Eigenvectors

Eigenvectors

1. Definition

For a square matrix A and an eigenvalue λ, an eigenvector is a non-zero column vector X satisfying

So the eigenvectors of λ are exactly the non-zero vectors of the null space of (A − λI) — that null space (together with 0) is called the eigenspace of λ. Two consequences follow immediately:

• an eigenvector is fixed only up to a scalar multiple (if X works, so does kX for any k ≠ 0), so answers may be quoted as any convenient non-zero multiple, or normalised to unit length;
• X = 0 is never called an eigenvector, even though it always solves the equation.

2. Method — Step by Step

1. Find the eigenvalues from det(A − λI) = 0.
2. Take one eigenvalue λ and form the matrix (A − λI).
3. Solve the homogeneous system (A − λI)X = 0 by row reduction (the rows are always dependent — that is why λ is an eigenvalue — so you will lose at least one row).
4. Write the general solution in terms of free variables; pick a non-zero convenient set of values for the free variables.
5. Verify by substituting into AX = λX.
6. Repeat for each remaining eigenvalue, including repeated ones (they may give one or more independent eigenvectors).

Count check: for each λ the number of independent eigenvectors is $n - ρ(A - λI)$ — the geometric multiplicity of λ. It is always ≤ the algebraic multiplicity (the number of times λ repeats as a root of the characteristic equation). Equality for every λ means the matrix is diagonalizable (later topic).

        [ 2  1  1 ]
   A =  [ 1  2  1 ]        eigenvalues (from the previous topic): λ = 1, 1, 4
        [ 1  1  2 ]

   ── For λ = 4 :  A - 4I = [ -2   1   1 ]
                            [  1  -2   1 ]
                            [  1   1  -2 ]
        Reduce:  R1 → R1 + 2R2  ⇒  [ 0  -3   3 ]   ⇒  -x2 + x3 = 0  ⇒  x2 = x3
                 R3 - R2        ⇒  [ 0   3  -3 ]       same equation
                 Row 2 remains :  x1 - 2x2 + x3 = 0  ⇒  x1 = 2x3 - x3 = x3
        ⇒  X = (x3, x3, x3) = x3(1, 1, 1)
        ⇒  EIGENVECTOR  X₁ = (1, 1, 1)ᵀ     Check: A(1,1,1) = (4,4,4) = 4(1,1,1) ✓

   ── For λ = 1 :  A - I = [ 1  1  1 ]   (all three rows identical)
                           [ 1  1  1 ]
                           [ 1  1  1 ]
        Single equation:  x1 + x2 + x3 = 0, so two of the variables are free.
        Choices:  x2 = 1, x3 = 0  ⇒  x1 = -1  ⇒  X₂ = (-1, 1, 0)ᵀ
                  x2 = 0, x3 = 1  ⇒  x1 = -1  ⇒  X₃ = (-1, 0, 1)ᵀ
        ⇒  TWO independent eigenvectors for the repeated eigenvalue λ = 1
        Count check: n - ρ(A - I) = 3 - 1 = 2 ✓ (= algebraic multiplicity 2)
        Check: A(-1,1,0) = (-1, 1, 0) ✓      A(-1,0,1) = (-1, 0, 1) ✓

6. Properties Worth Remembering

1. Eigenvectors corresponding to distinct eigenvalues are linearly independent.
2. For a symmetric matrix, eigenvectors belonging to distinct eigenvalues are mutually orthogonal (X₁ᵀX₂ = 0) — a very handy check.
3. One eigenvector cannot belong to two different eigenvalues.
4. Sums and multiples of eigenvectors of the same eigenvalue are again eigenvectors of that eigenvalue (the eigenspace is a subspace).
5. Number of independent eigenvectors for λ = n − ρ(A − λI); if this equals the algebraic multiplicity for every λ, the matrix has n independent eigenvectors and is diagonalizable.
6. A real symmetric matrix always has a full set of n mutually orthogonal real eigenvectors — hence it is always orthogonally diagonalizable.

7. Exam Points

• Define an eigenvector and derive (A − λI)X = 0 — 5 marks, almost certain.
• Find the eigenvalues and the corresponding eigenvectors of a 2 × 2 matrix — 5 marks, very frequent.
• Find the eigenvalues and eigenvectors of a 3 × 3 matrix (including a repeated eigenvalue) — 7 marks, very frequent.
• Show that eigenvectors of distinct eigenvalues are linearly independent — 5 marks.
• Show that eigenvectors of a symmetric matrix for distinct eigenvalues are orthogonal — 5 marks.
• Find the geometric multiplicity and decide whether a given matrix is diagonalizable — 5 marks.
• Verify that a given vector is an eigenvector and find its eigenvalue (apply AX and read off the multiple) — 3 marks.