Problem 43 - Advanced Engineering Mathematics

Question

In Problems 41-46, find the eigenvalues and corresponding eigenvectors of the given matrix. $$ \left(\begin{array}{lll} 3 & 2 & 4 \\ 2 & 0 & 2 \\ 4 & 2 & 3 \end{array}\right) $$

Step-by-Step Solution

Answer

Eigenvalues: 0, 3; Eigenvectors: \( t(-2,0,1), s(1,0,-1), u(-2,1,2) \).

Step 1: Matrix Setup and Characteristic Equation

To find the eigenvalues, we first set up the matrix equation \(A - \lambda I = 0\), where \(A\) is the given matrix and \(I\) is the identity matrix of the same size. The matrix subtraction results in: \[\begin{pmatrix} 3-\lambda & 2 & 4 \2 & -\lambda & 2 \4 & 2 & 3-\lambda\end{pmatrix}\]Next, we find the determinant of this matrix and set it equal to zero to find \(\lambda\).

Step 2: Calculate Determinant

To find the determinant, expand along the first row: \[(3-\lambda)((-\lambda)(3-\lambda) -4) - 2(2(3-\lambda) - 8) + 4(4 + 2\lambda)\]Simplifying this will lead to a characteristic polynomial.

Step 3: Solve Characteristic Polynomial

The expanded determinant equation simplifies to the polynomial: \[-\lambda^3 + 6\lambda^2 - 9\lambda = 0\]Factor this equation to solve for \(\lambda\): \[ -\lambda(\lambda^2 - 6\lambda + 9) = 0 \]This gives us \(\lambda(\lambda-3)^2 = 0\).

Step 4: Find Eigenvalues

From the equation in Step 3, the eigenvalues are found by solving: \[-\lambda = 0 \Rightarrow \lambda_1 = 0\]And for: \[(\lambda - 3)^2 = 0 \Rightarrow \lambda_2 = 3\] (which is a repeated eigenvalue). Thus, the eigenvalues are \(0\) and \(3\).

Step 5: Find Eigenvectors for \(\lambda_1 = 0\)

Substitute \(\lambda_1 = 0\) into \(A - \lambda I\) and solve: \[\begin{pmatrix} 3 & 2 & 4 \2 & 0 & 2 \4 & 2 & 3\end{pmatrix} \begin{pmatrix}x \y \z\end{pmatrix} = \begin{pmatrix} 0 \0 \0\end{pmatrix}\]Solving gives the eigenvector \(\mathbf{v}_1 = t\begin{pmatrix}-2 \0 \1 \end{pmatrix}\) for any scalar \(t\).

Step 6: Find Eigenvectors for \(\lambda_2 = 3\)

Substitute \(\lambda_2 = 3\) into \(A - \lambda I\) and solve: \[\begin{pmatrix} 0 & 2 & 4 \2 & -3 & 2 \4 & 2 & 0\end{pmatrix} \begin{pmatrix}x \y \z\end{pmatrix} = \begin{pmatrix} 0 \0 \0\end{pmatrix}\]Solving, we find two linearly independent eigenvectors: \(\mathbf{v}_{2a} = s\begin{pmatrix} 1 \0 \-1 \end{pmatrix}\) and \(\mathbf{v}_{2b} = u\begin{pmatrix} -2 \1 \2 \end{pmatrix}\), where \(s\) and \(u\) are scalars.

Key Concepts

Characteristic Equation

In the world of linear algebra, the characteristic equation is a crucial concept used to find the eigenvalues of a matrix. Consider a square matrix \( A \) of size \( n \times n \). The characteristic equation is derived from the matrix equation \( A - \lambda I = 0 \), where \( I \) is the identity matrix of the same size, and \( \lambda \) represents the eigenvalues. To obtain this characteristic equation, determine the determinant of the matrix \( A - \lambda I \) and set it equal to zero. This gives us a polynomial known as the characteristic polynomial. Solving this characteristic polynomial for \( \lambda \) will yield the eigenvalues of matrix \( A \). For example, the characteristic polynomial for the given matrix is \(-\lambda^3 + 6\lambda^2 - 9\lambda = 0\). Its roots \( \lambda = 0 \) and \( \lambda = 3 \) are the eigenvalues we seek.

Determinant of a Matrix

The determinant of a matrix is a special number calculated from the elements of a square matrix. It's a key element when creating the characteristic equation for eigenvalue problems. To compute it in this context, take \( A - \lambda I \) and perform determinant calculations. For a matrix like the given Example \[\begin{pmatrix}3-\lambda & 2 & 4 \2 & -\lambda & 2 \4 & 2 & 3-\lambda\end{pmatrix}\]the determinant can be found by expanding along any row or column. Expanding along the first row, we find the determinant is:\[(3-\lambda)((-\lambda)(3-\lambda) - 4) - 2(2(3-\lambda) - 8) + 4(4 + 2\lambda)\]Simplifying this yields the characteristic polynomial: \(-\lambda^3 + 6\lambda^2 - 9\lambda = 0\). This process is crucial for finding eigenvalues and understanding the matrix's properties.

Repeated Eigenvalue

Sometimes, an eigenvalue may appear more than once as a root within the characteristic polynomial. Such eigenvalues are known as repeated eigenvalues or degenerate eigenvalues. Repeated eigenvalues mean that the characteristic polynomial will have a root with a multiplicity greater than one. This doesn't change the outcome of finding eigenvectors, but it can impact the number of linearly independent eigenvectors available. In our example, \( \lambda = 3 \) appears twice, a phenomenon which often affects the types of eigenvectors that exist. The repeated nature is indicated by the square term in the polynomial:\[(\lambda - 3)^2 = 0\]. To find corresponding eigenvectors, you proceed in the same way as you would with non-repeated eigenvalues, but it's essential to ensure you find a sufficient number of linearly independent vectors to span the eigenspace.

Linear Independence

Linear independence is a central topic in vector spaces that deals with the relation between vectors. A set of vectors is said to be linearly independent if no vector in the set can be written as a combination of the others. This concept is crucial when finding eigenvectors, especially when dealing with repeated eigenvalues. For the matrix under study, solving for eigenvectors corresponding to the repeated eigenvalue \( \lambda = 3 \) involves finding vectors that are not just solutions, but linearly independent solutions. From the solution steps, we found two linearly independent eigenvectors \( \mathbf{v}_{2a} = s\begin{pmatrix} 1 \ 0 \ -1 \end{pmatrix} \) and \( \mathbf{v}_{2b} = u\begin{pmatrix} -2 \ 1 \ 2 \end{pmatrix} \). This establishes the basis of the eigenspace associated with the repeated eigenvalue, ensuring that we have enough vectors to span the eigenspace of the given eigenvalue.