Question

a. If \(A=\left[\begin{array}{ll}1 & 3 \\ 0 & 2\end{array}\right]\) and \(B=\left[\begin{array}{ll}2 & 0 \\ 0 & 1\end{array}\right]\) verify that \(A\) and \(B\) are diagonalizable, but \(A B\) is not. b. If \(D=\left[\begin{array}{rr}1 & 0 \\ 0 & -1\end{array}\right]\) find a diagonalizable matrix \(A\) such that \(D+A\) is not diagonalizable.

Short answer

Matrix A and B are diagonalizable; A*B is not. A can be \( \begin{bmatrix}-1 & 1\\ 0 & 0\end{bmatrix} \) for D+A non-diagonalizability.

Step-by-step solution

Verify if matrix A is diagonalizable

A matrix is diagonalizable if it has enough linearly independent eigenvectors to form a basis. Calculate the characteristic polynomial of matrix A:\[\text{det}(A - \lambda I) = \begin{vmatrix} 1-\lambda & 3 \ 0 & 2-\lambda \end{vmatrix} = (1-\lambda)(2-\lambda)\].The eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = 2\).Now, find the eigenvectors:For \(\lambda_1 = 1\), \((A - I)v = 0\) gives the eigenvector \(v = \begin{bmatrix}1\0\end{bmatrix}\).For \(\lambda_2 = 2\), \((A - 2I)v = 0\) gives the eigenvector \(v = \begin{bmatrix}3/2\1\end{bmatrix}\).Since we have two independent eigenvectors, matrix A is diagonalizable.

Verify if matrix B is diagonalizable

Similar to A, calculate the characteristic polynomial of matrix B:\[\text{det}(B - \lambda I) = \begin{vmatrix} 2-\lambda & 0 \ 0 & 1-\lambda \end{vmatrix} = (2-\lambda)(1-\lambda)\].The eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = 1\).For \(\lambda_1 = 2\), \((B - 2I)v = 0\) gives eigenvector \(v = \begin{bmatrix}1\0\end{bmatrix}\).For \(\lambda_2 = 1\), \((B - I)v = 0\) gives eigenvector \(v = \begin{bmatrix}0\1\end{bmatrix}\).As matrix B has two distinct eigenvectors, it is diagonalizable.

Check if AB is diagonalizable

Find the product of matrices A and B:\[AB = \begin{bmatrix}1 & 3 \ 0 & 2\end{bmatrix} \begin{bmatrix}2 & 0 \ 0 & 1\end{bmatrix} = \begin{bmatrix}2 & 3 \ 0 & 2\end{bmatrix}\].Next, calculate its characteristic polynomial:\[\text{det}(AB - \lambda I) = \begin{vmatrix} 2-\lambda & 3 \ 0 & 2-\lambda \end{vmatrix} = (2-\lambda)^2\].The only eigenvalue is \(\lambda = 2\), with algebraic multiplicity 2.Check its geometric multiplicity; for \(\lambda = 2\), we solve \((AB - 2I)v = 0\) which gives eigenvector \(v=\begin{bmatrix}1 \ 0\end{bmatrix}\) only, having geometric multiplicity 1.Since the geometric and algebraic multiplicities differ, AB is not diagonalizable.

Define Requirements for D + A to be Non-Diagonalizable

Given matrix \(D = \begin{bmatrix}1 & 0\ 0 & -1\end{bmatrix}\), for \(D+A\) to be non-diagonalizable, choose \(A\) such that the sum results in a matrix with eigenvalues that don't yield enough eigenvectors for diagonalization.A simple choice is to have \(A = \begin{bmatrix}-1 & 1\ 0 & 0\end{bmatrix}\) so that \(D+A\) becomes:\[(D+A) = \begin{bmatrix}0 & 1\ 0 & -1\end{bmatrix}\].

Confirm Non-Diagonalizability of D + A

Compute the characteristic polynomial of \(D+A\):\[\text{det}(D+A - \lambda I) = \begin{vmatrix}0-\lambda & 1 \ 0 & -1-\lambda\end{vmatrix} = \lambda(1+\lambda)\].Eigenvalues are \(\lambda_1 = 0\), \(\lambda_2 = -1\).For \(\lambda_1 = 0\), solve: \(v = \begin{bmatrix}1 \ 0\end{bmatrix}\).But for \(\lambda_2 = -1\), solve: \((D+A + I)v = 0\) yielding \(v = \begin{bmatrix}0 \ 1\end{bmatrix}\).These yield the same vector, proving geometric multiplicity less than algebraic multiplicity. Therefore, \(D+A\) is not diagonalizable.

Key concepts

Eigenvectors

Eigenvectors are crucial in understanding the diagonalization of matrices. An eigenvector of a matrix is a non-zero vector that, when the matrix is applied to it, only scales that vector by a certain factor, called an eigenvalue. The relationship is summarized by the equation \(A\vec{v} = \lambda\vec{v}\), where \(A\) is a matrix, \(\vec{v}\) is the eigenvector, and \(\lambda\) is the eigenvalue.

To find eigenvectors, we usually start by determining the eigenvalues of the matrix, and then solve the equation \((A - \lambda I)\vec{v} = \vec{0}\). Here, \(I\) is the identity matrix, and we are solving for the vector \(\vec{v}\) that satisfies the equation.

Understanding eigenvectors is vital because they give insights into the matrix’s structure. For example, in diagonalization, we use eigenvectors to form a basis that can transform the matrix into a diagonal form. A matrix is diagonalizable if there are enough linearly independent eigenvectors to span the space. This connection highlights the importance of identifying and understanding eigenvectors in solving problems related to matrix diagonalization.

Eigenvalues

Eigenvalues are intrinsic to the process of matrix diagonalization. They tell us how much an eigenvector is stretched or squished in the transformation described by a matrix. Mathematically, the eigenvalues of a matrix \(A\) are found from the characteristic equation \(\text{det}(A - \lambda I) = 0\), where \(I\) is the identity matrix.

This equation results from taking the determinant of the matrix \(A - \lambda I\) after substituting a placeholder for \(\lambda\). Solving the equation gives us the eigenvalues, \(\lambda\). Each eigenvalue corresponds to a set of eigenvectors.

In practice, eigenvalues help determine if a matrix is diagonalizable. A key criterion is that the algebraic multiplicity of each eigenvalue should equal its geometric multiplicity (the number of linearly independent eigenvectors associated with that eigenvalue). If they differ, as with certain matrices like \(AB\) from the exercise, the matrix is not diagonalizable. This makes understanding eigenvalues a central piece of analyzing matrix behavior.

Linear Independence

Linear independence is a fundamental concept in linear algebra, especially when dealing with eigenvectors. A set of vectors is linearly independent if no vector in the set can be written as a linear combination of the others. This property is essential for diagonalization, as having a full set of linearly independent eigenvectors means a matrix can be transformed into a diagonal form.

In the context of diagonalization, you check for linear independence to ensure you have enough eigenvectors to span the vector space. If you have \(n\) eigenvectors for an \(n \times n\) matrix, and they are linearly independent, the matrix is diagonalizable.

This is because you can form an invertible matrix \(P\) where its columns are the eigenvectors, allowing you to express the original matrix \(A\) as \(PDP^{-1}\), where \(D\) is diagonal. However, when eigenvectors are not linearly independent, as seen in the example of \(AB\), the transformation is not possible. Thus, linear independence is necessary for a matrix to be diagonalizable.