Recipe

Find the eigenvectors for a given eigenvalue

Use when: you already have a matrix $A$ and one of its eigenvalues $\lambda$ , and need the eigenvectors that go with it

Goal: a basis for the eigenspace: every eigenvector belonging to $\lambda$ , described as a span

Taught in Eigenspaces

Quick recipe

Form $A-\lambda I$ using the given eigenvalue
Row reduce $A-\lambda I$
Identify the free variable or variables
Write the solution as a span
Check one vector against the original equation

1
Subtract $\lambda$ from each diagonal entry of $A$ , leaving the off-diagonal entries untouched.

Example

Take $A = \begin{bmatrix} 5 & -2 \\ 3 & 0 \end{bmatrix}$ with the eigenvalue $\lambda = 3$ found earlier (the same matrix the widget above used). Subtract $3$ from each diagonal entry only: $A - 3I = \begin{bmatrix} 5-3 & -2 \\ 3 & 0-3 \end{bmatrix} = \begin{bmatrix} 2 & -2 \\ 3 & -3 \end{bmatrix}$ The off-diagonal $-2$ and $3$ are untouched — that's the whole step.

Why: an eigenvector for $\lambda$ is, by definition, a nonzero solution of $(A-\lambda I)\mathbf v=\mathbf 0$ , so this matrix is the actual object being solved.
2
Row reduce $A-\lambda I$ the same way as any homogeneous system, expecting at least one row to collapse to all zeros.

Example

Row 2 of $\begin{bmatrix} 2 & -2 \\ 3 & -3 \end{bmatrix}$ is $\frac32$ times row 1 ( $3 = \frac32\cdot 2$ and $-3 = \frac32\cdot(-2)$ ), so $R_2 \to R_2 - \frac32R_1$ clears it entirely: $\begin{bmatrix} 2 & -2 & \big| & 0 \\ 3 & -3 & \big| & 0 \end{bmatrix} \xrightarrow{R_2 - \frac32 R_1} \begin{bmatrix} 2 & -2 & \big| & 0 \\ \cancel{0} & \cancel{0} & \big| & \cancel{0} \end{bmatrix}$ One pivot survives, one row vanishes to zero — exactly the collapse a singular matrix guarantees.

Why: $\lambda$ is only an eigenvalue in the first place because $A-\lambda I$ is singular, and a singular matrix is guaranteed to produce at least one free variable when row reduced.
3
Read off which columns have no pivot; those columns' variables are free.

Example

The surviving row's pivot sits in column 1 ( $x_1$ ); column 2 ( $x_2$ ) never gets a pivot, so $x_2$ is the free variable. Set $x_2 = t$ .

Why: the number of free variables is the eigenspace's geometric multiplicity, and it can be one (a single line) or more (a plane or higher), never zero.
4
Solve the surviving equations for the pivot variables in terms of the free ones, then write $\mathbf v$ as a combination of one vector per free variable.

Example

The one surviving equation is $2x_1 - 2x_2 = 0$ , i.e. $x_1 = x_2$ . With $x_2 = t$ free, $x_1 = t$ as well: $\mathbf v = t\begin{bmatrix} 1 \\ 1 \end{bmatrix}$

Why: this span is exactly the eigenspace, the complete set of eigenvectors for $\lambda$ (plus the zero vector every subspace must contain).
5
Pick a nonzero vector from the span and confirm $A\mathbf v=\lambda\mathbf v$ directly.

Example

Pick $t=1$ , so $\mathbf v = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ , and check it against $A$ itself, not $A-3I$ : $A\mathbf v = \begin{bmatrix} 5 & -2 \\ 3 & 0 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \end{bmatrix} = 3\begin{bmatrix} 1 \\ 1 \end{bmatrix} = 3\mathbf v \ \checkmark$

Why: this is the definition being verified, independent of whether the row reduction was carried out correctly.

Common mistakes

Row reducing $A$ itself instead of $A-\lambda I$ .
Reporting only one eigenvector when the eigenspace is actually two- or three-dimensional, missing a free variable during row reduction.
Including the zero vector as "an eigenvector" when stating the answer, rather than describing the eigenspace as its span.
Skipping the final check and trusting the row reduction without confirming $A\mathbf v=\lambda\mathbf v$ on an actual vector.