How to Find Eigenvalues and Eigenvectors of a 3×3 Matrix
Master linear algebra with this complete walkthrough of finding eigenvalues and eigenvectors of a 3×3 matrix. Learn the characteristic polynomial method, how to solve cubic equations, and find eigenvectors through row reduction.
📹 Video Walkthrough: This Exact Problem
Watch the full solution for finding eigenvalues and eigenvectors of a 3×3 matrix step-by-step using the characteristic polynomial method.
The Problem
Find all eigenvalues and their corresponding eigenvectors for the matrix:
| 2 | -1 | 0 |
| -1 | 2 | -1 |
| 0 | -1 | 2 |
This is a classic symmetric matrix problem that appears frequently in linear algebra courses. The matrix has a special structure (tridiagonal and symmetric) that makes it particularly interesting. We'll solve this completely, step by step.
💡 Why this problem matters: Eigenvalues and eigenvectors are fundamental to understanding linear transformations, diagonalization, and many applications in physics, engineering, and data science.
This problem tests your ability to: (1) compute determinants, (2) solve cubic equations, (3) perform row reduction, and (4) understand the geometric meaning of eigenvectors.
Understanding Eigenvalues and Eigenvectors
Before we dive into the solution, let's make sure we understand what we're looking for.
Definition:
For a square matrix A, an eigenvalue λ is a scalar such that there exists a nonzero vector v (called an eigenvector) satisfying:
Av = λv
This means that when A acts on the eigenvector v, it only scales it by the factor λ, without changing its direction.
Key Insight:
The equation Av = λv can be rewritten as:
(A - λI)v = 0
where I is the identity matrix. For this to have a nonzero solution, the matrix (A - λI) must be singular (non-invertible), which means:
det(A - λI) = 0
This equation is called the characteristic equation, and the polynomial det(A - λI) is called the characteristic polynomial.
The trick here is recognizing that finding eigenvalues is really about finding the roots of a polynomial. For a 3×3 matrix, this means solving a cubic equation. Once you have the eigenvalues, finding eigenvectors is a matter of solving a homogeneous system of linear equations.
Step 1: Finding the Characteristic Polynomial
We need to compute det(A - λI). First, let's construct (A - λI):
| 2-λ | -1 | 0 |
| -1 | 2-λ | -1 |
| 0 | -1 | 2-λ |
Now we compute the determinant. For a 3×3 matrix, we can use the formula or expand along a row/column. Let's expand along the first row:
Computing the Determinant:
det(A - λI) = (2-λ) · det([2-λ -1
-1 2-λ]) - (-1) · det([-1 -1
0 2-λ]) + 0 · det([...])
The third term is zero, so we only need to compute the first two:
= (2-λ)[(2-λ)² - 1] - (-1)[-1(2-λ) - 0]
= (2-λ)[(2-λ)² - 1] + (2-λ)
= (2-λ)[(2-λ)² - 1 + 1]
= (2-λ)(2-λ)² = (2-λ)³
Wait, let's recalculate more carefully:
Detailed Calculation:
det(A - λI) = (2-λ) · [(2-λ)(2-λ) - (-1)(-1)] - (-1) · [(-1)(2-λ) - (-1)(0)] + 0
= (2-λ) · [(2-λ)² - 1] - (-1) · [-1(2-λ)]
= (2-λ) · [(2-λ)² - 1] + (2-λ)
= (2-λ) · [(2-λ)² - 1 + 1]
= (2-λ) · (2-λ)²
= (2-λ)³
Actually, let me verify this step by step more carefully. Expanding the first minor:
Minor₁₁ = (2-λ)(2-λ) - (-1)(-1) = (2-λ)² - 1 = 4 - 4λ + λ² - 1 = λ² - 4λ + 3
Minor₁₂ = (-1)(2-λ) - (-1)(0) = -2 + λ
det = (2-λ)(λ² - 4λ + 3) - (-1)(-2 + λ) + 0
= (2-λ)(λ² - 4λ + 3) - (2 - λ)
= (2-λ)(λ² - 4λ + 3 - 1)
= (2-λ)(λ² - 4λ + 2)
Let me recalculate this more systematically using the standard 3×3 determinant formula:
det(A - λI) = (2-λ)[(2-λ)² - 1] - (-1)[(-1)(2-λ) - (-1)(0)] + 0[(-1)(-1) - (2-λ)(0)]
= (2-λ)[(2-λ)² - 1] + [(-1)(2-λ)]
= (2-λ)[(2-λ)² - 1] - (2-λ)
= (2-λ)[(2-λ)² - 1 - 1]
= (2-λ)[(2-λ)² - 2]
= (2-λ)[4 - 4λ + λ² - 2]
= (2-λ)(λ² - 4λ + 2)
So our characteristic polynomial is p(λ) = (2-λ)(λ² - 4λ + 2). Setting this equal to zero gives us the characteristic equation.
