \)).

For our matrix problem, \( A = \left(\begin{array}{rr}{-2} & {1} \ {1} & {-2}\end{array}\right) \), substituting \( \lambda \) to form\( A - \lambda I \) gives:\[\left(\begin{array}{rr}{-2-\lambda} & {1} \ {1} & {-2-\lambda}\end{array}\right)\]The determinant calculation goes as follows:\[(-2-\lambda)(-2-\lambda) - (1)(1)\]Simplifying that,\( \lambda^2 + 4\lambda + 3 \). By equating this to zero, the characteristic equation emerges, allowing for eigenvalue extraction.

Linear Systems

Linear systems are closely tied to finding eigenvectors once eigenvalues are known.
By substituting each eigenvalue back into the equation\( (A - \lambda I)\mathbf{x} = 0 \), we aim to solve for the eigenvector, \( \mathbf{x} \).
This procedure reveals the vector combinations that remain invariant (aside from scaling) under the transformation represented by the matrix.

• The matrix \( A - \lambda I \) will have rows that are linearly dependent.
• This allows the system to have a non-trivial solution, where solutions can be expressed in terms of a free variable, often denoted as \( t \).

For the eigenvalue \( \lambda = -1 \), solving the linear system yielded the eigenvector\( \mathbf{v_1} = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).
Similarly, for \( \lambda = -3 \), solving provided the eigenvector\( \mathbf{v_2} = \begin{pmatrix} 1 \ -1 \end{pmatrix} \).
Understanding linear systems allows one to fully appreciate how both simplicity and interdependencies in matrix rows determine the nature of eigenvectors.