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?\(A=\left[\begin{array}{lll}5 & 0 & 1 \\ 0 & 5 & 0 \\ 0 & 0 & 5\end{array}\right]\)

Short answer

Based on the given matrix \(A\), we found that the only eigenvalue is \(\lambda_1=5\) with an algebraic multiplicity of 3 and a geometric multiplicity of 2. The corresponding eigenvectors for this eigenvalue are \(v_1=[1,0,0]^T\) and \(v_2=[0,1,0]^T\). However, the matrix \(A\) does not have a full set of eigenvectors.

Step-by-step solution

Find the characteristic equation of A

First, let's find the characteristic equation of matrix \(A\). The characteristic equation can be found by calculating the determinant of \((A - \lambda I)\), where \(\lambda\) is an eigenvalue, and \(I\) is the identity matrix of the same order as \(A\): \((A - \lambda I) = \begin{bmatrix}5-\lambda & 0 & 1\\0 & 5-\lambda & 0\\0 & 0 & 5-\lambda\end{bmatrix}\). The determinant of \((A - \lambda I)\) is then: \(|(A - \lambda I)|=(5-\lambda)^3\).

Calculate the eigenvalues

By solving the characteristic equation we obtained in Step 1, we can determine the eigenvalues: \((5-\lambda)^3=0\). It's clear that we have only one eigenvalue here: \(\lambda_1=5\).

Find the eigenvectors corresponding to each eigenvalue

Now, let's find the eigenvectors corresponding to the eigenvalue \(\lambda_1=5\). To do this, we'll plug this eigenvalue into the equation \((A-\lambda I)X = 0\), where \(X\) is the eigenvector we are trying to find: \((A-\lambda_1 I) = \begin{bmatrix}0 & 0 & 1\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}X=0\). The eigenvectors are the null space of \((A-\lambda_1 I)\). From the above matrix, we can see that there are two free variables (for the first and second components of X), so we can set them arbitrarily: \(x_1=a\), \(x_2=b\). The last row of the matrix is just the equation \(0 = 0\), which doesn't help us in finding the eigenvectors. The first row of the matrix gives us only one useful equation: \(1x_3=0 \Rightarrow x_3=0\). Thus, the eigenvectors for \(\lambda_1=5\) are given by \(X=[a,b,0]^T\), where \(a\) and \(b\) are arbitrary constants. We can choose \(a=1\) and \(b=0\) for eigenvector \(v_1=[1,0,0]^T\) and \(a=0\) and \(b=1\) for eigenvector \(v_2=[0,1,0]^T\).

Determine the geometric and algebraic multiplicities

The geometric multiplicity of an eigenvalue is the number of linearly independent eigenvectors corresponding to that eigenvalue. In our case, for \(\lambda_1=5\), we have two linearly independent eigenvectors \(v_1\) and \(v_2\). Therefore, the geometric multiplicity of \(\lambda_1=5\) is 2. The algebraic multiplicity of an eigenvalue is the number of times it appears as a root of the characteristic equation. For \(\lambda_1=5\), its algebraic multiplicity is 3, since it is the root of the equation \((5-\lambda)^3=0\) with a multiplicity of 3.

Check if A has a full set of eigenvectors

Matrix \(A\) has a full set of eigenvectors if the sum of the geometric multiplicities of its eigenvalues is equal to the size of the matrix, which is 3 in our case. The geometric multiplicity of \(\lambda_1=5\) is 2. Therefore, the matrix \(A\) does not have a full set of eigenvectors. In summary, the eigenvalue of the matrix \(A\) is \(\lambda_1=5\) with an algebraic multiplicity of 3 and a geometric multiplicity of 2. The corresponding eigenvectors are \(v_1=[1,0,0]^T\) and \(v_2=[0,1,0]^T\), but the matrix \(A\) does not have a full set of eigenvectors.

Key concepts

Characteristic Equation

Understanding the characteristic equation is vital when studying eigenvalues and eigenvectors. It is the foundation on which we can determine the special values known as eigenvalues for a given square matrix. The characteristic equation is formulated by taking the determinant of the matrix subtracted by an eigenvalue scalar multiple of the identity matrix, making it \(A - \lambda I\). Here, \(\lambda\) represents an eigenvalue, and \(I\) is the identity matrix with the same dimensions as matrix \(A\).

For instance, when working with the matrix \(A\) provided in the exercise, the characteristic equation becomes \(\operatorname{det}(A - \lambda I) = (5-\lambda)^3\), which essentially expresses the values of \(\lambda\) that satisfy the equation \(\operatorname{det}(A - \lambda I) = 0\). Solving this gives us the eigenvalues, and in this specific problem, we find there is only one eigenvalue, \(\lambda_1=5\).

Geometric Multiplicity

Delving into the geometric multiplicity of an eigenvalue, it sheds light on the dimension of the eigenspace associated with that eigenvalue. In simpler terms, it's the total number of linearly independent eigenvectors that correspond to it. As such, it gives us insight into the structural framework of a matrix through its eigenvectors.

In the context of the matrix \(A\) from the exercise, we calculate the geometric multiplicity for the eigenvalue \(\lambda_1=5\) by determining the number of linearly independent solutions to the equation \( (A - \lambda_1 I)X = 0\). In this case, with 2 linearly independent eigenvectors found, the geometric multiplicity for \(\lambda_1=5\) turns out to be 2. This falls short of the matrix's size, indicating that \(A\) lacks a complete set of eigenvectors and, thus, may not be diagonalizable by a similarity transformation.

Algebraic Multiplicity

The algebraic multiplicity distinctly differs from its geometric counterpart by counting how many times an eigenvalue appears as a root in the characteristic equation. It's a numerical value that indicates the frequency of the eigenvalue in question within the context of the characteristic polynomial.

In reference to the provided matrix \(A\), the algebraic multiplicity of the eigenvalue \(\lambda_1=5\) is deduced directly from how the characteristic equation \( (5-\lambda)^3=0\) breaks down. Since \(\lambda_1=5\) is the only root and it occurs three times, its algebraic multiplicity is 3. Understanding that the algebraic multiplicity is at least as great as geometric multiplicity helps in grasping the deeper implications of the structure and properties of a matrix, such as its definiteness or its potential to be diagonalized.