Solving Matrix A: Characteristic Equation and Eigenvectors

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• wefweff
In summary, the matrix A is given by $A= \begin{bmatrix}7 & 0 & -3 \\ -9 & -2 & 3 \\ 10 & 0 & -8 \end{bmatrix}$ and the characteristic equation is $|A- \lambda I|= (-2- \lambda)(-2- \lambda)(1- \lambda)$, which is a cubic equation. The eigenvalues of A are -2, -2, and 1, and the associated eigenvectors can be found by solving the equation $(A- \lambda I)v= 0$.
wefweff
good evening everyone!
Decided to solve the problems from last year's exams. I came across this example. Honestly, I didn't understand it. Who can help a young student? :)
Find characteristic equation of the matrix A in the form of the polynomial of degree of 3 (you do not need to find eigenvalues) and associated eigenvectors of the matrix. Eigenvalues of the matrix: -2, -2, 1.
А= 7 0 -3
-9 -2 3
18 0 -8

$\lambda$ is an eigenvalue of matrix A if there exist some non-zero vector, v, such that $Av= \lambda v$. That is the same as $Av-\lambda v= 0$ or $(A- \lambda I)v= 0$. v= 0 is obviously a solution. In order that there be another solution $A- \lambda$ must not have an inverse. That requires that the determinant or $A- \lambda$ be 0.

Here $A= \begin{bmatrix}7 & 0 & -3 \\ -9 & -2 & 3 \\ 10 & 0 & -8 \end{bmatrix}$ so $A- \lambda I=\begin{bmatrix}7- \lambda & 0 & 3 \\ -9 & -2- \lambda & 3 \\ 18 & 0 & -8- \lambda \end{bmatrix}$.

The determinant is $\left|\begin{array}{ccc}7-\lambda & 0 & -3 \\ -9 & -2-\lambda & 3 \\ 18 & 0 & -8-\lambda \end{array}\right|$ so the characteristic equation is $|A- \lambda I|=\left|\begin{array}{ccc}7-\lambda & 0 & -3 \\ -9 & -2-\lambda & 3 \\ 18 & 0 & -8-\lambda \end{array}\right|= 0$. Since this is a 3 by 3 matrix, that will be a cubic equation.

Last edited:
To calculate $\left|\begin{array}{ccc} 7- \lambda & 0 & -3 \\ -9 & -2- \lambda & 3 \\ 18 & 0 & -8- \lambda \end{array}\right|$ expand on the middle column: $(-2- \lambda)\left|\begin{array}{cc} 7- \lambda & -3 \\ 18 & -8- \lambda \end{array}\right|= (-2- \lambda)((7- \lambda)(-8- \lambda)+ 54)= (-2- \lambda)(-56+ \lambda+ \lambda^2+ 54)= (-2- \lambda)(\lambda^2+ \lambda- 2)= (-2- \lambda)(-2- \lambda)(1- \lambda)$.

1. What is a characteristic equation?

A characteristic equation is a polynomial equation that is used to find the eigenvalues of a matrix. It is formed by setting the determinant of the matrix minus a scalar multiple of the identity matrix equal to zero.

2. How do you solve a characteristic equation?

To solve a characteristic equation, you first need to find the determinant of the matrix and set it equal to zero. Then, you can use techniques such as factoring or the quadratic formula to find the roots of the equation, which will be the eigenvalues of the matrix.

3. What are eigenvectors?

Eigenvectors are special vectors that are associated with eigenvalues of a matrix. They represent the directions in which the matrix only stretches or compresses, without changing direction. They are used to understand the behavior of a matrix in certain transformations.

4. How do you find eigenvectors?

To find the eigenvectors of a matrix, you first need to find the eigenvalues using the characteristic equation. Then, for each eigenvalue, you can solve the system of equations (A - λI)x = 0, where A is the matrix, λ is the eigenvalue, and x is the eigenvector. The solutions to this system will be the eigenvectors.

5. Why are characteristic equations and eigenvectors important in matrix operations?

Characteristic equations and eigenvectors are important in matrix operations because they provide valuable information about the behavior of a matrix. They can help us understand how a matrix will transform a vector, and they are also used in many applications such as data analysis, computer graphics, and physics.

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