Eigenvalues, eigenvectors, and the Cayley-Hamilton theorem close the linear algebra module in VTU Engineering Mathematics 1 (BMATS101 under the 2022 scheme, 18MAT11 under 2018). They are procedural, but students who only memorise the steps get stuck when a question is worded differently or the matrix is larger. This guide builds the method from the definition, works through 2 × 2 and 3 × 3 examples, and shows how Cayley-Hamilton gives shortcuts for powers and inverses.
The Definition
For a square matrix A, a non-zero vector X is an eigenvector with eigenvalue λ if:
AX = λX
In words, multiplying X by A only stretches or shrinks it; the direction stays the same. Eigenvalues are the scaling factors, and eigenvectors are the directions that the transformation leaves unchanged.
Finding Eigenvalues: The Characteristic Equation
Rewrite AX = λX as (A − λI)X = 0. For a non-zero X to exist, the matrix A − λI must be singular, so its determinant must be zero:
det(A − λI) = 0
This is the characteristic equation. For an n × n matrix it is a polynomial of degree n in λ, so there are n eigenvalues, counted with multiplicity.
Worked Example 1: A 2 × 2 Matrix
Find the eigenvalues and eigenvectors of A = [[4, 1], [2, 3]].
Characteristic equation.
det(A − λI) = det [[4 − λ, 1], [2, 3 − λ]] = (4 − λ)(3 − λ) − 2 = λ² − 7λ + 12 − 2 = λ² − 7λ + 10.
Setting this to zero: λ² − 7λ + 10 = 0, so (λ − 5)(λ − 2) = 0. The eigenvalues are λ = 5 and λ = 2.
Eigenvector for λ = 5. Solve (A − 5I)X = 0: [[−1, 1], [2, −2]] (x, y) = (0, 0). The first row gives −x + y = 0, so y = x. Taking x = 1, the eigenvector is (1, 1).
Eigenvector for λ = 2. Solve (A − 2I)X = 0: [[2, 1], [2, 1]] (x, y) = (0, 0). The row gives 2x + y = 0, so y = −2x. Taking x = 1, the eigenvector is (1, −2).
Check. A(1, −2) = (4 − 2, 2 − 6) = (2, −4) = 2(1, −2) ✓.
Worked Example 2: A 3 × 3 Matrix
Find the eigenvalues of A = [[2, 0, 0], [0, 3, 4], [0, 4, 9]].
The matrix is block-diagonal: the first row and column are separate from the 2 × 2 block below. So one eigenvalue is 2 (from the entry in position (1,1)), and the others come from the 2 × 2 block [[3, 4], [4, 9]].
For the block, the characteristic equation is (3 − λ)(9 − λ) − 16 = λ² − 12λ + 27 − 16 = λ² − 12λ + 11 = 0. Factoring: (λ − 1)(λ − 11) = 0.
The eigenvalues of A are 2, 1, and 11.
Recognising block structure saves a great deal of work. In exams, look for zero entries before writing a full characteristic polynomial.
The Cayley-Hamilton Theorem
The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation. If the characteristic equation is λ² − 7λ + 10 = 0, then:
A² − 7A + 10I = 0
The theorem is powerful because it lets you replace high powers of A with lower powers. It also gives a formula for the inverse, as the next examples show.
Check: Cayley-Hamilton for the Matrix in Example 1
Compute A² for A = [[4, 1], [2, 3]]:
A² = [[4·4 + 1·2, 4·1 + 1·3], [2·4 + 3·2, 2·1 + 3·3]] = [[18, 7], [14, 11]].
Now A² − 7A + 10I = [[18, 7], [14, 11]] − [[28, 7], [14, 21]] + [[10, 0], [0, 10]] = [[0, 0], [0, 0]] ✓.
Worked Example 3: Finding the Inverse Using Cayley-Hamilton
From A² − 7A + 10I = 0, factor A: A(A − 7I) = −10I. Therefore:
A⁻¹ = −(A − 7I)/10 = (7I − A)/10
For A = [[4, 1], [2, 3]]: 7I − A = [[3, −1], [−2, 4]], so A⁻¹ = [[3/10, −1/10], [−2/10, 4/10]].
Check: A · A⁻¹ = [[4, 1], [2, 3]] · [[3/10, −1/10], [−2/10, 4/10]]. The top-left entry is 12/10 − 2/10 = 1. The top-right is −4/10 + 4/10 = 0. The bottom-left is 6/10 − 6/10 = 0. The bottom-right is −2/10 + 12/10 = 1. So A · A⁻¹ = I ✓.
Worked Example 4: Computing A⁵ Without Full Multiplication
Using A² = 7A − 10I, express higher powers in terms of A and I.
A³ = A · A² = A(7A − 10I) = 7A² − 10A = 7(7A − 10I) − 10A = 39A − 70I.
Continuing this way, A⁵ = 7A⁴ − 10A³ and so on, but a faster route is to note that the powers follow a linear recurrence. Compute it step by step: multiply A⁴ = 7A³ − 10A² = 7(39A − 70I) − 10(7A − 10I) = 273A − 490I − 70A + 100I = 203A − 390I, then A⁵ = A · A⁴ = 203A² − 390A = 203(7A − 10I) − 390A = 1421A − 2030I − 390A = 1031A − 2030I.
