Question

Determine by inspection whether or not the matrix is diagonalizable. Give a reason that supports your conclusion. (a) \(A_{1}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\) (b) \(A_{2}=\left[\begin{array}{rr}1 & 1 \\ 0 & -1\end{array}\right]\) (c) \(A_{3}=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]\)

Short answer

In conclusion, by inspection, we determined that matrix A1 and A3 are not diagonalizable since they have only one distinct eigenvalue, which leads to having fewer than 2 linearly independent eigenvectors. On the other hand, matrix A2 is diagonalizable because it has two distinct eigenvalues, resulting in 2 linearly independent eigenvectors.

Step-by-step solution

Matrix A1: \(\left[\begin{array}{ll}1 & 1 \\\ 0 & 1\end{array}\right]\)

By inspection, we can see that matrix A1 is an upper-triangular matrix with the same numbers on the diagonal (1,1). Generally, diagonal matrices have distinct eigenvalues for each of their diagonal elements. However, since both diagonal elements are 1, A1 has only one distinct eigenvalue, which is 1. As A1 is a 2x2 matrix, it should have 2 linearly independent eigenvectors to be diagonalizable. Nonetheless, it has only a single distinct eigenvalue, meaning it has fewer than 2 eigenvectors. Therefore, matrix A1 is not diagonalizable.

Matrix A2: \(\left[\begin{array}{rr}1 & 1 \\\ 0 & -1\end{array}\right]\)

Matrix A2 is also an upper-triangular matrix, but this time the diagonal elements are different (1,-1). Since the diagonal elements are different, A2 has two distinct eigenvalues, which are 1 and -1. Since A2 is a 2x2 matrix, to be diagonalizable, it should have 2 linearly independent eigenvectors. As it has two distinct eigenvalues, A2 has 2 linearly independent eigenvectors. Therefore, matrix A2 is diagonalizable.

Matrix A3: \(\left[\begin{array}{ll}1 & 1 \\\ 1 & 1\end{array}\right]\)

Matrix A3 is a square matrix with identical rows. Just like in the case of A1, both diagonal elements are 1. This means A3 has only one distinct eigenvalue, which is 1. A3 is a 2x2 matrix, so it should have 2 linearly independent eigenvectors to be diagonalizable. However, it only has a single distinct eigenvalue, which implies it has fewer than 2 eigenvectors. Therefore, matrix A3 is not diagonalizable.

Key concepts

Eigenvalues

Eigenvalues are important characteristics of a matrix used in various mathematical and engineering applications. They are scalars that indicate how much a particular vector is stretched or compressed during a linear transformation. To find the eigenvalues of a matrix, we can solve the characteristic equation, which is derived from the determinant of \(A - \lambda I\), where \(A\) is the matrix, \(\lambda\) is the eigenvalue, and \(I\) is the identity matrix.
In many cases, matrices have distinct eigenvalues, which means each eigenvalue is different from the others. If a matrix has distinct eigenvalues, it is usually diagonalizable, which greatly simplifies many calculations.
In our original exercise, for matrix \(A_1\), both diagonal elements are 1, which means it has only one distinct eigenvalue. This leads to it not being diagonalizable. For matrix \(A_2\), the diagonal elements are 1 and -1, providing two distinct eigenvalues, making it diagonalizable.

Eigenvectors

Eigenvectors are non-zero vectors that, when a matrix is applied to them, only get scaled by the corresponding eigenvalue instead of changing direction. They are crucial in understanding linear transformations.
To find eigenvectors, you solve the equation \(A\mathbf{v} = \lambda \mathbf{v}\), where \(\mathbf{v}\) is the eigenvector corresponding to eigenvalue \(\lambda\). This leads you to solve the system \(A - \lambda I\mathbf{v} = 0\).
In the context of diagonalization, a matrix requires a full set of linearly independent eigenvectors equal to the matrix's size. Without this, diagonalization is not possible. In our exercise, matrices with identical eigenvalues, like \(A_1\) and \(A_3\), have fewer independent eigenvectors, hence they cannot be diagonalized. Matrix \(A_2\) has distinct eigenvalues with corresponding independent eigenvectors, allowing for diagonalization.

Diagonalizable matrix

A diagonalizable matrix is one that can be converted into a diagonal matrix through a similarity transformation. Diagonal matrices are simpler to handle as operations like powers and exponentials are easier to compute for them.
To determine if a matrix is diagonalizable, it needs to have a sufficient set of linearly independent eigenvectors, equivalent to its size. When a matrix has distinct eigenvalues, it guarantees the availability of a full set of independent eigenvectors and is thus diagonalizable.
In the exercise, matrix \(A_2\) is diagonalizable because it has distinct eigenvalues, leading to two independent eigenvectors. The matrices \(A_1\) and \(A_3\) fail this requirement as their identical eigenvalues do not allow for a complete set of independent eigenvectors.

Upper-triangular matrix

An upper-triangular matrix is a matrix where all elements below the main diagonal are zero. These matrices are significant in linear algebra due to their simplicity in eigenvalue estimation. The eigenvalues in an upper-triangular matrix are the values located on its main diagonal.
Upper-triangular matrices are an essential step in many linear algebra algorithms, such as in solutions of systems of linear equations, eigenvalue algorithms, and matrix decompositions.
In the specific exercise, the concept of upper-triangular matrices is visible in all three matrices \(A_1\), \(A_2\), and \(A_3\). For matrix \(A_2\), the distinct diagonal elements help in establishing its eigenvalues directly, ensuring it is diagonalizable. While in matrices \(A_1\) and \(A_3\), identical diagonal elements restrict them to fewer independent eigenvectors, making them not diagonalizable.