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Find the eigenvalues of the given matrix. For each eigenvalue, give an eigenvector. $$\left[\begin{array}{ccc}2 & -1 & 1 \\ 0 & 3 & 6 \\ 0 & 0 & 7\end{array}\right]$$

Short Answer

Expert verified
Eigenvalues are 2, 3, 7. Corresponding eigenvectors are [0, 1, 1], [1, 1, 2], [1, -5, 4].

Step by step solution

01

Understand the Problem

You're asked to find the eigenvalues and corresponding eigenvectors of the matrix:\[\begin{bmatrix} 2 & -1 & 1 \ 0 & 3 & 6 \ 0 & 0 & 7 \end{bmatrix}\]Eigenvalues are values of \( \lambda \) such that the matrix \( A - \lambda I \) has a determinant of zero, where \( I \) is the identity matrix of the same order as \( A \).
02

Find the Characteristic Polynomial

To find eigenvalues, compute the determinant of \( A - \lambda I \) and set it equal to zero. The matrix \( A - \lambda I \) for the given matrix is:\[\begin{bmatrix} 2-\lambda & -1 & 1 \ 0 & 3-\lambda & 6 \ 0 & 0 & 7-\lambda \end{bmatrix}\]The determinant is calculated by multiplying the diagonal elements since the matrix is upper triangular:\[(2-\lambda)(3-\lambda)(7-\lambda)\]Set this equal to zero to get the characteristic equation:\[(2-\lambda)(3-\lambda)(7-\lambda) = 0\]
03

Solve for Eigenvalues

Solve the characteristic equation:\[\begin{align*}2 - \lambda &= 0, \3 - \lambda &= 0, \7 - \lambda &= 0\end{align*}\]Thus, the eigenvalues are:\[\lambda_1 = 2, \quad \lambda_2 = 3, \quad \lambda_3 = 7\]
04

Find Eigenvectors for λ = 2

For \( \lambda = 2 \), substitute into \( A - \lambda I \):\[\begin{bmatrix} 0 & -1 & 1 \ 0 & 1 & 6 \ 0 & 0 & 5 \end{bmatrix}\]The first row gives \(-y + z = 0\). Set \( z = t \), then \( y = t \), and \( x \) can be any value; choose \( x = 0 \) for simplicity. An eigenvector is \( \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} \).
05

Find Eigenvectors for λ = 3

For \( \lambda = 3 \), substitute into \( A - \lambda I \):\[\begin{bmatrix} -1 & -1 & 1 \ 0 & 0 & 6 \ 0 & 0 & 4 \end{bmatrix}\]The first row gives \(-x - y + z = 0\). Set \( x = y = t \), then \( z = 2t \). An eigenvector is \( \begin{bmatrix} 1 \ 1 \ 2 \end{bmatrix} \).
06

Find Eigenvectors for λ = 7

For \( \lambda = 7 \), substitute into \( A - \lambda I \):\[\begin{bmatrix} -5 & -1 & 1 \ 0 & -4 & 6 \ 0 & 0 & 0 \end{bmatrix}\]The third row is all zeros, indicating infinite solutions. Solve the system:- From \(-5x - y + z = 0\): Set \( x = t \), then \( y = -5t \), \( z = 4t \).An eigenvector is \( \begin{bmatrix} 1 \ -5 \ 4 \end{bmatrix} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Characteristic Polynomial
The characteristic polynomial is a key concept in determining the eigenvalues of a matrix. To find the eigenvalues, we need the matrix's characteristic polynomial, which is derived from the equation \( ext{det}(A - \lambda I) = 0 \). Here, \( A \) is our original matrix and \( \lambda \) represents the eigenvalues we are looking for.

For an upper triangular matrix, like the one in the given problem, finding the characteristic polynomial is straightforward. We subtract \( \lambda \) from each of the diagonal elements of the matrix and calculate the determinant of this matrix, called \( A - \lambda I \).

In this exercise, the matrix \( A - \lambda I \) becomes upper triangular as well:
  • \( (2-\lambda) \)
  • \( (3-\lambda) \)
  • \( (7-\lambda) \)
The product of these diagonal elements gives the characteristic polynomial:
\[(2-\lambda)(3-\lambda)(7-\lambda) = 0\]
This equation is then solved to find the eigenvalues.
Matrix Algebra
Matrix algebra is a set of mathematical rules that allow us to perform operations on matrices, such as addition, multiplication, and finding determinants. Understanding these operations is essential when working with concepts like eigenvalues and eigenvectors.

In the context of the exercise, we start with a 3x3 matrix. To solve for eigenvalues, we modify this matrix by subtracting \( \lambda \times I \) from it, where \( I \) is the identity matrix of the same order. This operation is part of matrix algebra and transforms the matrix such that the difference keeps its structure but shifts the diagonal elements.

These operations ensure that we're able to maintain the properties of a matrix while allowing us to explore its deeper characteristics, such as eigenvalues, through its algebraic structures. Remember, understanding these operations is crucial for simplifying and solving matrix-related problems.
Upper Triangular Matrix
An upper triangular matrix is a special type of square matrix where all the elements below the main diagonal are zero. These matrices have several interesting properties that simplify computations, one of which is the ease of determining eigenvalues.

For an upper triangular matrix like the one in this one:
  • The eigenvalues can be directly read from the diagonal elements.
  • It is computationally less intensive to calculate the determinant for such matrices, as the determinant is simply the product of the diagonal elements.
In this exercise, the matrix is already in an upper triangular form, which simplifies finding its determinant and hence obtaining the characteristic polynomial. This makes it straightforward to calculate the eigenvalues by observing that the diagonal elements \((2 - \lambda), (3 - \lambda), (7 - \lambda)\) solve the characteristic equation directly.
Determinant
The determinant is a special number that can be calculated from a square matrix. It offers insights into various properties of the matrix, such as whether a matrix is invertible, the volume change during linear transformations, and crucially, it helps in finding eigenvalues.

When finding the eigenvalues of a matrix, we compute the determinant of \( A - \lambda I \), and set this determinant to zero. For our matrix which is upper triangular, the determinant is
\[(2-\lambda)(3-\lambda)(7-\lambda)\]
This form easily allows us to solve the characteristic equation and find eigenvalues because each factor corresponds to a shift of a single diagonal element by \( \lambda \).

Understanding how to compute and interpret the determinant is an essential skill in matrix algebra as it can also reveal other properties such as invertibility and the geometric interpretations of transformations performed by the matrix.

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