Question

If \(A\) is \(2 \times 2\) and has an eigenvalue \(c\) of multiplicity two, then the characteristic polynomial is \(p_{A}(\lambda)=(\lambda-c)^{2}\). Because \(p_{A}(A)=0\) (see the Cayley-Hamilton theorem), \((A-c I)^{2}=0\). Each of the \(2 \times 2\) matrices in Exercises 58 \(-61\) has an eigenvalue \(c\) of multiplicity two. Find that eigenvalue \(c\) and verify that \((A-c I)^{2}=0\). $$ \left[\begin{array}{rr} -6 & 1 \\ -1 & -4 \end{array}\right] $$

Short answer

The eigenvalue is \(-5\), and \((A-cI)^2 = 0\) is verified.

Step-by-step solution

Identify the Matrix

We are given the matrix \( A = \begin{pmatrix} -6 & 1 \ -1 & -4 \end{pmatrix} \). We need to find its eigenvalue of multiplicity two.

Find the Eigenvalue

Since \( A \) is a \( 2 \times 2 \) matrix with an eigenvalue of multiplicity two, the characteristic polynomial is \( p_A(\lambda) = (\lambda - c)^2 \). We find \( c \) by solving \( \text{trace}(A) = 2c \) since both eigenvalues are equal. The trace of \( A \) is the sum of its diagonal elements: \[ \text{trace}(A) = -6 + (-4) = -10 \]Equating yields:\[ 2c = -10 \Rightarrow c = -5 \] So, the eigenvalue \( c = -5 \).

Form the Matrix \( A - cI \)

Compute \( A - cI \) where \( c = -5 \) and \( I \) is the identity matrix. \[ I = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \]\( A - cI = \begin{pmatrix} -6 & 1 \ -1 & -4 \end{pmatrix} - (-5) \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} -6+5 & 1 \ -1 & -4+5 \end{pmatrix} = \begin{pmatrix} -1 & 1 \ -1 & 1 \end{pmatrix} \)

Compute \((A - cI)^2\)

Now, we need to verify \((A - cI)^2 = 0\). Calculate:\[(A - cI)^2 = \begin{pmatrix} -1 & 1 \ -1 & 1 \end{pmatrix} \begin{pmatrix} -1 & 1 \ -1 & 1 \end{pmatrix} \]Performing matrix multiplication:\[= \begin{pmatrix} (-1)(-1) + (1)(-1) & (-1)(1) + (1)(1) \ (-1)(-1) + (1)(-1) & (-1)(1) + (1)(1) \end{pmatrix} \]\[= \begin{pmatrix} 1 - 1 & -1 + 1 \ 1 - 1 & -1 + 1 \end{pmatrix} \]\[= \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix}\]Thus, \((A - cI)^2 = 0\), verifying the result.

Key concepts

Characteristic Polynomial

To determine the eigenvalues of a matrix, we often use the characteristic polynomial. This polynomial is derived from the matrix in question, and it provides a central role in understanding the behavior of linear transformations.
The characteristic polynomial of a matrix \(A\) is defined as \(p_A(\lambda) = \det(A - \lambda I)\), where \(\lambda\) is a scalar and \(I\) is the identity matrix of the same order as \(A\).
For a \(2 \times 2\) matrix, the polynomial is typically quadratic. In the case where an eigenvalue \(c\) has a multiplicity of two, the characteristic polynomial simplifies to \((\lambda - c)^2\), meaning both roots correspond to the eigenvalue \(c\). In our example with the matrix \(\begin{pmatrix} -6 & 1 \ -1 & -4 \end{pmatrix}\), the characteristic polynomial reveals that the eigenvalue \(c\) is \(-5\).

Cayley-Hamilton Theorem

The Cayley-Hamilton theorem is a fundamental theorem in linear algebra. It states that every square matrix satisfies its own characteristic equation.
For a matrix \(A\), this means if the characteristic polynomial is \(p_A(\lambda)\), then \(p_A(A) = 0\).
What this implies is that if \(p_A(\lambda) = (\lambda - c)^2\), then substituting \(A\) into this polynomial gives \((A - cI)^2 = 0\).
As demonstrated in the solution, for our given matrix \(A\), substituting \(c = -5\) derived from the characteristic polynomial, ensures that \((A + 5I)^2 = 0\), confirming the theorem holds for \(A\). This is a key result in verifying the multiplicity of eigenvalues.

Matrix Trace

The trace of a matrix is the sum of its diagonal elements. This seemingly simple property has significant implications, particularly with eigenvalues.
For a \(2 \times 2\) matrix, if it has an eigenvalue \(c\) with multiplicity two, the trace can be used to quickly find this value. It is shown that the trace \(\text{trace}(A) = 2c\), due to the eigenvalues being equal.
In our exercise with \(A = \begin{pmatrix} -6 & 1 \ -1 & -4 \end{pmatrix}\), the trace is \(-6 + (-4) = -10\). Solving for \(c\), we find \(2c = -10\), so \(c = -5\).
Thus, the trace helps not only verify eigenvalues but also determines them succinctly when their multiplicities are understood.

Multiplicity of Eigenvalues

The multiplicity of an eigenvalue refers to how many times it appears as a root of the characteristic polynomial. There are two types: algebraic and geometric multiplicity.
**Algebraic Multiplicity**
This is the number of times an eigenvalue appears in the characteristic polynomial. In our matrix \(A\)'s polynomial, \((\lambda + 5)^2\), the eigenvalue \(-5\) appears twice.
**Geometric Multiplicity**
This counts the number of linearly independent eigenvectors associated with an eigenvalue. Ideally, it is equal or less than the algebraic multiplicity.
In the specific exercise, the eigenvalue \(-5\) has both algebraic and geometric multiplicity of two, aligning perfectly which assures consistency and confirms (A - cI)^2 = 0". This concept is crucial for understanding not just the existence of eigenvalues, but how they interact in a matrix context.