Eigenvalues, Eigenvectors, And Characteristic Equation Of Matrix A

Oct 19, 2025 by SLV Team 67 views

Finding Eigenvalues, Eigenvectors, and the Characteristic Equation of Matrix A

Hey guys! Today, we're diving deep into linear algebra to tackle a classic problem: finding the characteristic equation, eigenvalues, and eigenvectors of a given matrix. This stuff might seem a bit abstract at first, but trust me, it's super useful in many areas, from physics and engineering to computer graphics and data science. We will solve it step by step, with a detailed explanation to make it as clear as possible. So, let's jump right in!

The Matrix We're Working With

Let's start by defining the matrix we'll be working with. This is the matrix A:

A = | 2 -1 1 |
    | -1 2 -1 |
    | 1 -1 2 |

Our goal is to find:

The characteristic equation of A.
The eigenvalues of A.
The eigenvectors corresponding to each eigenvalue.

What are Eigenvalues and Eigenvectors?

Before we dive into the calculations, let's quickly recap what eigenvalues and eigenvectors actually are. Eigenvectors are special vectors that, when multiplied by a matrix, don't change direction; they only get scaled. The factor by which they are scaled is called the eigenvalue. In other words, if v is an eigenvector of matrix A and λ is its corresponding eigenvalue, then:

A**v** = λ**v**

The characteristic equation is a polynomial equation derived from the matrix that helps us find the eigenvalues. Alright, with the basics covered, let's get to work!

Step 1: Finding the Characteristic Equation

The characteristic equation is found using the determinant of (A - λI), where λ (lambda) represents the eigenvalues and I is the identity matrix. The equation is given by:

det(A - λI) = 0

First, we need to find (A - λI). If A is our matrix

A = | 2 -1 1 |
    | -1 2 -1 |
    | 1 -1 2 |

and I is the 3x3 identity matrix

I = | 1 0 0 |
    | 0 1 0 |
    | 0 0 1 |

Then λI is:

λI = | λ 0 0 |
     | 0 λ 0 |
     | 0 0 λ |

Now, subtract λI from A:

A - λI = | 2-λ  -1    1    |
        | -1   2-λ  -1    |
        | 1    -1   2-λ  |

Next, we need to calculate the determinant of this matrix. Remember, for a 3x3 matrix, the determinant is calculated as follows:

det(M) = a(ei - fh) - b(di - fg) + c(dh - eg)

Where M is the matrix:

M = | a b c |
    | d e f |
    | g h i |

Applying this to our (A - λI) matrix:

det(A - λI) = (2-λ)[(2-λ)(2-λ) - (-1)(-1)] - (-1)[(-1)(2-λ) - (-1)(1)] + 1[(-1)(-1) - (2-λ)(1)]

Let's break this down step by step:

(2-λ)[(2-λ)(2-λ) - (-1)(-1)]
- First, expand (2-λ)(2-λ) which gives us 4 - 4λ + λ²
- Then, subtract (-1)(-1) = 1 from it: 4 - 4λ + λ² - 1 = λ² - 4λ + 3
- Multiply (2-λ) by (λ² - 4λ + 3): (2-λ)(λ² - 4λ + 3) = 2λ² - 8λ + 6 - λ³ + 4λ² - 3λ = -λ³ + 6λ² - 11λ + 6
-(-1)[(-1)(2-λ) - (-1)(1)]
- Simplify (-1)(2-λ) to -2 + λ
- Simplify (-1)(1) to -1, and then subtract it (so it becomes +1): -2 + λ + 1 = λ - 1
- Multiply by -(-1) which is just 1: 1 * (λ - 1) = λ - 1
1[(-1)(-1) - (2-λ)(1)]
- Simplify (-1)(-1) to 1
- Simplify (2-λ)(1) to 2 - λ
- Subtract: 1 - (2 - λ) = 1 - 2 + λ = λ - 1
- Multiply by 1: 1 * (λ - 1) = λ - 1

Now, let's add these parts together:

(-λ³ + 6λ² - 11λ + 6) + (λ - 1) + (λ - 1) = -λ³ + 6λ² - 9λ + 4

Setting this equal to zero gives us the characteristic equation:

-λ³ + 6λ² - 9λ + 4 = 0

To make it look a bit nicer, we can multiply the entire equation by -1:

λ³ - 6λ² + 9λ - 4 = 0

Great! We've found the characteristic equation.

