Question

a. Show that the only diagonalizable matrix \(A\) that has only one eigenvalue \(\lambda\) is the scalar matrix \(A=\lambda I\) b. Is \(\left[\begin{array}{cc}3 & -2 \\ 2 & -1\end{array}\right]\) diagonalizable?

Short answer

Only scalar matrices with a single eigenvalue are diagonalizable. The given matrix is not diagonalizable.

Step-by-step solution

Define Diagonalizable Matrix

A matrix is diagonalizable if there exists an invertible matrix \(P\) and a diagonal matrix \(D\) such that \(A = PDP^{-1}\). For a matrix to be diagonalizable, it must have eigenvectors that form a basis of the vector space.

Analyze Matrix with One Eigenvalue

Let matrix \(A\) have only one eigenvalue \(\lambda\). For \(A\) to be diagonalizable with a single eigenvalue, the matrix must be a scalar multiple of the identity matrix. This means \(A = \lambda I\), where \(I\) is the identity matrix. Any other structure, due to Jordan form, would not remain diagonalizable unless \(A = \lambda I\).

Check Eigenvalues of Given Matrix

Compute the characteristic polynomial of matrix \(B = \begin{bmatrix} 3 & -2 \ 2 & -1 \end{bmatrix}\): \(\text{det}(B - \lambda I) = \begin{vmatrix} 3-\lambda & -2 \ 2 & -1-\lambda \end{vmatrix}\). This evaluates to \((3-\lambda)(-1-\lambda) - (-2)(2) = \lambda^2 - 2\lambda + 1\).

Solve the Characteristic Equation

The characteristic equation is \(\lambda^2 - 2\lambda + 1 = 0\). This can be simplified to \((\lambda - 1)^2 = 0\), giving the repeated eigenvalue \(\lambda = 1\).

Determine Diagonalizability

Since the eigenvalue \(\lambda = 1\) has algebraic multiplicity 2, check the geometric multiplicity by finding eigenvectors: \((B-I)\mathbf{v} = 0\), which translates to \(\begin{bmatrix} 2 & -2 \ 2 & -2 \end{bmatrix}\begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 0 \ 0 \end{bmatrix}\). Simplifying, we find the single linearly independent eigenvector, indicating the geometric multiplicity is 1, not equal to 2, hence \(B\) is not diagonalizable.

Key concepts

Eigenvalues

Eigenvalues are fundamental in understanding how a matrix behaves when it is transformed or "stretched" along its own directions. In simple terms, when you apply a matrix to a vector, the result can be seen as stretching the vector by a certain factor. This stretch factor is what we call an eigenvalue.
Eigenvalues are not just random numbers; they are intimately related to the matrix itself. To find them, we examine the characteristic polynomial of the matrix, which emerges from the determinant of the matrix minus a scalar multiplied by the identity matrix, often denoted as \( (A - \lambda I) \).

• Eigenvalues determine stability: For real-world systems described by matrices, eigenvalues can indicate stability or predict oscillatory behavior.
• In diagonalization, distinct eigenvalues suggest that a matrix is diagonalizable, but repeated eigenvalues require further inspection of eigenvectors.

Understanding eigenvalues is crucial for matrix diagonalization, as they reveal if and how a matrix can be converted into a simpler, diagonal form.

Eigenvectors

While eigenvalues tell us how a matrix stretches a vector, eigenvectors show the direction in which this stretching occurs. They are the vectors that remain unchanged in direction during the application of the matrix transformation. Hence, an eigenvector corresponds to each eigenvalue.
To find an eigenvector corresponding to a matrix eigenvalue, we solve the equation \( (A - \lambda I) \mathbf{v} = 0 \), where \( \mathbf{v} \) is the eigenvector. The solution to this equation provides us with the basis vectors that span the eigenspace of the matrix.

• For a matrix to be diagonalizable, its eigenvectors must form a basis; in other words, they should be linearly independent and fill the entire vector space.
• If the number of linearly independent eigenvectors is less than the dimension of the matrix, then the matrix is not diagonalizable.

In cases where a matrix has repeated eigenvalues, special attention is needed to check if enough linearly independent eigenvectors exist.

Invertible Matrix

An invertible matrix, also known as a non-singular matrix, is one that can be "reversed", meaning that there exists another matrix which, when multiplied with the original, results in the identity matrix \( I \). For a matrix \( P \), being invertible implies there exists a matrix \( P^{-1} \) such that \( PP^{-1} = I \).
In the context of matrix diagonalization, an invertible matrix \( P \) is crucial. It is used to relate the original matrix \( A \) to its diagonal form \( D \) through the transformation \( A = PDP^{-1} \).

• The presence of \( P^{-1} \) ensures that the transformation can be "undone" to retrieve the original matrix \( A \).
• An invertible matrix has non-zero determinant, further implying that all its eigenvalues are non-zero.

Thus, invertibility is a necessary condition to employ the diagonalization process, ensuring that the matrix representation can be reversed.

Characteristic Polynomial

The characteristic polynomial provides a crucial insight into the matrix by revealing its eigenvalues. Constructed from the matrix \( A \), the characteristic polynomial is obtained by computing the determinant of \( (A - \lambda I) \).
For instance, if we have a matrix \( B \), its characteristic polynomial is derived by determining \( \det(B - \lambda I) \). The solutions to the equation set by the characteristic polynomial, \( \det(B - \lambda I) = 0 \), are the eigenvalues of \( B \).

• This polynomial has a degree equal to the size of the matrix, indicating the maximum number of eigenvalues the matrix can have.
• The roots of the characteristic polynomial, in terms of algebraic multiplicity, indicate the number of times each eigenvalue appears.

Thus, the characteristic polynomial serves as a vital tool in diagonalization, identifying eigenvalues that need further verification for geometric properties.