Eigenvalues of a 2×2 matrix

What the eigenvalues of a 2×2 matrix measure

Write Av = λv, with A the given matrix and v a nonzero column vector. On the left you transform v by the matrix. On the right you multiply both coordinates of the same vector by the scalar λ. An eigenvector is any nonzero v for which the two results agree. The corresponding eigenvalue is λ. There is no requirement that the vector have length one.

If v works, every nonzero multiple of v works for the same λ. For example, if multiplying v by A gives 3v, then multiplying 2v by A gives 6v, which is 3 times 2v. A drawing might show several arrows along one line, but those arrows represent one independent direction. Two vectors on the same line cannot serve as two independent directions even if they look different or point in opposite directions.

The zero vector is excluded because A0 = λ0 holds for every λ. It tells you nothing about how the matrix acts on a direction. A zero eigenvalue, by contrast, can tell you something. It means a nonzero vector is sent to zero. If you encounter a zero on the list of roots, test the system in the same way as for any other root. Keep the distinction between a number and a vector clear on the page.

Subtract λ from the diagonal before taking the determinant

Move λv to the left to write (A − λI)v = 0, where I is the identity matrix. Multiplying I by λ puts λ on each diagonal entry and zero everywhere else. In our example the rows of A − λI are (5 − λ, 2) and (−2, 1 − λ). The off-diagonal entries keep their original signs and values.

For a nonzero solution to exist, A − λI must be singular. If it were invertible, multiplying the equation by its inverse would force v = 0. This explains why the determinant must be zero. The Harvard notes on two-by-two matrices express the same determinant in terms of the trace and determinant of the original matrix. That gives a useful check after you expand it directly.

For our matrix, the determinant is (5 − λ)(1 − λ) − 2(−2). The second term adds four because the off-diagonal product is negative. Expanding gives λ² − 6λ + 9 = (λ − 3)². The only root is λ = 3, with multiplicity two. As a quick check, the coefficient of λ is minus the trace, −(5 + 1), and the constant is the original determinant, 5 × 1 − 2 × (−2) = 9. Calculating the two-by-two determinant carefully saves you from carrying a sign mistake into the vector calculation.

For an upper or lower triangular matrix, one off-diagonal factor is zero, so the determinant factors into the two shifted diagonal entries. That explains when reading the diagonal works. Our original matrix has two nonzero off-diagonal entries, and the shortcut misses the extra four.

Find a matching vector after you find the root

Substitute λ = 3 into A − λI. The rows are now (2, 2) and (−2, −2). Writing v as (x, y) in column form gives the equations 2x + 2y = 0 and −2x − 2y = 0. These are the same condition, x + y = 0. Choose x = 1 and y = −1. Every nonzero multiple of (1, −1) is also an eigenvector for 3.

Check using the original matrix. Its first row applied to (1, −1) gives 5 − 2 = 3. Its second row gives −2 − 1 = −3. The output (3, −3) is 3(1, −1). Both coordinates agree. Checking just one can miss a mismatched pair because a vector might accidentally satisfy one row while failing the other.

The root appears twice in the polynomial, yet the eigenvectors occupy only one line. You can see that from the single independent condition x + y = 0. Compare with the matrix 3I, which has the same characteristic polynomial. For λ = 3 its shifted matrix is entirely zero. It imposes no condition on either coordinate, so every nonzero vector works. The polynomial records that the root repeats. The shifted system reveals how many independent directions are available.

If you are checking a solution someone else wrote, compare the vector and eigenvalue as a pair. A different nonzero multiple of (1, −1) is fine. A vector with the same components in a different order may represent a different direction or a negative multiple. The original matrix product is the reliable test. In larger problems, writing the row-by-column products explicitly can make the check easier to audit.

Use Eqora to examine the step that changed the answer

Capture the worksheet in Eqora with both matrix rows, the brackets and the negative sign in the second row visible. Before reading any explanation, compare the recognized matrix with your paper. A missing minus sign would change the determinant and could produce different roots. Also check whether the task asks for eigenvalues alone or for their eigenvectors as well.

Compare the determinant line with your own. Make sure λ is subtracted only from the diagonal. Check the parentheses around both shifted diagonal entries. Then ask a question about the specific line that gave you trouble, such as “Why does subtracting the off-diagonal product add four?” If the roots match but the vectors are unclear, ask “Why does the repeated root produce only one independent direction in this matrix?” Look for a response that explains the row relation in A − 3I. Verify the row relation yourself.

Use Eqora as a learning aid. Its explanation can help you locate a step to revisit, but the input, notation, assumptions and result need your own check. It cannot guarantee a correct answer or a grade or replace your work in an assessment. When you can explain the determinant and verify the vector with the original matrix, put the explanation away and try the next example independently.

Solve a second matrix on paper

Use B with rows (4, 3) and (2, 5). Before checking the answer below, write B − λI and compute its determinant. Factor the resulting polynomial, then use a separate shifted system for each root. Do not carry the vector from the first root into the second calculation without checking it.

The determinant is (4 − λ)(5 − λ) − 3 × 2 = λ² − 9λ + 14. It factors as (λ − 2)(λ − 7), so the roots are 2 and 7. At λ = 2, the shifted rows are (2, 3) and (2, 3). The condition is 2x + 3y = 0, and one solution is (3, −2). Multiplying by B gives (6, −4), which is twice the vector.

At λ = 7, the rows are (−3, 3) and (2, −2). They give x = y, so use (1, 1). Multiplying by B gives (7, 7), which is seven times the vector. These two eigenvectors are not multiples of each other. Each root gave a different line. Notice how both checks use B itself. The shifted matrix is the device for finding the vector, while the original product verifies its meaning.

When you get stuck, ask which of the two problems you are solving. If your determinant does not match, check the matrix you copied and the products you subtracted. If your roots match but your vector fails, check the shifted system for the particular root and test both coordinates with the original matrix. This separates the algebra for the factor from the linear equations for the direction and makes it easier to find the first line that needs correction.

You can also check the two roots against the trace and determinant before solving the systems of linear equations. Their sum is 9, which is 4 + 5, and their product is 14, which is 4 × 5 − 3 × 2. These two checks test the polynomial coefficients. The separate vector products still test which direction belongs to each factor.

Test what a negative factor means

Try C with rows (1, 2) and (2, 1). Its determinant equation is (1 − λ)² − 4 = 0, so the roots are 3 and −1. The vector (1, 1) gives (3, 3). The vector (1, −1) gives (−1, 1). For λ = −1, the output is on the original line and points in the opposite direction. That still satisfies the definition. This is a good test of whether you have confused keeping a line with keeping an arrow's orientation. It also gives you a simple matrix you can use to explain why the eigenvalue and its vector always have to be checked together.