Step 2: Finding the Eigenvalues

Now that we have the characteristic equation, we need to find its roots. These roots are the eigenvalues of matrix A. Our characteristic equation is a cubic equation:

λ³ - 6λ² + 9λ - 4 = 0

Solving cubic equations can be tricky, but often we can find at least one root by trying out some simple integer values (like 1, -1, 2, -2, etc.). This is based on the Rational Root Theorem. Let's try λ = 1:

(1)³ - 6(1)² + 9(1) - 4 = 1 - 6 + 9 - 4 = 0

Bingo! λ = 1 is a root. This means (λ - 1) is a factor of our cubic equation. We can now perform polynomial division (or synthetic division) to find the other factor. Let's perform polynomial division:

        λ²  - 5λ  + 4
λ - 1 | λ³ - 6λ² + 9λ - 4
      - (λ³ - λ²)
      -------------
           -5λ² + 9λ
           -(-5λ² + 5λ)
           -------------
                4λ - 4
                -(4λ - 4)
                ---------
                     0

So, dividing λ³ - 6λ² + 9λ - 4 by (λ - 1) gives us λ² - 5λ + 4. Now we have factored our cubic equation into:

(λ - 1)(λ² - 5λ + 4) = 0

The quadratic factor can be further factored easily. We're looking for two numbers that multiply to 4 and add up to -5. Those numbers are -1 and -4. So,

λ² - 5λ + 4 = (λ - 1)(λ - 4)

Thus, our fully factored characteristic equation is:

(λ - 1)(λ - 1)(λ - 4) = 0

This gives us the eigenvalues:

λ₁ = 1 (with a multiplicity of 2)
λ₂ = 4

Excellent! We've found the eigenvalues of matrix A.

Step 3: Finding the Eigenvectors

Now for the final step: finding the eigenvectors corresponding to each eigenvalue. For each eigenvalue, we need to solve the equation:

(A - λI)v = 0

where v is the eigenvector we're trying to find.

Eigenvector for λ₁ = 1

First, let's plug λ₁ = 1 into (A - λI):

A - 1I = | 2-1  -1    1    | = | 1 -1  1 |
         | -1   2-1  -1    |   | -1 1 -1 |
         | 1    -1   2-1  |   | 1 -1  1 |

So, we need to solve the system:

| 1 -1  1 | | x | = | 0 |
| -1 1 -1 | | y | = | 0 |
| 1 -1  1 | | z | = | 0 |

This gives us the following equations:

x - y + z = 0
-x + y - z = 0
x - y + z = 0

Notice that all three equations are essentially the same. We have one independent equation with three variables. Let's express x in terms of y and z:

x = y - z

Now, let y = a and z = b, where a and b are arbitrary constants. Then, x = a - b. So our eigenvector v₁ can be written as:

v₁ = | a - b |
     | a     |
     | b     |

We can rewrite this as a linear combination of two vectors:

v₁ = a | 1 | + b | -1 |
     | 1 |   | 0  |
     | 0 |   | 1  |

This means we have two linearly independent eigenvectors for λ₁ = 1:

v₁₁ = | 1 |
      | 1 |
      | 0 |

and

v₁₂ = | -1 |
      | 0  |
      | 1  |

These two vectors span the eigenspace corresponding to λ₁ = 1. It makes sense that we have two eigenvectors for this eigenvalue since it had a multiplicity of 2.

Eigenvector for λ₂ = 4

Now, let's find the eigenvector for λ₂ = 4. Plug λ₂ = 4 into (A - λI):

A - 4I = | 2-4  -1    1    | = | -2 -1  1 |
         | -1   2-4  -1    |   | -1 -2 -1 |
         | 1    -1   2-4  |   | 1 -1  -2 |

We need to solve the system:

| -2 -1  1 | | x | = | 0 |
| -1 -2 -1 | | y | = | 0 |
| 1 -1  -2 | | z | = | 0 |

This gives us the following equations:

-2x - y + z = 0
-x - 2y - z = 0
x - y - 2z = 0

Let's solve this system of equations. From the first equation, we can express z as:

z = 2x + y

Substitute this into the second equation:

-x - 2y - (2x + y) = 0
-3x - 3y = 0
x = -y

Now substitute x = -y into the expression for z:

z = 2(-y) + y
z = -y

So, we have x = -y and z = -y. Let y = c, where c is an arbitrary constant. Then x = -c and z = -c. Our eigenvector v₂ can be written as:

v₂ = | -c |
     | c  |
     | -c |

We can factor out c:

v₂ = c | -1 |
     | 1  |
     | -1 |

So, we have one eigenvector for λ₂ = 4:

v₂ = | -1 |
     | 1  |
     | -1 |

And there you have it! We have successfully found the eigenvector corresponding to λ₂ = 4.

Conclusion

Okay, guys, that was quite the journey! We've successfully found the characteristic equation, eigenvalues, and eigenvectors of the given matrix A. To recap:

The characteristic equation is: λ³ - 6λ² + 9λ - 4 = 0
The eigenvalues are: λ₁ = 1 (multiplicity 2) and λ₂ = 4

The eigenvectors are:

For λ₁ = 1:

v₁₁ = | 1 | and v₁₂ = | -1 |
     | 1 |         | 0  |
     | 0 |         | 1  |

For λ₂ = 4:
```
v₂ = | -1 |
     | 1  |
     | -1 |
```

I hope this step-by-step guide has made the process clearer for you. Remember, practice makes perfect, so try tackling similar problems to solidify your understanding. Until next time, happy calculating!