Eigenvalues and Eigenvectors — Step-by-Step Guide

Eigenvalues and eigenvectors tell you which directions a square matrix only scales instead of turning. For a square matrix $A$, an eigenvector is a nonzero vector $v$ such that

$$Av = \lambda v$$

for some scalar $\lambda$. The number $\lambda$ is the eigenvalue: eigenvectors keep their line, while eigenvalues tell you the scale factor on that line.

When this method applies

Use the eigenvalue procedure on a square matrix whenever you want the special directions that survive a transformation without turning away from their original line. Think of a matrix as a transformation that usually rotates, shears, stretches, or mixes directions. An eigenvector does not turn; it may get stretched, shrunk, or reversed if $\lambda < 0$, but it stays on the same line. The eigenvalue tells you what happens along that direction:

• If $\lambda > 1$, the vector is stretched.
• If $0 < \lambda < 1$, the vector is shrunk.
• If $\lambda < 0$, the vector is scaled and reversed.
• If $\lambda = 0$, the matrix sends that eigenvector to the zero vector.

The zero vector is never counted as an eigenvector. If it were allowed, every matrix would have it, and the idea would lose its meaning. This method applies to square matrices; if the matrix is not square, the standard eigenvalue problem is not defined in this form.

The procedure, step by step

1. Set up the equation. Start from $Av = \lambda v$ and move everything to one side so you get $(A - \lambda I)v = 0$.
2. Find the eigenvalues. For a nonzero solution $v$, the matrix $A - \lambda I$ must be singular, so solve $\det(A - \lambda I) = 0$ to find the allowed values of $\lambda$.
3. Find each eigenvector. Plug one eigenvalue back into $(A - \lambda I)v = 0$ and solve for nonzero vectors.
4. Check the pair. Verify that multiplying $A$ by your vector really gives $\lambda v$.

A full example: a 2x2 matrix

Let

$$A = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix}.$$

Step 1 and 2: compute $\det(A - \lambda I)$

First form

$$A - \lambda I = \begin{bmatrix} 2 - \lambda & 1 \\ 0 & 3 - \lambda \end{bmatrix},$$

so

$$\det(A - \lambda I) = (2 - \lambda)(3 - \lambda).$$

Setting this equal to zero gives the eigenvalues $\lambda = 2$ and $\lambda = 3$.

Step 3: eigenvectors for $\lambda = 2$

Substitute $\lambda = 2$:

$$A - 2I = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}.$$

With $v = \begin{bmatrix} x \\ y \end{bmatrix}$, the system gives $y = 0$. The variable $x$ is free, so every nonzero vector on the $x$-axis works. A simple choice is

$$\begin{bmatrix} 1 \\ 0 \end{bmatrix}$$

and any nonzero multiple of it is also an eigenvector for $\lambda = 2$.

Step 3 again: eigenvectors for $\lambda = 3$

Now use $\lambda = 3$:

$$A - 3I = \begin{bmatrix} -1 & 1 \\ 0 & 0 \end{bmatrix}.$$

This gives $-x + y = 0$, so $y = x$. A simple choice is

$$\begin{bmatrix} 1 \\ 1 \end{bmatrix}$$

and any nonzero multiple of it is also an eigenvector for $\lambda = 3$.

Step 4: check one pair

Take $v = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ for $\lambda = 3$:

$$Av = \begin{bmatrix} 2 & 1 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \end{bmatrix}.$$

So the check works: $Av = 3v$. Geometrically, if you picture the plane being transformed by $A$, most arrows tilt into new directions, and eigenvectors are the rare arrows that stay on their own line. That is why they matter: they reveal the simple directions hidden inside the transformation.

Where students get stuck, and how to check each step

1. Forgetting that eigenvectors must be nonzero.
2. Solving $\det(A - \lambda I) = 0$ incorrectly, especially the determinant step.
3. Finding the eigenvalues but not solving for the corresponding eigenvectors.
4. Assuming every square matrix has enough independent eigenvectors to form a basis. Some do not.
5. Assuming every real matrix has real eigenvalues. That depends on the matrix.

The Step 4 check is your safety net: any eigenpair you report should satisfy $Av = \lambda v$ by direct multiplication. They show up wherever a linear process has preferred directions or natural modes, including differential equations, vibration problems, dynamical systems, Markov models, and principal component analysis.

Practice this procedure

Run all four steps for

$$\begin{bmatrix} 4 & 0 \\ 1 & 1 \end{bmatrix}.$$

Find the eigenvalues first, then solve for the eigenvectors, and check one pair by direct multiplication. To go one step further, run your own version in a solver and compare the eigenpairs, not just the final numbers.