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Chapter 6: Problem 2

Show that the eigenvalues of a triangular matrix are the diagonal elements of the matrix.

Short Answer

Expert verified
In order to show that the eigenvalues of a triangular matrix are the diagonal elements of the matrix, we first calculate the determinant of the matrix (A - 位I) and form the characteristic equation. Since (A - 位I) is a triangular matrix, its determinant is the product of its diagonal elements: \(\det(A - 位I) = (a_{11} - 位)(a_{22} - 位) \cdots (a_{nn} - 位)\). Solving the characteristic equation, the n eigenvalues 位 correspond to the diagonal elements of matrix A: \(位_1 = a_{11}\), \(位_2 = a_{22}\), ... , \(位_n = a_{nn}\). This holds for both upper and lower triangular matrices.

Step by step solution

01

Define a Triangular Matrix

A triangular matrix is a square matrix in which all the entries above (or below) the main diagonal are zero. Here, we can consider an upper triangular matrix as an example, but the same method applies to a lower triangular matrix. Let A be an upper triangular matrix of order n. \[ A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ 0 & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \end{bmatrix} \]
02

Calculate the Characteristic Equation

The eigenvalues of a matrix can be found by solving its characteristic equation. The characteristic equation is given by the determinant of the matrix (A - 位I), where 位 is an eigenvalue and I is the identity matrix of the same order as A. The matrix (A - 位I) is: \[(A - 位I) = \begin{bmatrix} a_{11} - 位 & a_{12} & \cdots & a_{1n} \\ 0 & a_{22} - 位 & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} - 位 \end{bmatrix} \] Now, we will calculate the determinant of (A - 位I).
03

Calculate the Determinant

Since the matrix (A - 位I) is a triangular matrix, its determinant is the product of the diagonal elements: \(\det(A - 位I) = (a_{11} - 位)(a_{22} - 位) \cdots (a_{nn} - 位)\)
04

Solve the Characteristic Equation

Now, we need to solve the characteristic equation for 位, which is: \(\det(A - 位I) = (a_{11} - 位)(a_{22} - 位) \cdots (a_{nn} - 位) = 0\) From this equation, we can observe that the n eigenvalues 位 are the diagonal elements of the matrix A, i.e., \(位_1 = a_{11}\), \(位_2 = a_{22}\), ... , \(位_n = a_{nn}\).
05

Conclusion

We have shown that the eigenvalues of a triangular matrix are the diagonal elements of the matrix by solving its characteristic equation and finding its determinant. This is valid for both upper and lower triangular matrices.

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Most popular questions from this chapter

Given \\[ \mathbf{Y}=c_{1} e^{\lambda_{1} t} \mathbf{x}_{1}+c_{2} e^{\lambda_{2} t} \mathbf{x}_{2}+\cdots+c_{n} e^{\lambda_{n} t} \mathbf{x}_{n} \\] is the solution to the initial value problem: \\[ \mathbf{Y}^{\prime}=A \mathbf{Y}, \quad \mathbf{Y}(0)=\mathbf{Y}_{0} \\] (a) Show that \\[ \mathbf{Y}_{0}=c_{1} \mathbf{x}_{1}+c_{2} \mathbf{x}_{2}+\cdots+c_{n} \mathbf{x}_{n} \\] (b) Let \(X=\left(\mathbf{x}_{1}, \ldots, \mathbf{x}_{n}\right)\) and \(\mathbf{c}=\left(c_{1}, \ldots, c_{n}\right)^{T}\) Assuming that the vectors \(\mathbf{x}_{1}, \ldots, \mathbf{x}_{n}\) are linearly independent, show that \(\mathbf{c}=X^{-1} \mathbf{Y}_{0}\)

It follows from Exercise 14 that, for a diagonalizable matrix, the number of nonzero eigenvalues (counted according to multiplicity) equals the rank of the matrix. Give an example of a defective matrix whose rank is not equal to the number of nonzero eigenvalues.

Show that a nonzero nilpotent matrix is defective.

Prove that if \(A\) is a symmetric matrix with eigenvalues \(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n},\) then the singular values of \(A\) are \(\left|\lambda_{1}\right|,\left|\lambda_{2}\right|, \ldots,\left|\lambda_{n}\right|\)

Let \(p(\lambda)=(-1)^{n}\left(\lambda^{n}-a_{n-1} \lambda^{n-1}-\cdots-a_{1} \lambda-a_{0}\right)\) be a polynomial of degree \(n \geq 1\), and let \\[ C=\left(\begin{array}{ccccc} a_{n-1} & a_{n-2} & \cdots & a_{1} & a_{0} \\ 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & & & \\ 0 & 0 & \cdots & 1 & 0 \end{array}\right) \\] (a) Show that if \(\lambda_{i}\) is a root of \(p(\lambda)=0,\) then \(\lambda_{i}\) is an eigenvalue of \(C\) with eigenvector \(\mathbf{x}=\) \(\left(\lambda_{i}^{n-1}, \lambda_{i}^{n-2}, \ldots, \lambda_{i}, 1\right)^{T}\) (b) Use part (a) to show that if \(p(\lambda)\) has \(n\) distinct roots, then \(p(\lambda)\) is the characteristic polynomial of \(C\). The matrix \(C\) is called the companion matrix of \(p(\lambda)\)
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