Skip to content
Benji your study buddy
Start learning free
← All articles

How to Understand Eigenvectors: A Worked Example

Illustration for How to Understand Eigenvectors: A Worked Example

An eigenvector is a nonzero vector that a square matrix sends to a multiple of itself. In a real two-dimensional picture, the vector stays on the same line through the origin. Its eigenvalue tells you the scaling: the vector may get longer, shorter, reverse direction, or become zero.

That is the idea behind the equation Av = λv. The matrix A acts on the vector v, and the result is the same as multiplying that vector by the number λ, pronounced "lambda." Start with that comparison before learning a method for finding the answers.

Understand the matrix's job first

Think of a two-dimensional vector as an arrow from the origin to a point. The vector (1, 1) points one unit right and one unit up. A matrix gives a rule for transforming that arrow.

If matrix multiplication is still confusing, work through the earlier steps in our linear algebra study plan. Eigenvectors depend on being able to calculate where a matrix sends a vector. You do not need advanced applications to begin, but you do need that calculation.

For example, consider a matrix that triples the horizontal coordinate and leaves the vertical coordinate unchanged:

D = [3  0]
    [0  1]

D(x, y) = (3x, y)

A horizontal arrow stays horizontal, and a vertical arrow stays vertical. A diagonal arrow will generally change its angle because its horizontal and vertical coordinates are scaled differently.

Compare the original vector with the result

Apply that rule to a few vectors:

Vector v Result Dv What happened?
(1, 0) (3, 0) The result is 3v, so v is an eigenvector with eigenvalue 3.
(0, 1) (0, 1) The result is 1v, so v is an eigenvector with eigenvalue 1.
(1, 1) (3, 1) There is no single number that scales both coordinates this way, so v is not an eigenvector.

The last row is the useful comparison. Multiplying (1, 1) by 3 would give (3, 3), not (3, 1). Multiplying it by 1 would leave (1, 1). Neither matches the result.

The question is whether the whole output vector is one scalar multiple of the input. A "scalar" here just means a number. You need the same number for every coordinate.

Georgia Tech's Interactive Linear Algebra textbook gives the formal definition and interactive pictures for this comparison. Try moving a vector in its figures, then explain why it does or does not remain on its original line.

A worked example with two eigenvectors

Now use a matrix whose eigenvectors are diagonal rather than horizontal and vertical:

A = [3  1]
    [1  3]

First try v = (1, 1). Multiply each row of the matrix by the vector:

First coordinate:  3 × 1 + 1 × 1 = 4
Second coordinate: 1 × 1 + 3 × 1 = 4

Av = (4, 4) = 4(1, 1)

So (1, 1) is an eigenvector, and its eigenvalue is 4. The arrow lies on the upward diagonal before and after the transformation.

Next try u = (1, -1):

First coordinate:  3 × 1 + 1 × (-1) = 2
Second coordinate: 1 × 1 + 3 × (-1) = -2

Au = (2, -2) = 2(1, -1)

This vector is also an eigenvector, but its eigenvalue is 2. The two diagonal lines have different scaling factors.

For a comparison that fails, try w = (1, 0). The result is Aw = (3, 1). Every multiple of (1, 0) has a second coordinate of zero, so (3, 1) cannot be one of those multiples. This vector is not an eigenvector of A.

How to find them instead of guessing

Checking a proposed vector is straightforward. Finding the vectors from a matrix alone requires another step.

Start with Av = λv and move everything to one side. This gives (A - λI)v = 0. The identity matrix I has ones on its diagonal and zeros elsewhere, so subtracting λI changes only the diagonal entries.

For this two-by-two matrix, we want A - λI to have a nonzero solution. That happens when its determinant is zero:

det(A - λI) = (3 - λ)(3 - λ) - 1
            = λ² - 6λ + 8
            = (λ - 4)(λ - 2)

The eigenvalues are therefore 4 and 2.

For λ = 4, solving (A - 4I)v = 0 gives y = x. Choose any nonzero vector on that line, such as (1, 1).

For λ = 2, the equations give x + y = 0, or y = -x. One choice is (1, -1).

Finish by substituting each answer into the original equation, as we did above. MIT OpenCourseWare's eigenvalues and eigenvectors session has a lecture, a summary, and problems with solutions if you want to continue practicing the method.

Mistakes that make the idea harder

An eigenvector is not usually one unique arrow. For the matrix above, (2, 2) also has eigenvalue 4, because it lies on the same line as (1, 1). A nonzero multiple of an eigenvector is another eigenvector with the same eigenvalue.

The zero vector is excluded from the definition. Every matrix sends zero to zero, so that vector cannot identify a particular eigenvalue.

A negative eigenvalue is allowed. For example, the matrix with diagonal entries -2 and 1 sends (1, 0) to (-2, 0). The arrow reverses direction while staying on the same line.

An eigenvalue can also be zero. The matrix with diagonal entries 1 and 0 sends the nonzero vector (0, 1) to zero. That vector is an eigenvector with eigenvalue zero.

Finally, do not assume every real matrix has a real eigenvector. A rotation through 90 degrees sends every nonzero real arrow off its original line. Complex numbers become relevant when you study examples like that.

Check your understanding with a fresh example

Try the matrix with diagonal entries 2 and 5 and zeros elsewhere. Decide whether (3, 0), (0, -2), and (1, 1) are eigenvectors. Calculate the outputs before checking the answers.

The first produces (6, 0), which is twice the original vector, so its eigenvalue is 2. The second produces (0, -10), which is five times the original, so its eigenvalue is 5. The third produces (2, 5), which is not a single multiple of (1, 1), so it is not an eigenvector.

Then explain why changing the length of an eigenvector does not change its eigenvalue. If you can calculate the result but cannot explain that relationship, use the Socratic self-study questions to work through your reasoning.

Keep the next study session focused

A useful goal is to check a proposed eigenvector, explain the scaling factor, and solve one small example without copying the steps. If something breaks, identify whether the problem is matrix multiplication, solving the equation, or interpreting the result. Our guide to finding study weak spots can help you choose the right earlier skill to revisit.

Benji can organize linear algebra into a saved learning path with lessons, practice, feedback, and checkpoints. Use it to work on a specific step, and check mathematical answers against your course materials. If you want a place to continue, start a linear algebra path in Benji and explain what Av = λv means before trying another calculation.