Question

Let \(A=\left[\begin{array}{ccc}2 & 3 & -3 \\ 1 & 0 & -1 \\ 1 & 1 & -2\end{array}\right]\) and \(B=\) \(\left[\begin{array}{lll}0 & 1 & 0 \\ 3 & 0 & 1 \\ 2 & 0 & 0\end{array}\right] .\) Show that \(c_{A}(x)=c_{B}(x)=(x+1)^{2}(x-\)2), but \(A\) is diagonalizable and \(B\) is not.

Short answer

\(A\) is diagonalizable with three independent eigenvectors; \(B\) is not due to insufficient dimension of eigenspaces.

Step-by-step solution

Calculate the Characteristic Polynomial of A

For a matrix \(A\), the characteristic polynomial is given by \(c_A(x) = \det(xI - A)\). First, we calculate \(xI - A\): \[\begin{bmatrix}x & 0 & 0 \ 0 & x & 0 \ 0 & 0 & x\end{bmatrix} - \begin{bmatrix}2 & 3 & -3 \ 1 & 0 & -1 \ 1 & 1 & -2\end{bmatrix} = \begin{bmatrix}x-2 & -3 & 3 \ -1 & x & 1 \ -1 & -1 & x+2\end{bmatrix}\]Now calculate the determinant of this matrix. The determinant is calculated to be \[ (x+1)^2(x-2) \].

Calculate the Characteristic Polynomial of B

Now calculate the characteristic polynomial for \(B\) in the same manner. We have:\[xI - B = \begin{bmatrix}x & 0 & 0 \ 0 & x & 0 \ 0 & 0 & x\end{bmatrix} - \begin{bmatrix}0 & 1 & 0 \ 3 & 0 & 1 \ 2 & 0 & 0\end{bmatrix} = \begin{bmatrix}x & -1 & 0 \ -3 & x & -1 \ -2 & 0 & x\end{bmatrix}\]The determinant, after expansion, results in \[ (x+1)^2(x-2) \].

Verify Diagonalizability of A

Matrix \(A\) is diagonalizable if there are three linearly independent eigenvectors or enough eigenvectors to have a basis of \(\mathbb{R}^3\). Given the distinct roots \(x = -1, -1, 2\), the algebraic multiplicity of \(x = -1\) is 2 and \(x = 2\) is 1. Calculate the eigenspaces to ensure there are a total of three independent eigenvectors. Since \(E_{-1}\) has dimension 2 and \(E_2\) has dimension 1, \(A\) is diagonalizable.

Check Non-diagonalizability of B

For \(B\), similar calculation shows the eigenvalues \(x = -1\) with algebraic multiplicity 2 and \(x = 2\) with algebraic multiplicity 1. However, calculating the eigenspaces, it turns out that the geometric multiplicity of the eigenvalue \(x = -1\) is less than 2, giving only one eigenvector. Thus, \(B\) doesn't have a complete set of eigenvectors, and hence it is not diagonalizable.

Key concepts

Characteristic Polynomial

The characteristic polynomial is a foundational concept in linear algebra, used to understand a matrix's properties. It gives insights into eigenvalues, which tell us about the matrix's behavior. For a matrix \(A\), the characteristic polynomial \(c_A(x)\) is derived using the determinant of \(xI - A\), where \(I\) is the identity matrix of the same size as \(A\).

• This expression \(xI - A\) transforms the matrix by scaling and shifting its values.
• The determinant of this transformation matrix, when set to zero, yields the roots which are the eigenvalues.

For matrices \(A\) and \(B\) from our example, both have the same characteristic polynomial \((x+1)^2(x-2)\), indicating that they share the same eigenvalues.

Eigenvectors

Eigenvectors are vectors associated with a specific eigenvalue, which signify directions that do not change under transformation by the matrix. They are crucial for understanding the structure of a matrix.

• Given an eigenvalue \(\lambda\), to find its eigenvectors for a matrix \(A\), solve \((A - \lambda I)\mathbf{v} = 0\), where \(\mathbf{v}\) is the eigenvector.
• This results in a system of linear equations, whose solutions give the set of eigenvectors.

In our example, finding the eigenvectors of \(A\) confirms whether it is diagonalizable. Eigenvectors form the basis for simplification in matrix transformation, aiding in tasks like diagonalization.

Eigenvalues

Eigenvalues are scalars that signify how an eigenvector is stretched during transformation by its matrix. They are critical for determining a matrix's properties, such as stability and behavior.

• An eigenvalue \(\lambda\) solves the equation \(\det(A - \lambda I) = 0\).
• Each root of the characteristic polynomial corresponds to an eigenvalue, indicating distinct effects on the transformation.

In both matrices \(A\) and \(B\), the eigenvalues identified are \(-1\) with multiplicity 2, and \(2\) with multiplicity 1. These values inform whether a matrix can be diagonalized based on the available eigenvectors.

Geometric Multiplicity

Geometric multiplicity refers to the number of linearly independent eigenvectors corresponding to a specific eigenvalue. It reveals whether the matrix can be fully diagonalized.

• It is the dimension of the eigenspace for a particular eigenvalue.
• The geometric multiplicity is always less than or equal to the algebraic multiplicity.

For matrix \(A\), each eigenvalue has enough eigenvectors to allow diagonalization with geometric multiplicities matching their algebraic counterparts. For matrix \(B\), however, the geometric multiplicity for \(\lambda = -1\) is less than its algebraic multiplicity, meaning insufficient independent eigenvectors for diagonalization.

Algebraic Multiplicity

Algebraic multiplicity measures how many times an eigenvalue appears as a root of the characteristic polynomial. It helps anticipate the number of potential eigenvectors for diagonalization processes.

• Given by the degree of the term \((\lambda - \text{root})^m\) in the factored form of the characteristic polynomial.
• A higher algebraic multiplicity usually suggests more eigenvectors, but physical interpretation confirms this via geometric multiplicity.

For both matrices \(A\) and \(B\), the algebraic multiplicity of eigenvalue \(-1\) is 2, and \(2\) is 1. \(A\) has a matching geometric multiplicity, allowing diagonalization, whilst \(B\) does not, restricting it from being diagonalizable.