Recipe

Find or verify a $2\times2$ matrix's eigenvalues using trace and determinant

Use when: you have a $2\times2$ matrix — or a claimed pair of numbers said to be its eigenvalues — and want the eigenvalues confirmed without expanding $\det(A-\lambda I)$ by hand

Goal: the confirmed pair of eigenvalues, either derived directly or cleared against a candidate guess

Taught in Trace, determinant, and the eigenvalues

Quick recipe

Compute the trace and determinant directly from $A$
If you have a candidate pair, check both the sum and the product
If there's no candidate, or the candidate failed, solve for $\lambda$ directly
Verify the final pair against both trace and determinant

1
Add the two diagonal entries for the trace; compute $ad-bc$ for the determinant.

Example

For $A = \begin{bmatrix} 6 & -2 \\ 4 & 0 \end{bmatrix}$ : $\operatorname{tr}(A) = 6 + 0 = 6, \qquad \det(A) = 6(0) - (-2)(4) = 8$

Why: every square matrix's eigenvalues sum to its trace and multiply to its determinant, straight from the characteristic polynomial's coefficients — this holds before any factoring is done.
2
Add the two candidate numbers and compare to $\operatorname{tr}(A)$ ; multiply them and compare to $\det(A)$ .

Example

A student claims the eigenvalues of $A = \begin{bmatrix} 5 & 3 \\ 1 & 3 \end{bmatrix}$ are $5$ and $3$ — "the diagonal entries." Sum check: $5+3=8=\operatorname{tr}(A)$ ✓. Product check: $5\times3=15$ , but $\det(A) = 5(3)-3(1)=12$ — $15 \ne 12$ , so the claim fails despite passing the first test.
- both the sum and the product match — the candidate pair is confirmed; stop here.
- the sum matches but the product doesn't (or vice versa) — the candidate is wrong; discard it and continue to Step 3.
Why: a matching sum alone proves nothing — the diagonal entries always sum to the trace by definition, whether or not they're the eigenvalues — so the product check is what actually catches an error, and together the two checks pin down a $2\times2$ pair uniquely.
3
Solve $\lambda^2 - \operatorname{tr}(A)\lambda + \det(A) = 0$ by factoring, or with the quadratic formula if it doesn't factor by inspection.

Example

For $A = \begin{bmatrix} 6 & -2 \\ 4 & 0 \end{bmatrix}$ : $\lambda^2-6\lambda+8=0$ factors as $(\lambda-4)(\lambda-2)=0$ , since $4+2=6$ and $4\cdot2=8$ , giving $\lambda=4$ or $\lambda=2$ .

Why: trace and determinant are exactly this quadratic's coefficients, so solving it is guaranteed to produce the true eigenvalues, not just a checked guess.
4
Add and multiply the pair you end up with and confirm both match $\operatorname{tr}(A)$ and $\det(A)$ .

Example

$4+2=6=\operatorname{tr}(A)$ and $4\cdot2=8=\det(A)$ — both checks hold, confirming $\lambda_1=4$ , $\lambda_2=2$ .

Why: this catches a factoring or arithmetic slip immediately, the same way the candidate check in Step 2 caught a wrong guess.

Common mistakes

Trusting a matching sum by itself — the diagonal entries of ANY matrix sum to the trace, whether or not they happen to be the eigenvalues.
Computing the determinant as $ad+bc$ instead of $ad-bc$ .
Forgetting that trace and determinant only pin the eigenvalues down uniquely for a $2\times2$ matrix — for $3\times3$ and larger, they're real information but not enough by themselves to name every eigenvalue.
Skipping the final verification and trusting the factored quadratic without re-checking it against both $\operatorname{tr}(A)$ and $\det(A)$ .

Find or verify a 2\times2 matrix's eigenvalues using trace and determinant

Find or verify a $2\times2$ matrix's eigenvalues using trace and determinant