Question

VFor each matrix \(A\), find the eigenvalues and eigenvectors. Give the geometric and algebraic multiplicity of each eigenvalue. Does \(A\) have a full set of eigenvectors?\(A=\left[\begin{array}{lll}5 & 0 & 0 \\ 1 & 5 & 0 \\ 1 & 0 & 5\end{array}\right]\)

Short answer

Explain your answer. Answer: No, the matrix A does not have a full set of eigenvectors. The sum of geometric multiplicities (2) is not equal to the size of the matrix (3).

Step-by-step solution

Compute the eigenvalues

We will find the eigenvalues by solving the characteristic equation: \(determinant(\lambda I - A) = 0\) $\left|\begin{array}{lll} \lambda-5 & 0 & 0 \\ 1 & \lambda-5 & 0 \\ 1 & 0 & \lambda-5 \end{array}\right|=0$ Solving for \(determinant(\lambda I-A)\), we have: \((\lambda-5)((\lambda-5)^2)=0\) So, the eigenvalues are: \(\lambda = 5\)

Find the eigenvectors

For each eigenvalue, we'll find the eigenvectors by solving the following equation: \((A-\lambda I)X=0\) For \(\lambda=5\): \((A - 5I)X = 0\) $\left[\begin{array}{lll} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 0 \end{array}\right]\left[\begin{array}{l} x_1 \\ x_2 \\ x_3 \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right]$ From the second and third rows, we have one equation: \(x_1 = 0\) We can let \(x_2\) and \(x_3\) be free variables. Therefore, eigenvectors for \(\lambda=5\) are in the form: $X=\left[\begin{array}{l} 0 \\ x_2 \\ x_3 \end{array}\right]\( where \)x_2\( and \)x_3$ are arbitrary constants.

Calculate geometric multiplicities

The geometric multiplicity of an eigenvalue is the number of linearly independent eigenvectors associated with that eigenvalue. Since there are two eigenvectors for the eigenvalue \(\lambda = 5\), the geometric multiplicity of 5 is 2.

Calculate algebraic multiplicities

The algebraic multiplicity of an eigenvalue is the number of times it appears as a root of the characteristic equation. In our case, the eigenvalue 5 appears once. So its algebraic multiplicity is 1.

Determine if A has a full set of eigenvectors

To have a full set of eigenvectors, the sum of geometric multiplicities must be equal to the size of the matrix. In this case, the size of the matrix is 3, and the sum of geometric multiplicities is 2. Therefore, the matrix A does not have a full set of eigenvectors.

Key concepts

Geometric Multiplicity

Geometric multiplicity is an essential concept when discussing eigenvalues and eigenvectors of a matrix. It refers to the number of linearly independent eigenvectors associated with a particular eigenvalue. In simpler terms, it tells us how many distinct directions in the vector space are "eigendirections" for a given eigenvalue.

If you have an eigenvalue, \( \lambda \), of a matrix, its geometric multiplicity is determined by finding how many solutions exist for the equation \((A - \lambda I)X = 0\).

• For example, in our exercise, the eigenvalue \( \lambda = 5 \) has a geometric multiplicity of 2.
• This is because there are two linearly independent vectors fitting the condition (where \( x_1 = 0 \), and \( x_2 \) and \( x_3 \) are arbitrary).

Geometric multiplicity is crucial as it helps indicate whether a matrix can be diagonalized. A matrix can be diagonalized if the sum of geometric multiplicities equals the size of the matrix. In our example, the matrix cannot be fully diagonalized as the sum is 2, which is less than the size 3.

Algebraic Multiplicity

Algebraic multiplicity gives us insight into how often an eigenvalue appears as a root of the characteristic equation of a matrix. Essentially, it represents the number of times an eigenvalue is repeated.

To find the algebraic multiplicity, you first determine the characteristic equation from the determinant \(\det(\lambda I - A) = 0\), and then check how many times each eigenvalue is a solution.

• In the exercise, the eigenvalue \( \lambda = 5 \) appears exactly once as a root of the equation \((\lambda - 5)((\lambda - 5)^2) = 0\).
• Thus, it has an algebraic multiplicity of 3.

Understanding algebraic multiplicity is important as it often signals potential degeneracy in the eigenvalues. This repeats hints at certain constraints regarding the degree of freedom available within the system represented by the matrix.

Characteristic Equation

The characteristic equation is a central tool for finding eigenvalues of a matrix. It arises from the expression \(\det(\lambda I - A) = 0\), where \(A\) is the matrix, \(I\) is the identity matrix of the same size, and \(\lambda\) represents a scalar value that we're trying to find (the eigenvalue).

This determinant gives us a polynomial equation in \(\lambda\), known as the characteristic polynomial. The roots of this polynomial are the eigenvalues.

• In our example, by setting up the characteristic determinant for matrix \(A\), we determined \((\lambda - 5)((\lambda - 5)^2) = 0\).
• This yields an eigenvalue of \( \lambda = 5 \).

Solving the characteristic equation is often the first step in finding an eigenvalue, providing the basis for further exploration of its multiplicities and corresponding eigenvectors. This equation is vital as it lays down the mathematical roadmap for understanding the deeper structure of the matrix at hand. Knowing how to construct and solve it is crucial for advancing in linear algebra.