Question

For 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?\(A=\left[\begin{array}{lll}5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right]\)

Short answer

Answer: Yes, the given matrix A has a full set of eigenvectors. It has one eigenvalue, λ = 5, with algebraic and geometric multiplicities both equal to 3.

Step-by-step solution

Find the eigenvalues

To find the eigenvalues of A, we first need to set up and solve the characteristic equation. The characteristic equation of a matrix A is given by \(\textup{det}(\lambda\textit{I}-A)=0,\) where \(\lambda\) represents the eigenvalue and \(\textit{I}\) is the identity matrix. For the given matrix A, the characteristic equation is: \(\textup{det}\left(\lambda\textit{I}-\left[\begin{array}{lll}5 & 0 & 0 \\\ 0 & 5 & 0 \\\ 0 & 0 &5\end{array}\right]\right) = \textup{det}\left(\left[\begin{array}{lll}\lambda-5 & 0 & 0 \\\ 0 & \lambda-5 & 0 \\\ 0 & 0 & \lambda-5\end{array}\right]\right)=0\) Now we can calculate the determinant: \((\lambda-5)^3 = 0\) Solving for \(\lambda\), we get only one eigenvalue: \(\lambda = 5\)

Find the eigenvectors and algebraic/ geometric multiplicity

For the eigenvalue \(\lambda = 5\), we need to find the eigenvectors by solving the homogeneous system of equations \((A-\lambda\textit{I})\textbf{x}=\textbf{0}\). This gives us the following system of equations: \((A-5\textit{I})\textbf{x} = \left[\begin{array}{lll}0 & 0 & 0 \\\ 0 & 0 & 0 \\\ 0 & 0 & 0\end{array}\right]\textbf{x} = \textbf{0}\) Since this system has no unique solution, we can choose any non-zero vector to be an eigenvector of A. Let's choose a few eigenvectors to verify that they indeed are eigenvectors of A: \(\textbf{x}_1 = \left[\begin{array}{l}1 \\\ 0 \\\ 0\end{array}\right], \textbf{x}_2 = \left[\begin{array}{l}0 \\\ 1 \\\ 0\end{array}\right], \textbf{x}_3 = \left[\begin{array}{l}0 \\\ 0 \\\ 1\end{array}\right]\) Multiplying A with each of these eigenvectors, we see that: \(A\textbf{x}_1 = \left[\begin{array}{lll}5 & 0 & 0 \\\ 0 & 5 & 0 \\\ 0 & 0 & 5\end{array}\right]\left[\begin{array}{l}1 \\\ 0 \\\ 0\end{array}\right] = \left[\begin{array}{l}5 \\\ 0 \\\ 0\end{array}\right] = 5\textbf{x}_1\) \(A\textbf{x}_2 = \left[\begin{array}{lll}5 & 0 & 0 \\\ 0 & 5 & 0 \\\ 0 & 0 & 5\end{array}\right]\left[\begin{array}{l}0 \\\ 1 \\\ 0\end{array}\right] = \left[\begin{array}{l}0 \\\ 5 \\\ 0\end{array}\right] = 5\textbf{x}_2\) \(A\textbf{x}_3 = \left[\begin{array}{lll}5 & 0 & 0 \\\ 0 & 5 & 0 \\\ 0 & 0 & 5\end{array}\right]\left[\begin{array}{l}0 \\\ 0 \\\ 1\end{array}\right] = \left[\begin{array}{l}0 \\\ 0 \\\ 5\end{array}\right] = 5\textbf{x}_3\) So, indeed these are eigenvectors of A corresponding to the eigenvalue \(\lambda = 5\). The algebraic and geometric multiplicities for this eigenvalue are both equal to 3, because there are three linearly independent eigenvectors, and the eigenvalue appears only once in the characteristic equation.

Determine if the matrix has a full set of eigenvectors

Since our given matrix A is a 3x3 matrix, and we have found three linearly independent eigenvectors (\(\textbf{x}_1, \textbf{x}_2, \textbf{x}_3\)), we can conclude that A does indeed have a full set of eigenvectors. In summary, the matrix A has one eigenvalue, \(\lambda = 5\), with algebraic and geometric multiplicities both equal to 3, and A has a full set of eigenvectors.

Key concepts

Algebraic Multiplicity

In the context of eigenvalue and eigenvector analysis, algebraic multiplicity plays a crucial role. It is defined as the number of times an eigenvalue appears as a root of the characteristic equation.
For the given matrix, we find that the eigenvalue \(\lambda = 5\) is repeated in the characteristic equation \((\lambda-5)^3 = 0\).
This means that the algebraic multiplicity of the eigenvalue \(\lambda = 5\) in this problem is 3.

The algebraic multiplicity gives us insight into the behavior of a matrix because it indicates how many solutions (eigenvalues) we have in the characteristic equation. However, it does not directly tell us about the number of linearly independent eigenvectors, which is where geometric multiplicity comes in.

Geometric Multiplicity

Geometric multiplicity is a vital concept when analyzing matrices. It tells us about the number of linearly independent eigenvectors associated with a specific eigenvalue.
For our matrix, upon solving the system \((A-5I)\mathbf{x}=\mathbf{0}\), we identified three possible eigenvectors:

• \(\mathbf{x}_1 = \begin{bmatrix}1 \ 0 \ 0\end{bmatrix}\)
• \(\mathbf{x}_2 = \begin{bmatrix}0 \ 1 \ 0\end{bmatrix}\)
• \(\mathbf{x}_3 = \begin{bmatrix}0 \ 0 \ 1\end{bmatrix}\)

All of these are linearly independent, and when associated with the eigenvalue \(\lambda = 5\), they lead us to determine that the geometric multiplicity is 3.

Interestingly, in this exercise, the geometric multiplicity is equal to the algebraic multiplicity, which indicates the matrix has enough eigenvectors to form a basis of the vector space. This means the matrix does have a full set of eigenvectors.

Characteristic Equation

The characteristic equation is foundational in finding eigenvalues of a matrix. It is an essential tool used by setting up the determinant of \((\lambda I - A)\) equal to zero.
Essentially, you are looking at the "companion polynomial" of \(A\).
In this case, for matrix \(A\), the calculation gave us:\[\text{det}\left(\begin{bmatrix}\lambda-5 & 0 & 0 \ 0 & \lambda-5 & 0 \ 0 & 0 & \lambda-5\end{bmatrix}\right) = (\lambda-5)^3\]
Then solving \((\lambda-5)^3 = 0\), we find the eigenvalue \(\lambda = 5\) appears with a multiplicity of 3.

By setting up and solving this equation during the analysis, we not only uncover the eigenvalues but also start to build a picture of their multiplicities (both algebraic and geometric). It sets the stage for understanding other concepts, like the number of eigenvectors and whether a matrix is diagonalizable.