Eigenvalues of a 2x2 matrix from trace and determinant
BlinkNBuild practice problem · GATE standard, authored and verified in-house
The eigenvalues of $$A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$$ are:
Show the step-by-step derivation
Answer
(A) 2 and 5
Step-by-step derivation
- For any $2\times 2$ matrix the characteristic equation can be written directly from two invariants: $$\lambda^2 - (\text{tr}\,A)\lambda + \det A = 0.$$
- Compute the trace (sum of the diagonal): $$\text{tr}\,A = 4 + 3 = 7.$$
- Compute the determinant: $$\det A = (4)(3) - (1)(2) = 12 - 2 = 10.$$
- Form the characteristic equation: $$\lambda^2 - 7\lambda + 10 = 0.$$
- Factorise - look for two numbers summing to $7$ and multiplying to $10$: $$(\lambda - 2)(\lambda - 5) = 0 \;\Longrightarrow\; \lambda = 2,\ 5.$$ That is option (A).
- Verify against the invariants. The eigenvalues must satisfy $\lambda_1 + \lambda_2 = \text{tr}\,A$ and $\lambda_1\lambda_2 = \det A$. Here $2 + 5 = 7$ ✓ and $2 \times 5 = 10$ ✓. This check takes three seconds and catches nearly every sign error.
The trap this question is built around
Sign errors on the determinant of a matrix with negative entries. The trace-and-determinant check at the end costs nothing and catches them - which is why it is worth doing even when you are confident.
The idea behind this question
An eigenvalue $\lambda$ of a matrix $A$ satisfies $\det(A - \lambda I) = 0$. For a $2\times2$ matrix this is the quadratic $\lambda^2 - (\text{tr}\,A)\lambda + \det A = 0$, so the eigenvalues always add up to the trace and multiply to the determinant - a free check on any answer.
Try a variation
What are the eigenvalues of $\begin{bmatrix} 5 & 2 \\ 2 & 2 \end{bmatrix}$?
Show the answer
Answer: $1$ and $6$
Trace $7$, determinant $6$: $\lambda^2 - 7\lambda + 6 = 0$.
Other mistakes to avoid
- Subtracting $\lambda$ from every entry instead of only the diagonal.
- Solving the quadratic correctly but pairing it with the wrong matrix after a transcription slip.