Recipe

Find the eigenvalues and eigenvectors of a 2×2 matrix

Use when: you have a $2\times2$ matrix $A$ — often the coefficient matrix of a system $\mathbf x'=A\mathbf x$ — and need every $\lambda$ and matching $\mathbf v$ satisfying $A\mathbf v=\lambda\mathbf v$

Goal: every eigenvalue of $A$ , each paired with an eigenvector spanning its eigenspace

Taught in Eigenvalues and eigenvectors of a 2x2 matrix

Quick recipe

Compute the trace and determinant
Solve $\lambda^2-\operatorname{tr}(A)\lambda+\det(A)=0$ for $\lambda$
For each eigenvalue, solve $(A-\lambda I)\mathbf v=\mathbf 0$
Verify by substituting back into $A\mathbf v=\lambda\mathbf v$

1
Add the two diagonal entries to get $\operatorname{tr}(A)$ ; compute $\det(A)=ad-bc$ from all four entries.

Example

For $B=\begin{pmatrix}5&2\\2&2\end{pmatrix}$ : $\operatorname{tr}(B)=5+2=7$ , and $\det(B)=(5)(2)-(2)(2)=10-4=6$ .

Why: for a $2\times2$ matrix, these two numbers alone are enough to build the characteristic equation — there's no need to expand $\det(A-\lambda I)$ by hand.
2
Substitute the two numbers from Step 1 into this quadratic and solve it — factor if it factors, otherwise fall back to the quadratic formula.

Example

$\lambda^2-7\lambda+6=0$ factors as $(\lambda-6)(\lambda-1)=0$ , so $\lambda=6$ or $\lambda=1$ .
- two distinct real roots — find a separate eigenvector for each one in Step 3.
- one repeated real root — Step 3 may turn up only one independent eigenvector direction instead of two; that's a genuine possibility for a repeated eigenvalue, not a sign of a mistake.
Why: the characteristic equation's roots are exactly the values of $\lambda$ where $A-\lambda I$ fails to be invertible, so they're the only candidates worth testing in Step 3.
3
Substitute $\lambda$ , form $A-\lambda I$ , then solve the resulting homogeneous system for a nonzero $\mathbf v$ .

Example

For $\lambda=6$ : $B-6I=\begin{pmatrix}-1&2\\2&-4\end{pmatrix}$ . The first row gives $-v_1+2v_2=0$ , i.e. $v_1=2v_2$ ; choosing $v_2=1$ gives $\mathbf v=(2,1)$ . For $\lambda=1$ : $B-1I=\begin{pmatrix}4&2\\2&1\end{pmatrix}$ . The first row gives $4v_1+2v_2=0$ , i.e. $v_2=-2v_1$ ; choosing $v_1=1$ gives $\mathbf v=(1,-2)$ .

Why: rearranging $A\mathbf v=\lambda\mathbf v$ gives exactly $(A-\lambda I)\mathbf v=\mathbf 0$ , so its nonzero solutions are the eigenvectors.
4
Multiply $A$ by each eigenvector found and confirm it comes back as $\lambda$ times that same vector.

Example

$B(2,1)=(5\cdot2+2\cdot1,\ 2\cdot2+2\cdot1)=(12,6)=6(2,1)$ . ✓ $B(1,-2)=(5\cdot1+2\cdot(-2),\ 2\cdot1+2\cdot(-2))=(1,-2)=1(1,-2)$ . ✓

Why: this catches an arithmetic slip in solving the homogeneous system before it carries into a system's general solution or phase portrait built on that eigenvector.

Common mistakes

Treating the trace and determinant themselves as the eigenvalues, instead of solving the quadratic they build for $\lambda$ .
Substituting the wrong $\lambda$ into $A-\lambda I$ partway through the row reduction.
Reporting only one eigenvector for a repeated eigenvalue that genuinely has two independent ones, or forcing a second independent one where only a single direction actually exists.
Skipping the Step 4 check and carrying an eigenvector into later work that doesn't actually satisfy $A\mathbf v=\lambda\mathbf v$ .