Step 2: Solving the Characteristic Equation
We need to solve (2-λ)(λ² - 4λ + 2) = 0. This gives us:
Finding the Roots:
Either (2-λ) = 0 → λ₁ = 2
Or (λ² - 4λ + 2) = 0
For the quadratic, we use the quadratic formula:
λ = [4 ± √(16 - 8)] / 2 = [4 ± √8] / 2 = [4 ± 2√2] / 2 = 2 ± √2
So we have three eigenvalues:
- λ₁ = 2
- λ₂ = 2 + √2
- λ₃ = 2 - √2
💡 Key Insight: Notice that all three eigenvalues are real. This is guaranteed for symmetric matrices (which our matrix A is). Symmetric matrices always have real eigenvalues.
Also notice that λ₂ and λ₃ are irrational numbers. This is fine—eigenvalues don't have to be nice integers.
Step 3: Finding the Eigenvectors
For each eigenvalue, we need to solve (A - λI)v = 0 to find the corresponding eigenvector(s). Let's do this for each eigenvalue.
For λ₁ = 2:
We solve (A - 2I)v = 0:
[0 -1 0
-1 0 -1
0 -1 0] [v₁
v₂
v₃] = [0
0
0]
This gives us the system:
- -v₂ = 0 → v₂ = 0
- -v₁ - v₃ = 0 → v₁ = -v₃
- -v₂ = 0 → v₂ = 0
So v₂ = 0 and v₁ = -v₃. Letting v₃ = t (a free parameter), we get:
v₁ = t[1, 0, -1]ᵀ
Taking t = 1, we get the eigenvector: v₁ = [1, 0, -1]ᵀ
For λ₂ = 2 + √2:
We solve (A - (2+√2)I)v = 0:
[-√2 -1 0
-1 -√2 -1
0 -1 -√2] [v₁
v₂
v₃] = [0
0
0]
From the first equation: -√2 v₁ - v₂ = 0 → v₂ = -√2 v₁
From the third equation: -v₂ - √2 v₃ = 0 → v₃ = -v₂/√2 = v₁
The second equation should be automatically satisfied. Letting v₁ = t:
v₂ = t[1, -√2, 1]ᵀ
Taking t = 1, we get: v₂ = [1, -√2, 1]ᵀ
For λ₃ = 2 - √2:
We solve (A - (2-√2)I)v = 0:
[√2 -1 0
-1 √2 -1
0 -1 √2] [v₁
v₂
v₃] = [0
0
0]
From the first equation: √2 v₁ - v₂ = 0 → v₂ = √2 v₁
From the third equation: -v₂ + √2 v₃ = 0 → v₃ = v₂/√2 = v₁
v₃ = t[1, √2, 1]ᵀ
Taking t = 1, we get: v₃ = [1, √2, 1]ᵀ
✅ Summary of Results:
- λ₁ = 2, with eigenvector v₁ = [1, 0, -1]ᵀ
- λ₂ = 2 + √2, with eigenvector v₂ = [1, -√2, 1]ᵀ
- λ₃ = 2 - √2, with eigenvector v₃ = [1, √2, 1]ᵀ
Step 4: Verifying Your Answer
Always verify that Av = λv for each eigenvalue-eigenvector pair. Let's check:
Verification for λ₁ = 2:
Av₁ = [2 -1 0
-1 2 -1
0 -1 2] [1, 0, -1]ᵀ
= [2(1) + (-1)(0) + 0(-1), -1(1) + 2(0) + (-1)(-1), 0(1) + (-1)(0) + 2(-1)]ᵀ
= [2, 0, -2]ᵀ = 2[1, 0, -1]ᵀ = 2v₁ ✓
You should verify the other two as well. This step is crucial—it catches calculation errors and gives you confidence in your answer.