Check the arithmetic: the coefficient of A in A⁵ is 1031 and the constant term is −2030. Compare with the eigenvalues 5 and 2: the scalars must satisfy 5⁵ = 3125 and 2⁵ = 32, and 1031·5 − 2030 = 5155 − 2030 = 3125 ✓, and 1031·2 − 2030 = 2062 − 2030 = 32 ✓. Both checks pass.
This check against the eigenvalues is a reliable way to verify any power computed with Cayley-Hamilton, and it is worth doing in exams when time allows.
How to Structure an Eigenvalue Question
- Write A − λI. Subtract λ from each diagonal entry.
- Compute the determinant and set it to zero. Expand carefully, especially for 3 × 3 matrices.
- Solve for λ. Factor the polynomial, or use the quadratic formula.
- For each λ, solve (A − λI)X = 0. Write the eigenvector with a convenient scaling.
- Check with AX = λX. Multiply out and confirm the result.
Common Mistakes That Cost Marks
Mistake 1: Forgetting the identity matrix. The characteristic matrix is A − λI, not A − λ. Subtract λ only from the diagonal, and make sure each diagonal entry is adjusted.
Mistake 2: Sign errors in the determinant. For 3 × 3 matrices, the cofactor expansion has alternating signs. Recheck each term.
Mistake 3: Writing the wrong eigenvector. Substitute the eigenvalue back into the matrix, not the original matrix. Check AX = λX every time.
Mistake 4: Using Cayley-Hamilton with the wrong coefficients. The coefficients come from the characteristic equation with its sign convention. For λ² − 7λ + 10 = 0, the theorem reads A² − 7A + 10I = 0, not A² + 7A + 10I.
Practice Questions
- Find the eigenvalues of [[2, 1], [1, 2]] and an eigenvector for each.
- Verify the Cayley-Hamilton theorem for [[1, 2], [3, 4]].
- Using Cayley-Hamilton, find A⁻¹ for A = [[2, 1], [1, 2]].
Checks: (1) det(A − λI) = (2 − λ)² − 1 = 0, so λ = 1 or 3. For λ = 3, eigenvector (1, 1). For λ = 1, eigenvector (1, −1). (2) Characteristic equation λ² − 5λ − 2 = 0, so A² − 5A − 2I = 0. Check by computing A² = [[7, 10], [15, 22]], then 5A = [[5, 10], [15, 20]], and A² − 5A − 2I = [[7 − 5 − 2, 10 − 10], [15 − 15, 22 − 20 − 2]] = 0 ✓. (3) From A² − 4A + 3I = 0, A⁻¹ = (4I − A)/3 = [[2/3, −1/3], [−1/3, 2/3]].
Orthogonal Eigenvectors of a Symmetric Matrix
Find the eigenvalues and eigenvectors of A = [[2, 1], [1, 2]], and check whether the eigenvectors are perpendicular.
Characteristic equation. det(A − λI) = (2 − λ)² − 1 = 0, so λ² − 4λ + 3 = 0 and (λ − 1)(λ − 3) = 0. The eigenvalues are 3 and 1.
Eigenvector for λ = 3. Solve (A − 3I)X = 0: [[−1, 1], [1, −1]], which gives x = y. The eigenvector is (1, 1).
Eigenvector for λ = 1. Solve (A − I)X = 0: [[1, 1], [1, 1]], which gives x = −y. The eigenvector is (1, −1).
The dot product of (1, 1) and (1, −1) is 1 − 1 = 0, so the eigenvectors are perpendicular. This is not a coincidence. A real symmetric matrix always has eigenvectors for distinct eigenvalues that are orthogonal. Recognising a symmetric matrix in a question is therefore a useful signal: you can expect orthogonal eigenvectors, and you can check your answer by testing the dot product.
Diagonalisation in Brief
If the eigenvectors are collected as the columns of a matrix P, then P⁻¹AP is a diagonal matrix with the eigenvalues on the diagonal. For the example above, P = [[1, 1], [1, −1]] gives P⁻¹AP = [[3, 0], [0, 1]]. This is useful in engineering because powers of a diagonal matrix are easy to compute: Aⁿ = P Dⁿ P⁻¹. Your course may ask you to diagonalise a matrix, so it is worth working through one example by hand.
Connecting This Topic to the Full Module
Eigenvalues, eigenvectors, and the Cayley-Hamilton theorem complete the linear algebra module of VTU Engineering Mathematics 1. For the full module outline, the topics most often tested, and the places where marks are lost, see our VTU Engineering Maths 1 Notes — Complete Module-Wise Guide (BMATS101).
Frequently Asked Questions
Q: Can eigenvalues be complex?
Yes, for some real matrices. The characteristic equation may have complex roots, and the corresponding eigenvectors are complex too. Check whether your course covers this case.
Q: Do eigenvectors need to be unique?
No. Any non-zero multiple of an eigenvector is also an eigenvector. Give the simplest form in your answer.
Q: Where does this fit in the module?
It is the last topic in the linear algebra module, and it builds on rank and determinants. The full module outline is in our VTU Engineering Maths 1 notes guide.
For more solved eigenvalue problems, see the Eigenvalues and Eigenvectors Solved Examples page and the Engineering Math Survival Kit. To discuss batch options, use the contact page.