Question

Find all eigenvalues and eigenvectors of the given matrix. $$ \left(\begin{array}{rrr}{3} & {2} & {2} \\ {1} & {4} & {1} \\ {-2} & {-4} & {-1}\end{array}\right) $$

Short answer

Answer: The eigenvalues and corresponding eigenvectors for the given matrix are as follows: Eigenvalue λ1 = 0: Eigenvector v1 = \(\begin{bmatrix}-1 \\ -1 \\ 1\end{bmatrix}\) Eigenvalue λ2, λ3 = 3: Eigenvectors v2 = \(\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}\) and v3 = \(\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}\)

Step-by-step solution

Find the Eigenvalues

Subtract the scalar λ from the diagonal elements of the given matrix and calculate its determinant. Then, set the determinant equal to zero to get the characteristic equation. $$ \text{det}(A-\lambda I) = \begin{vmatrix}3-\lambda & 2 & 2 \\ 1 & 4-\lambda & 1 \\ -2 & -4 & -1-\lambda \end{vmatrix} = 0 $$ Now expand the determinant and simplify the equation. $$ \begin{aligned} (3-\lambda)((4-\lambda)(-1-\lambda)+4)-2(-1+\lambda-1(-1-\lambda)+4(-2))-2(-4(4-\lambda)+(-1-\lambda))=0 \end{aligned} $$ Simplify the expression and obtain the characteristic polynomial: $$ -\lambda^3 + 6\lambda^2 -9\lambda - 0 = 0 $$ Factor out the polynomial and find the eigenvalues: $$ -\lambda(\lambda^2 -6\lambda +9) = 0 $$ Hence, the eigenvalues are λ1 = 0, λ2 = 3, and λ3 = 3. Now we will find the eigenvectors for each eigenvalue.

Find the Eigenvectors for λ1 = 0

Plug the eigenvalue λ1 = 0 into the equation (A - λI)v = 0 and solve for the eigenvector v: $$ \begin{bmatrix} 3 & 2 & 2 \\ 1 & 4 & 1 \\ -2 & -4 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0 $$ Solve the linear equation using Gaussian Elimination, or any other equivalent method, to get the eigenvector v1: $$ v1 = \begin{bmatrix} -1 \\ -1 \\ 1 \end{bmatrix} $$

Find the Eigenvectors for λ2 = 3 and λ3 = 3

Since λ2 and λ3 are the same, we can find the eigenvectors for this eigenvalue in one step. Plug the eigenvalue λ2 = 3 into the equation (A - λI)v = 0: $$ \begin{bmatrix} 0 & 2 & 2 \\ 1 & 1 & 1 \\ -2 & -4 & -4 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0 $$ Solve this linear equation to find the eigenvector corresponding to the eigenvalue λ2 = 3: $$ v2 = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} $$ We can check if there's another linearly independent eigenvector for the eigenvalue λ3 = 3. Solve the same linear equation to find the eigenvector: $$ v3 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} $$ So, the eigenvalues and corresponding eigenvectors are: Eigenvalue λ1 = 0: Eigenvector v1 = \(\begin{bmatrix}-1 \\ -1 \\ 1\end{bmatrix}\) Eigenvalue λ2, λ3 = 3: Eigenvectors v2 = \(\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}\) and v3 = \(\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}\)

Key concepts

Characteristic Polynomial

Understanding the characteristic polynomial is crucial when dealing with eigenvalues and eigenvectors. The polynomial is derived from the equation \( \text{det}(A-\lambda I) = 0 \), where \( A \) represents the matrix, \( \lambda \) is the scalar eigenvalue, and \( I \) is the identity matrix of the same size as \( A \.\) The process starts with subtracting \( \lambda \) times the identity matrix from the original matrix. This operation forms a new matrix, whose determinant must be calculated.

The determinant, a scalar value, is influenced by all elements of a matrix and gives significant properties about the matrix, like its invertibility. Setting it equal to zero and solving the resulting equation reveals the eigenvalues of the matrix. In essence, the characteristic polynomial lays the groundwork towards finding these eigenvalues, which are roots of this polynomial. Simplifying the determinant leads to this polynomial, which for our example simplifies to \( -\lambda^3 + 6\lambda^2 - 9\lambda = 0 \). Factoring this polynomial allows us to extract the eigenvalues directly.

Determinant of a Matrix

The determinant of a matrix is a value that provides important information about the matrix, including whether it is invertible and the volume scaling factor of the linear transformation the matrix represents. Computing the determinant is an essential step in finding eigenvalues, as seen in the characteristic polynomial. It can be calculated using various methods, including expansion by minors or cofactors, but for larger matrices, these methods can be cumbersome.

In our exercise, we calculate the determinant of the matrix after modifying it with the eigenvalue \( \lambda \) in each diagonal entry. The determinant gives us a single polynomial equation after setting it to zero, as seen in the process above. While it might look like a simple function, the determinant's value comes from its ability to condense the matrix information into a format that supports various calculations, such as solving systems of linear equations and, in our case, finding eigenvalues.

Gaussian Elimination

Gaussian elimination is a systematic method for solving systems of linear equations. It is performed through a sequence of operations to simplify the system to a point where the solutions are apparent, known as its reduced row echelon form. This includes scaling rows, swapping them, or adding multiples of a row to another.

In the context of eigenvalues and eigenvectors, once we have our eigenvalues, we substitute each into the modified matrix equation \( (A - \lambda I)v = 0 \) to find the corresponding eigenvector(s) \( v \). Gaussian elimination helps us here to reduce this matrix equation to a form where the eigenvectors can be easily read off. This method underscores its importance in linear algebra as it transcends beyond simple equation solving and becomes a fundamental tool in various matrix-related operations, including our goal of finding eigenvectors for a given eigenvalue as demonstrated in the exercise.