Common Mistakes to Avoid
Here are the mistakes that cost students the most points. Learn them now so you don't make them on test day. If you want personalized help avoiding these errors, create a custom study video for your specific problem.
❌ Mistake #1: Forgetting the Negative Sign in (A - λI)
Writing A - λI incorrectly, especially the diagonal entries. It should be (aᵢᵢ - λ), not (λ - aᵢᵢ).
Fix: Always write A - λI carefully. The diagonal entries become (aᵢᵢ - λ), not (λ - aᵢᵢ).
❌ Mistake #2: Errors in Computing the 3×3 Determinant
Making arithmetic mistakes when expanding the determinant, especially with the signs in the cofactor expansion.
Fix: Use the checkerboard pattern for signs: +, -, + for the first row. Double-check your arithmetic, especially when dealing with (2-λ) terms.
❌ Mistake #3: Not Simplifying the Characteristic Polynomial
Leaving the polynomial in an unfactored form, making it harder to find roots. Always try to factor when possible.
Fix: Look for common factors. In our case, we could factor out (2-λ). Always check if the polynomial factors nicely.
❌ Mistake #4: Errors in the Quadratic Formula
Making mistakes when applying the quadratic formula, especially with the discriminant or simplifying √8.
Fix: Remember: λ = [-b ± √(b² - 4ac)] / 2a. Simplify radicals: √8 = 2√2. Double-check your arithmetic.
❌ Mistake #5: Row Reduction Errors When Finding Eigenvectors
Making mistakes in Gaussian elimination when solving (A - λI)v = 0. This is the most common source of errors.
Fix: Work slowly and carefully. Check each row operation. Remember you're solving a homogeneous system, so there should be at least one free variable.
❌ Mistake #6: Not Normalizing or Choosing a Convenient Eigenvector
Leaving eigenvectors with fractions or messy coefficients when they could be simplified. Or forgetting that any scalar multiple is also an eigenvector.
Fix: Choose the simplest form. Multiply by constants to clear denominators. Remember: if v is an eigenvector, so is cv for any c ≠ 0.
❌ Mistake #7: Not Verifying Your Answer
Skipping the verification step. This is a huge mistake—verification catches errors and is often worth partial credit even if your method was wrong.
Fix: Always check that Av = λv for each pair. This takes 30 seconds and can save you from losing all points.
Practice Problems with Video Solutions
Best way to get good at this? Practice. Try these similar problems and check your work with video solutions. You can also generate instant video explanations for any linear algebra problem you're working on.
Practice Problem 1: Different Matrix
Find the eigenvalues and eigenvectors of A = [3 1 0
1 3 1
0 1 3]
Hint: This has a similar structure. The eigenvalues should be 3, 3+√2, and 3-√2.
Get instant video solution on Torial →Practice Problem 2: Matrix with Repeated Eigenvalue
Find the eigenvalues and eigenvectors of A = [4 2 0
2 4 2
0 2 4]
Hint: Watch out for repeated eigenvalues. You may need to find multiple linearly independent eigenvectors.
Get instant video solution on Torial →Practice Problem 3: Non-Symmetric Matrix
Find the eigenvalues and eigenvectors of A = [1 2 1
0 2 1
0 0 3]
Hint: This is upper triangular, so the eigenvalues are just the diagonal entries. But you still need to find eigenvectors.
Get instant video solution on Torial →When to Use This Method
The characteristic polynomial method is the standard approach for finding eigenvalues and eigenvectors of any square matrix. Here's when it's the right choice:
Use the characteristic polynomial method when:
- You need exact eigenvalues (not approximations)
- The matrix is small (2×2, 3×3, maybe 4×4)
- You need all eigenvalues and eigenvectors
- The problem asks for the characteristic polynomial explicitly
Consider numerical methods when:
- The matrix is large (5×5 or bigger)
- You only need the largest or smallest eigenvalue
- Approximations are acceptable
- The characteristic polynomial doesn't factor nicely
For this problem? The characteristic polynomial method is perfect. The matrix is 3×3, and we can solve the cubic equation exactly. The symmetric structure also guarantees real eigenvalues, which makes the problem more tractable. When you're juggling determinants, polynomial roots, and row reduction, it helps to have a step-by-step video walkthrough that shows exactly how to organize your work.
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