Worked examples · Sheet 2

Eigenvalues and eigenspaces of matrices of order 3

Theory used: Characteristic polynomial Triangular matrices and expansion along a row Eigenspace Computing an eigenspace Algebraic and geometric multiplicity Trace, determinant and eigenvalues

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Determine the eigenvalues and eigenspaces of the matrix

A = 121031002 .

p_{A} (λ) = (1 - λ) (3 - λ) (2 - λ) ⟹ λ_{1} = 1, λ_{2} = 2, λ_{3} = 3

A - I_{3} = 021021001 ⟹ {2 x + 2 y = 0 x + y + z = 0 ⟹ {y = - x z = 0 ⟹ V_{1} = ⟨(1, - 1, 0)⟩

A - 2 I_{3} = - 1 21 011000 ⟹ {- x = 0 2 x + y = 0 ⟹ V_{2} = ⟨(0, 0, 1)⟩

A - 3 I_{3} = - 2 21 001 00 - 1 ⟹ {x = 0 x + y - z = 0 ⟹ V_{3} = ⟨(0, 1, 1)⟩

The matrix is lower triangular, therefore $A - λ I_{3}$ is also triangular and its determinant is the product of the diagonal entries: the eigenvalues are the diagonal entries $1$ , $3$ , $2$ , all simple. For $λ = 1$ the first row of $A - I_{3}$ is zero and two independent equations remain; from the first one gets $y = - x$ , and substituting into the second one obtains $z = 0$ , so with $x = 1$ one finds the generator $(1, - 1, 0)$ . For $λ = 2$ the third column of $A - 2 I_{3}$ is zero, therefore $z$ does not appear in the equations and is free, while the first two equations give $x = 0$ and $y = 0$ . For $λ = 3$ the first two rows both give $x = 0$ , and the third becomes $y = z$ . Each eigenspace has dimension $1$ , in agreement with the fact that a simple eigenvalue has $m_{g} = m_{a} = 1$ .

Exercise 2

Determine the eigenvalues, the eigenspaces and the multiplicities of the matrix

A = 201010102 .

p_{A} (λ) = 2 - λ 01 0 1 - λ 0 10 2 - λ = (1 - λ) 2 - λ 1 1 2 - λ = (1 - λ) [(2 - λ)^{2} - 1] = (1 - λ) (1 - λ) (3 - λ) = (1 - λ)^{2} (3 - λ)

m_{a} (1) = 2, m_{a} (3) = 1

A - I_{3} = 101000101, rk (A - I_{3}) = 1, m_{g} (1) = 3 - 1 = 2

x + z = 0 ⟹ (x, y, z) = (- z, y, z) = y (0, 1, 0) + z (- 1, 0, 1) ⟹ V_{1} = ⟨(0, 1, 0), (- 1, 0, 1)⟩

A - 3 I_{3} = - 1 01 0 - 2 0 10 - 1 ⟹ {- x + z = 0 - 2 y = 0 ⟹ V_{3} = ⟨(1, 0, 1)⟩

The second row of $A - λ I_{3}$ has a single nonzero entry, therefore the Laplace expansion along that row is convenient: the polynomial is obtained with the factor $1 - λ$ already factored out. The difference of squares $(2 - λ)^{2} - 1$ factors as $(2 - λ - 1) (2 - λ + 1) = (1 - λ) (3 - λ)$ , so the eigenvalue $1$ is double and the eigenvalue $3$ is simple. For $λ = 1$ the matrix $A - I_{3}$ has equal first and third rows and a zero second row, therefore it has rank $1$ and the eigenspace has dimension $3 - 1 = 2$ : it is the plane $x + z = 0$ , and the free parameters $y$ and $z$ give the two generators. For $λ = 3$ the equations give $y = 0$ and $x = z$ , therefore the eigenspace is the line spanned by $(1, 0, 1)$ . In this case $m_{g} = m_{a}$ for both eigenvalues.

Exercise 3

Determine the eigenvalues and eigenspaces of the matrix

A = 011101110 .

p_{A} (λ) = - λ 11 1 - λ 1 11 - λ = - λ^{3} + 1 + 1 - (- λ - λ - λ) = - λ^{3} + 3 λ + 2

p_{A} (- 1) = 1 - 3 + 2 = 0 ⟹ - λ^{3} + 3 λ + 2 = - (λ + 1) (λ^{2} - λ - 2) = - (λ + 1)^{2} (λ - 2)

m_{a} (- 1) = 2, m_{a} (2) = 1

A + I_{3} = 111111111 ⟹ x + y + z = 0 ⟹ V_{- 1} = ⟨(1, - 1, 0), (1, 0, - 1)⟩

A - 2 I_{3} = - 2 11 1 - 2 1 11 - 2 R_{1} \leftrightarrow R_{2} 1 - 2 1 - 2 11 11 - 2 R_{2} \to R_{2} + 2 R_{1} 101 - 2 - 3 1 13 - 2

R_{3} \to R_{3} - R_{1} 100 - 2 - 3 3 13 - 3 R_{3} \to R_{3} + R_{2} 100 - 2 - 3 0 130

{x - 2 y + z = 0 - 3 y + 3 z = 0 ⟹ {y = z x = z ⟹ V_{2} = ⟨(1, 1, 1)⟩

No row has zeros, therefore the determinant is computed with the rule of Sarrus: the three products of the main diagonals give $- λ^{3} + 1 + 1$ , and the three products of the anti-diagonals are each equal to $- λ$ . The polynomial $- λ^{3} + 3 λ + 2$ has the integer root $- 1$ , sought among the divisors of the constant term, therefore with Ruffini’s rule one divides by $λ + 1$ and the quotient $λ^{2} - λ - 2$ has roots $- 1$ and $2$ . The eigenvalue $- 1$ is therefore double and the eigenvalue $2$ is simple; the check with the trace confirms the computation, since $- 1 - 1 + 2 = 0 = tr A$ . The matrix $A + I_{3}$ has all rows equal, therefore it has rank $1$ and the eigenspace $V_{- 1}$ is the plane $x + y + z = 0$ , of dimension $2$ ; setting $y = - 1, z = 0$ and then $y = 0, z = - 1$ one obtains the two generators. For $λ = 2$ row reduction gives rank $2$ and the eigenspace is the line spanned by $(1, 1, 1)$ , as confirmed by the computation $A (1, 1, 1) = (2, 2, 2)$ .