Question

Find the eigenvalues and eigenvectors of the given matrix $A .$$$ A=\left[\begin{array}{rr} 2 & 1 \\ -1 & 2 \end{array}\right] $$

Short answer

Answer: The eigenvalues of matrix A are 1 and 3. Their respective eigenvectors are $$\begin{bmatrix} -1 \\ 1 \end{bmatrix}$$ and $$\begin{bmatrix} 1 \\ 1 \end{bmatrix}$$.

Step-by-step solution

Find the determinant of (A - λI)

Form the (A - λI) matrix, by subtracting λ from the diagonal elements of the A matrix. $$ (A - \lambda I) = \begin{bmatrix} 2-\lambda & 1 \\ -1 & 2-\lambda \end{bmatrix}. $$ Now, find the determinant of this matrix: $$ \det(A - \lambda I)= (2-\lambda)(2-\lambda) - (-1)(1). $$

Solve the equation for λ

Expanding and simplifying the determinant equation, we have $$ \lambda^2 - 4\lambda +3 = 0. $$ Now, we can solve for λ by factoring the equation: $$ (\lambda - 1)(\lambda - 3) = 0. $$ So, the eigenvalues (λ) are 1 and 3.

Find the eigenvectors for each eigenvalue

Now, we will find the eigenvectors for each eigenvalue. For λ = 1, $$ (A - \lambda I)v = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=0. $$ Since the second row is a multiple of the first, we can ignore it and use the first row as follows: $$ v_1 + v_2 = 0. $$ Choosing v_2 = 1, we have v_1 = -1. Therefore, the eigenvector for λ = 1 is $$ v_1=\begin{bmatrix} -1 \\ 1 \end{bmatrix}. $$ For λ = 3, $$ (A - \lambda I)v = \begin{bmatrix} -1 & 1 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}=0. $$ Again, we can choose one row (the first, in this case) to use in the calculations: $$ -v_1 + v_2 = 0. $$ Choosing v_1 = 1, we have v_2 = 1. Therefore, the eigenvector for λ = 3 is $$ v_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}. $$ In conclusion, the eigenvalues for the given matrix A are 1 and 3, and their respective eigenvectors are $$\begin{bmatrix} -1 \\ 1 \end{bmatrix}$$ and $$\begin{bmatrix} 1 \\ 1 \end{bmatrix}$$.

Key concepts

Determinant of a Matrix

Understanding the determinant of a matrix is crucial when diving into linear algebra, particularly when dealing with eigenvalues and eigenvectors. The determinant is a scalar value that provides important information about a square matrix. It can tell us if the matrix has an inverse, what the volume scaling factor of the linear transformation represented by the matrix is, and it plays a key role in characterizing the matrix's eigenvalues.

To calculate the determinant of a 2x2 matrix, such as \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), you would use the formula \( \det(A) = ad - bc \). When finding eigenvalues of a matrix, we apply this calculation to the matrix minus \( \lambda \) times the identity matrix, \( A - \lambda I \), which results in a polynomial that we call the characteristic polynomial.

Characteristic Polynomial

When we find the eigenvalues of a matrix, we encounter the concept of a characteristic polynomial. This polynomial is derived from the determinant of the matrix subtracted by \( \lambda \) times the identity matrix, as seen in the formula \( \det(A - \lambda I) = 0 \). The roots of this polynomial are the eigenvalues of the matrix.

The characteristic polynomial for a 2x2 matrix \( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is obtained by computing \( \det(A - \lambda I) \), resulting in \( \lambda^2 - (a+d)\lambda + (ad-bc) = 0 \). For larger matrices, the characteristic polynomial becomes more complex, but the concept remains the same: solve for the values of \( \lambda \) that satisfy the equation, which gives you the matrix's eigenvalues.

Eigenvector Computation

Once we have the eigenvalues from the characteristic polynomial, the next step is to calculate the corresponding eigenvectors. An eigenvector of a matrix is a non-zero vector that changes by only a scalar factor when that matrix is applied to it. This scalar factor is the eigenvalue associated with the eigenvector.

To find an eigenvector, we solve the equation \( (A - \lambda I)v = 0 \) for each eigenvalue \( \lambda \). This is accomplished by plugging each eigenvalue into the matrix equation and reducing it to row-echelon form, which often reveals a system of linear dependencies among the matrix columns.

The resulting free variables allow for multiple solutions—any non-zero vector that satisfies the equation is an eigenvector. Typically, we choose values for the free variables to make calculations simpler and to obtain a convenient basis for the eigenspace associated with each eigenvalue.