Problem 13 - Differential Equations and Linear Algebra

Question

Determine whether the given matrix \(A\) is diagonalizable. Where possible, find a matrix \(S\) such that $$S^{-1} A S=\operatorname{diag}\left(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}\right).$$ $$A=\left[\begin{array}{rrr}1 & -2 & 0 \\\\-2 & 1 & 0 \\\0 & 0 & 3\end{array}\right]$$

Step-by-Step Solution

Answer

The given matrix \(A\) is diagonalizable. We determined the eigenvalues \(\lambda_1 = 3\) and two complex eigenvalues. We constructed the matrix S = \(\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}\) using the eigenvector corresponding to \(\lambda_1\). We then multiplied \(S^{-1} A S\) and obtained the diagonal matrix \(\operatorname{diag}(3)\).

Step 1: Find the eigenvalues of A

To do this, we need to solve the following equation for the eigenvalues, \(\lambda\): \(|\lambda I - A| = 0\) where \(I\) is the identity matrix of the same order as matrix \(A\). In our case: \(|\lambda I - A| = \begin{vmatrix} \lambda - 1 & 2 & 0 \\ 2 & \lambda - 1 & 0 \\ 0 & 0 & \lambda - 3 \end{vmatrix}\) Next, we expand the determinant and solve for the values of \(\lambda\).

Step 2: Calculate the determinant

In our case, the determinant is calculated as follows: \((\lambda - 1)((\lambda - 1)(\lambda - 3) - 0) - 2((2)(\lambda - 3) - 0) + 0\) Expanding the determinant and solving for \(\lambda\): \(\lambda^3 - 4\lambda^2 + 5\lambda - 8 = 0\) We can factor this equation as: \((\lambda - 3)(\lambda^2 - \lambda + 2) = 0\) This gives us three eigenvalues: \(\lambda_1 = 3\), and the quadratic equation \(\lambda^2 - \lambda + 2 = 0\) has two complex eigenvalues \(\lambda_2\) and \(\lambda_3\) (the exact values are not needed).

Step 3: Find the eigenvectors

We first find the eigenvector corresponding to \(\lambda_1 = 3\). We need to solve the following equation: \((A - \lambda I)v = 0\) where \(v\) is the eigenvector. For \(\lambda_1 = 3\), we get: \(\left[\begin{array}{rrr} -2 & 2 & 0 \\ 2 & -2 & 0 \\ 0 & 0 & 0 \end{array}\right]v = 0\) Row reduce the augmented matrix to echelon form: \(\left[\begin{array}{rrr|r} -2 & 2 & 0 & 0 \\ 2 & -2 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \rightarrow \left[\begin{array}{rrr|r} 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]\) The parameterization of the eigenvector becomes: \(v_1 = t\left[\begin{array}{r} 1 \\ 1 \\ 0 \end{array}\right]\)

Step 4: Construct the matrix S and calculate S^(-1)AS

Construct the matrix S using the eigenvector found: \(S = \left[\begin{array}{r} 1 \\ 1 \\ 0 \end{array}\right]\) Now we multiply \(S^{-1} A S\): \(S^{-1} A S = S^{-1}AS = \operatorname{diag}\left(\lambda_{1}\right) = \operatorname{diag}(3)\) Since we've found a matrix \(S\) which diagonalizes \(A\), the given matrix \(A\) is diagonalizable. The diagonal matrix is given as \(\operatorname{diag}(3)\).

Key Concepts

Eigenvalues

Eigenvalues are a fundamental concept in linear algebra, closely tied to the behavior of linear transformations represented by matrices. In essence, an eigenvalue is a special scalar associated with a linear transformation that describes the factor by which the transformation stretches or shrinks vectors along a certain direction. For a matrix, these eigenvalues characterize the scaling factors applied to vectors that are not reoriented by the transformation.

To find the eigenvalues of a matrix, we solve the characteristic equation, which is formulated as the determinant of \(\lambda I - A\) set to zero, where \(\lambda\) is an eigenvalue, \(^A\) is the matrix in question, and \(I\) is the identity matrix of the same size. The resulting polynomial equation in \(\lambda\) will yield the eigenvalues upon solving. In our exercise, after computing and factoring the determinant, we retrieved the eigenvalues of the matrix.

Eigenvectors

Closely related to eigenvalues, eigenvectors are vectors that only get scaled (not rotated) when a matrix transformation is applied to them. Each eigenvalue has at least one corresponding eigenvector, and the pair describes an invariant line under the transformation. To find an eigenvector for a given eigenvalue, we solve the equation \(\left(A - \lambda I\right)v = 0\), which essentially seeks vectors \(v\) that remain parallel to themselves after the transformation by \(A\) is applied.

In our step by step solution, we used this method to find the eigenvectors for the matrix \(A\). By setting up and reducing the augmented matrix, we identified a vector that satisfies the equation for \(\lambda_1 = 3\), indicating the direction along which \(A\) will stretch the vector by a factor of 3.

Determinant

The determinant is a scalar value that can be computed from the elements of a square matrix. It gives important information about the matrix, such as whether it is invertible or not (a determinant of zero indicates a matrix is singular and thus not invertible), and it is also used in volume scaling, finding eigenvalues, and solving systems of linear equations.

In the exercise we processed, the determinant was instrumental in finding the eigenvalues. By expanding the determinant \(\lambda I - A\) and solving for \(\lambda\), we obtained a polynomial that factored to reveal the eigenvalues of matrix \(A\). This illustrates one of the determinant's key roles in matrix analysis and its importance in understanding the properties of linear transformations.

Identity Matrix

An identity matrix, denoted as \(I\), plays a crucial role in matrix algebra. It is a square matrix in which all the elements on the main diagonal are 1, and all other elements are 0. This matrix acts as the multiplicative identity in the matrix world, much like the number 1 does in regular multiplication. Therefore, multiplying any matrix \(A\) by \(I\) leaves \(A\) unchanged.

In the context of diagonalization, the identity matrix is used to construct the characteristic equation. When we compute \(\lambda I - A\) for a given matrix \(A\), we are setting the stage to find \(A\)'s eigenvalues and subsequently its eigenvectors. The clarity of the identity matrix's role in these processes is vital for fully grasping the concept of diagonalization and its implications for simplifying matrix operations.