Question

Let \(A=\left[\begin{array}{ll}a & b \\ b & c\end{array}\right]\) be a \((2 \times 2)\) real symmetric matrix. In Exercise 28 of Section \(4.4\), it was shown that such a matrix has only real eigenvalues. We now show that \(A\) has a full set of eigenvectors. Note, by Exercise 30 of Section \(4.4\), that if \(A\) has distinct eigenvalues, then \(A\) has a full set of eigenvectors. Thus, the only case to consider is the case where \(A\) has repeated eigenvalues, \(\lambda_{1}=\lambda_{2}\). (a) If \(\lambda_{1}=\lambda_{2}\), show that \(a=c, b=0\), and therefore \(A=a I\). (b) Exhibit a pair of linearly independent eigenvectors in this case.

Short answer

Based on the step-by-step solution: Show that if a given 2x2 real symmetric matrix A has repeated eigenvalues λ₁=λ₂, it will always have a full set of linearly independent eigenvectors. Answer: Yes, a 2x2 real symmetric matrix with repeated eigenvalues will always have a full set of linearly independent eigenvectors. In this case, the matrix A becomes aI, with two linearly independent eigenvectors X1=[1,0] and X2=[0,1].

Step-by-step solution

Show that a=c and b=0 if λ₁=λ₂

To solve this problem, we need to calculate the eigenvalues of matrix A using the characteristic equation det(A-λI)=0. The matrix A-λI will be: Λ\(A-\lambda I=\begin{bmatrix}a-\lambda & b \\ b & c-\lambda\end{bmatrix}\Λ\) Now, we'll compute the determinant: Λ\(\det(A-\lambda I)=(a-\lambda)(c-\lambda)-b^2\Λ\) When λ₁=λ₂, the determinant should give us a single solution. Therefore we can set it to λ: Λ\(\lambda^2 - (a+c)\lambda + (ac-b^2)=0\Λ\) If λ₁=λ₂, the discriminant of this quadratic equation should be equal to zero, so: Λ\((a+c)^2 - 4(ac-b^2)=0\Λ\) Expanding this equation, we have: Λ\(a^2 + 2ac + c^2 - 4ac + 4b^2 = 0\Λ\) Simplifying the equation, we get: Λ\(a^2 - 2ac + c^2 = 4b^2\Λ\) This equation represents the square of the differences a-c and b: Λ\((a-c)^2 = 4b^2\Λ\) Now, since a, b, and c are real and symmetric, it's impossible for a-c to be a complex number. Therefore, (a-c)^2 must be a non-negative real number. The only way for the equation to hold true is if a=c and b=0. Thus, the matrix A becomes: Λ\(A=\begin{bmatrix}a & 0 \\ 0 & a\end{bmatrix}\Λ\) So, A = aI.

Find a pair of linearly independent eigenvectors

Now that we have established that A = aI, let's find the eigenvectors. We start by finding the eigenvectors corresponding to the repeated eigenvalue λ: Λ\((A-\lambda I)X=0\Λ\) We already have A=aI, so substituting: Λ\(\begin{bmatrix}(a-\lambda) & 0 \\ 0 & (a-\lambda)\end{bmatrix} \begin{bmatrix}x_1 \\ x_2\end{bmatrix}= \begin{bmatrix}0 \\ 0\end{bmatrix}\Λ\) So we have the system of equations: Λ\((a-\lambda)x_1=0\Λ\) Λ\((a-\lambda)x_2=0\Λ\) Since λ=a, these equations simplify to: Λ\(0x_1=0\Λ\) Λ\(0x_2=0\Λ\) Any non-zero value of x₁ and x₂ will satisfy these equations. Therefore, we can choose two linearly independent eigenvectors, for example: Λ\(X_1=\begin{bmatrix}1 \\ 0\end{bmatrix}\Λ\) and Λ\(X_2=\begin{bmatrix}0 \\ 1\end{bmatrix}\Λ\) These two eigenvectors are linearly independent and form a full set of eigenvectors for the matrix A when λ₁=λ₂.

Key concepts

Symmetric Matrices

Symmetric matrices are key objects in linear algebra, characterized by their mirror-image symmetry across the diagonal. This means the entries of the matrix are symmetric with respect to the main diagonal, so if you have a matrix A, it satisfies the condition A = A^T, where A^T is the transpose of A.

In the context of eigenvalues and eigenvectors, symmetric matrices possess some very nice properties. One of the most interesting properties is that they only have real eigenvalues, as opposed to complex ones that can arise from non-symmetric matrices. Moreover, symmetric matrices have orthonormal eigenvectors, meaning that the eigenvectors are perpendicular to one another and each have unit length. This property simplifies many problems in mathematics and physics, where symmetric matrices commonly occur. For instance, in problems involving quadratic forms or optimization, symmetric matrices play a significant role.

Characteristic Equation

The characteristic equation is a fundamental concept when exploring the eigenvalues of a matrix. It is formed by taking the determinant of the matrix [A] subtracted by a scalar λ times the identity matrix [I] and setting it equal to zero, written as det(A - λI) = 0. The roots of this polynomial equation in λ are the eigenvalues of the matrix A.

The significance of the characteristic equation lies in its ability to consolidate a matrix's essential behavior into a single, solvable equation. Once the eigenvalues are found using this method, we can then proceed to discover the eigenvectors associated with each eigenvalue by plugging each eigenvalue back into the equation (A - λI)x = 0 and solving for the vector x. The characteristic equation is a cornerstone in understanding the spectral properties of matrices.

Eigenvalues

Eigenvalues are scalars associated with a matrix that provide deep insights into the matrix's properties. When a non-zero vector x is multiplied by the matrix A, the vector changes direction, length, or both. If the vector's direction doesn't change, only the length is scaled by a factor, then this factor is known as an eigenvalue, and the vector is called an eigenvector.

In the original exercise, the characteristic equation leads to finding the eigenvalues of a symmetric matrix. These eigenvalues are always real numbers for symmetric matrices, as opposed to potentially being complex numbers for non-symmetric matrices. This real quality is critical in many applications, such as in stability analysis and solutions to differential equations, where real eigenvalues translate into real and interpretable behaviors of systems.

Linearly Independent Vectors

Linearly independent vectors are the building blocks of vector spaces. A set of vectors is linearly independent if no vector in the set can be expressed as a linear combination of the others. For instance, in two-dimensional space, any pair of non-parallel vectors are linearly independent.

In the context of eigenvectors, a full set of linearly independent eigenvectors for a matrix corresponds to the basis of the space spanned by these vectors. This concept becomes especially important when dealing with repeated eigenvalues because it guarantees that there are enough eigenvectors to span the entire space, allowing for diagonalization of the matrix. As shown in the exercise, even with repeated eigenvalues, as long as we can find linearly independent eigenvectors, the matrix possesses a complete set of eigenvectors to fully describe the space.