Problem 39 - Differential Equations and Linear Algebra

Question

Prove that the eigenvalues of an upper (or lower) triangular matrix are just the diagonal elements of the matrix.

Step-by-Step Solution

Answer

In conclusion, the eigenvalues \(\lambda_i\) of an upper (or lower) triangular matrix are just the diagonal elements of the matrix. For an upper triangular matrix A, the eigenvalues are \(\lambda_1 = a_{11}\), \(\lambda_2 = a_{22}\), and \(\lambda_3 = a_{33}\). Similarly, for a lower triangular matrix B, the eigenvalues are \(\lambda_1 = b_{11}\), \(\lambda_2 = b_{22}\), and \(\lambda_3 = b_{33}\). This is proven by calculating the characteristic equation and finding that the determinant of the matrix subtraction (\(M - \lambda I\)) is equal to zero, which results in the eigenvalues being equal to the diagonal elements.

Step 1: Definition of Upper and Lower Triangular Matrix

An upper triangular matrix is a square matrix with all the elements below its main diagonal (bottom-left to top-right diagonal) equal to zero. Similarly, a lower triangular matrix is a square matrix with all the elements above its main diagonal equal to zero. For example, consider the following upper triangular matrix A: \[A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a_{22} & a_{23} \\ 0 & 0 & a_{33} \end{pmatrix}\] And the following lower triangular matrix B: \[B =\begin{pmatrix} b_{11} & 0 & 0 \\ b_{21} & b_{22} & 0 \\ b_{31} & b_{32} & b_{33} \end{pmatrix}\]

Step 2: Definition of Eigenvalue and Eigenvector

An eigenvalue \(\lambda\) and the corresponding eigenvector \(x \neq 0\) of a square matrix \(M\) satisfy the following equation: \[Mx = \lambda x\] Here, \(M\) is the given matrix, \(x\) is the eigenvector, and \(\lambda\) is the eigenvalue.

Step 3: Calculate the Characteristic Equation

To find the eigenvalues, we first need to compute the determinant of the matrix subtraction (\(M - \lambda I\)), where \(I\) is the identity matrix of the same size as \(M\): \[\det(M - \lambda I) = 0\]

Step 4: Calculate the Eigenvalues for Upper Triangular Matrix

For an upper triangular matrix A, the characteristic equation is: \[\det (A-\lambda I) = \det \begin{pmatrix} a_{11} - \lambda & a_{12} & a_{13} \\ 0 & a_{22} - \lambda & a_{23} \\ 0 & 0 & a_{33} - \lambda \end{pmatrix} = 0\] The determinant of an upper (and lower) triangular matrix is the product of the diagonal elements. Therefore: \[(a_{11} - \lambda)(a_{22} - \lambda)(a_{33} - \lambda) = 0\] The eigenvalues are: \[\lambda_1 = a_{11}\] \[\lambda_2 = a_{22}\] \[\lambda_3 = a_{33}\]

Step 5: Calculate the Eigenvalues for Lower Triangular Matrix

For a lower triangular matrix B, the characteristic equation is: \[\det (B-\lambda I) = \det \begin{pmatrix} b_{11} - \lambda & 0 & 0 \\ b_{21} & b_{22} - \lambda & 0 \\ b_{31} & b_{32} & b_{33} - \lambda \end{pmatrix} = 0\] Similar to the upper triangular case, the determinant of a lower triangular matrix is the product of the diagonal elements: \[(b_{11} - \lambda)(b_{22} - \lambda)(b_{33} - \lambda) = 0\] The eigenvalues are: \[\lambda_1 = b_{11}\] \[\lambda_2 = b_{22}\] \[\lambda_3 = b_{33}\] In conclusion, the eigenvalues of an upper (or lower) triangular matrix are just the diagonal elements of the matrix.

Key Concepts

Upper Triangular Matrix

In linear algebra, an upper triangular matrix is a special kind of square matrix where all the elements below the main diagonal are zero. If you imagine a line from the top-left corner to the bottom-right corner (the main diagonal), every number below that line is a zero.

Here's a basic example with a 3x3 matrix:

• \[A = \begin{pmatrix}a_{11} & a_{12} & a_{13} \0 & a_{22} & a_{23} \0 & 0 & a_{33} \end{pmatrix}\]

This property makes computations like finding the determinant a lot easier, because you only have to multiply the diagonal elements together. Speaking of which, this feature actually leads us straight to the handy fact that the eigenvalues of such a matrix are simply the values on the main diagonal, which brings us to the characteristic equation.

Characteristic Equation

The characteristic equation is a must-know when you're looking to find out an important feature of matrices: their eigenvalues. It's a bit like a secret code that when solved, gives you the eigenvalues.

Let's break it down: To get the characteristic equation, you subtract lambda (\(\lambda\)), our placeholder for a potential eigenvalue, times the identity matrix (that's a matrix with 1s on the diagonal and 0s everywhere else) from your matrix. Then you work out the determinant and set that equal to zero:

• \[\det(M - \lambda I) = 0\]

For an upper triangular matrix, this equation simplifies dramatically because the determinant is just the product of the diagonal elements once the lambda values have been subtracted. This simplification is what lets you see that the eigenvalues are the diagonal elements themselves. It's a beautiful shortcut that saves a lot of time and headaches!

Eigenvector

Now, if the eigenvalues are the soul of the matrix, think of the eigenvector as its shadow; it gives you directional information about the matrix. An eigenvector is a non-zero vector that changes at most by a scalar factor when that linear transformation is applied.

In simpler words, if a matrix were a transformation that squishes, rotates, or flips space in any way, the eigenvector points in a direction that stays consistent after the transformation, just stretched or squished. The how much it gets stretched or squished is where the eigenvalue comes into play:

• \[Mx = \lambda x\]

In this equation, \(M\) is our matrix, \(x\) is the eigenvector, and \(\lambda\) is the eigenvalue acting as the stretch/squish factor. For triangular matrices, while eigenvalues are easy to spot, finding the eigenvectors can be a bit trickier, involving solving systems of equations. Knowing the eigenvalues from the get-go is super helpful because it's the starting point in unraveling the secrets of the eigenvectors. Each eigenvalue will have a corresponding eigenvector, and together they offer a deep insight into the nature and behavior of a matrix.