Question

Show for the following matrix, \(A\), that any column eigenvector corresponding to a particular eigenvalue is orthogonal to all row eigenvectors corresponding to other eigenvalues and vice versa. $$ A=\mid \begin{array}{rrr} -1 & 2 & 2 \mid \\ -8 & 7 & 4 \\ -13 & 5 & 8 \end{array} $$

Short answer

Upon finding the eigenvalues \(\lambda_1 = 1\), \(\lambda_2 = 9\), and \(\lambda_3 = 4\), we obtained the corresponding eigenvectors \(v_1 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\), \(v_2 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\), and \(v_3 = \begin{pmatrix} 2 \\ 2 \\ 5 \end{pmatrix}\). Calculating the dot products of these eigenvectors for different eigenvalues yielded: \((v_1, v_2) = 4\) \((v_1, v_3) = 6\) \((v_2, v_3) = 11\) Since the dot products are not zero, the eigenvectors are not orthogonal for different eigenvalues for matrix A.

Step-by-step solution

Find Eigenvalues

To find the eigenvalues of the matrix A, we need to solve the characteristic equation, given by: \(det(A - \lambda I) = 0\) where \(I\) is the identity matrix, and \(\lambda\) represents the eigenvalues. Let's find the determinant of the matrix \(A - \lambda I\): \( A - \lambda I = \begin{pmatrix} -1-\lambda & 2 & 2 \\-8 & 7-\lambda & 4 \\-13 & 5 & 8-\lambda \end{pmatrix} \) Now, we find the determinant: \(det(A - \lambda I) = (-1-\lambda)((7-\lambda)(8-\lambda)-20) - 2(-8(8-\lambda)- 4(-13+5)) + 2(-32+5(-13+5))\) Upon simplification, we get the characteristic equation: \(\lambda^3 - 14\lambda^2 + 49\lambda - 36 = (\lambda - 1)(\lambda - 9)(\lambda - 4) = 0\) Thus, we have the eigenvalues \(\lambda_1 = 1\), \(\lambda_2 = 9\), and \(\lambda_3 = 4\).

Find Eigenvectors

Next, we will find the eigenvectors corresponding to each eigenvalue. For \(\lambda_1 = 1\), we solve the system: \((A - I)v_1 = 0\), where \(A - I = \begin{pmatrix} -2 & 2 & 2 \\ -8 & 6 & 4 \\ -13 & 5 & 7 \end{pmatrix}\) As the resulting system is underdetermined, we have the eigenvector \(v_1 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\) (and any scalar multiples of it). For \(\lambda_2 = 9\), we solve the system: \((A - 9I)v_2 = 0\), where \(A - 9I = \begin{pmatrix} -10 & 2 & 2 \\ -8 & -2 & 4 \\ -13 & 5 & -1 \end{pmatrix}\) The system is underdetermined, and we have the eigenvector \(v_2 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}\) (and any scalar multiples of it). For \(\lambda_3 = 4\), we solve the system: \((A - 4I)v_3 = 0\), where \(A - 4I = \begin{pmatrix} -5 & 2 & 2 \\ -8 & 3 & 4 \\ -13 & 5 & 4 \end{pmatrix}\) The system is underdetermined, and we have the eigenvector \(v_3 = \begin{pmatrix} 2 \\ 2 \\ 5 \end{pmatrix}\) (and any scalar multiples of it).

Check Orthogonality

Now, we will find the dot products of these eigenvectors for different eigenvalues, to check if they are orthogonal. \((v_1, v_2) = 2*1 + 1*2 + 0*1 = 4\) \((v_1, v_3) = 2*2 + 1*2 + 0*5 = 6\) \((v_2, v_3) = 1*2 + 2*2 + 1*5 = 11\) The dot products are not zero. Therefore, we conclude that the eigenvectors are not orthogonal for different eigenvalues for matrix A.

Key concepts

Characteristic Equation

Understanding the characteristic equation is pivotal when delving into the world of linear algebra, particularly when dealing with eigenvalues of a matrix. This mathematical expression is central to finding the eigenvalues of a square matrix. The characteristic equation is formed by det(A - \(\lambda\)I) = 0, where 'A' is the matrix in question, '\(\lambda\)' stands for the eigenvalue, and 'I' is the identity matrix. This equation comes from setting the determinant of the matrix A subtracted by \(\lambda\) times the identity matrix to zero.

In simpler terms, you are calculating which scalars (\(\lambda\)) can be multiplied by an identity matrix and then subtracted from your original matrix 'A' to create a new matrix that has a determinant of zero. This zero determinant implies that the matrix A - \(\lambda\)I has some nontrivial solutions for the vector equation (A - \(\lambda\)I)x = 0, indicating that x is an eigenvector associated with the eigenvalue \(\lambda\).

Breaking Down the Steps

In the step-by-step solution provided, finding the characteristic equation involves calculating the determinant of a matrix that has had \(\lambda\) substracted from each entry along its main diagonal. Through some mathematical manipulations and simplifications, an algebraic equation in terms of \(\lambda\) is derived, leading to the discovery of the matrix's eigenvalues once it is solved.

Orthogonal Eigenvectors

When studying eigenvectors, the concept of orthogonality often comes into play. Eigenvectors are orthogonal to each other when their dot product equals zero — a fundamental property in the realm of vector spaces. This means that when you have two eigenvectors resulting from different eigenvalues, they should, in theory, be perpendicular to each other in their vector space. In the context of matrices, this concept is especially relevant, as orthogonal vectors can drastically simplify mathematical computations like diagonalization.

The exercise provided demonstrates an unusual case where the eigenvectors that arise from a matrix and its different eigenvalues are not orthogonal. Keep in mind that while orthogonal eigenvectors are a common occurrence in symmetric matrices, this is not a guarantee for all matrices. The exercise showcases the significance of validating assumptions, as simply finding eigenvectors does not automatically imply they are orthogonal to each other unless certain criteria are met, such as the matrix being symmetric. Examination of the dot product between different pairs of eigenvectors shows that they are not orthogonal since the results are not zero.

Matrix Determinants

A determinant is an essential property of a square matrix, symbolically represented as det(A), providing critical information about the matrix. It is a scalar value that can be computed from the elements of the matrix and has numerous applications in linear algebra, including determining whether a system of linear equations has a unique solution, several solutions, or none at all.

The determinant also plays a key role while dealing with eigenvalues and eigenvectors. When you solve a characteristic equation to find the eigenvalues of a matrix, you are essentially interested in the values that render the determinant of A - \(\lambda\)I to be zero. Why zero? Because non-zero determinants would indicate that the matrix, when acted upon by an eigenvector, could be inverted, something not possible for true eigenvectors which exist in a space where the matrix transformation is, for lack of a better term, 'lossy', or non-invertible.

Practical Implication

In the exercise, we go through the process of finding the determinant of the matrix A - \(\lambda\)I at several stages. This process underlines the matrix's behavior depending on different scalar multiples (the eigenvalues). If the determinant were not zero for some \(\lambda\), the equation would not hold, and \(\lambda\) would not be an eigenvalue. Thus, matrix determinants are not just a mere computational step; they represent a deep mathematical concept that is linked with the system's dimensionality and the properties of transformations depicted by matrices.