Worked examples · Sheet 1

Eigenvalues and eigenspaces of matrices of order 2

Theory used: Characteristic polynomial Eigenvalues as roots of the characteristic polynomial Trace, determinant and eigenvalues Eigenspace Computing an eigenspace Algebraic and geometric multiplicity

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Determine the eigenvalues and eigenspaces of the matrix

A = (1141) .

p_{A} (λ) = 1 - λ 1 4 1 - λ = (1 - λ)^{2} - 4 = λ^{2} - 2 λ - 3 = (λ - 3) (λ + 1)

λ_{1} = 3, λ_{2} = - 1

A - 3 I_{2} = (- 2 1 4 - 2) R_{1} \to R_{1} + 2 R_{2} (01 0 - 2) ⟹ x - 2 y = 0 ⟹ V_{3} = ⟨(2, 1)⟩

A + I_{2} = (2142) R_{1} \to R_{1} - 2 R_{2} (0102) ⟹ x + 2 y = 0 ⟹ V_{- 1} = ⟨(- 2, 1)⟩

A (21) = (63) = 3 (21), A (- 2 1) = (2 - 1) = - 1 \cdot (- 2 1)

y = \dfrac{x}{2}

y = \dfrac{-x}{2}

The eigenspaces

V_{3}

and

V_{- 1}

are the lines through the origin spanned by

u = (2, 1)

and by

w = (- 2, 1)

The characteristic polynomial is obtained by subtracting $λ$ from the diagonal entries and computing the determinant of order $2$ . The polynomial $λ^{2} - 2 λ - 3$ has roots $3$ and $- 1$ , and the check with trace and determinant confirms the result, since $3 + (- 1) = 2 = tr A$ and $3 \cdot (- 1) = - 3 = det A$ . For each eigenvalue one solves the homogeneous system with matrix $A - λ I_{2}$ : the two rows of this matrix are proportional, because its determinant is zero, therefore the system reduces to a single equation. From the equation $x - 2 y = 0$ one gets $x = 2 y$ , and with $y = 1$ one finds the generator $(2, 1)$ ; from the equation $x + 2 y = 0$ one gets the generator $(- 2, 1)$ . The final check $A v = λ v$ confirms that the two vectors are eigenvectors. Both eigenspaces have dimension $1$ .

Exercise 2

Determine the eigenvalues and eigenspaces of the matrix

A = (66 - 2 - 1) .

tr A = 5, det A = - 6 + 12 = 6

p_{A} (λ) = λ^{2} - 5 λ + 6 = (λ - 2) (λ - 3)

A - 2 I_{2} = (46 - 2 - 3) ⟹ 2 x - y = 0 ⟹ V_{2} = ⟨(1, 2)⟩

A - 3 I_{2} = (36 - 2 - 4) ⟹ 3 x - 2 y = 0 ⟹ V_{3} = ⟨(2, 3)⟩

A (12) = (24), A (23) = (69)

For a matrix of order $2$ the characteristic polynomial can be written directly as $λ^{2} - (tr A) λ + det A$ , therefore trace and determinant are sufficient. The roots of $λ^{2} - 5 λ + 6$ are $2$ and $3$ . The matrix $A - 2 I_{2}$ has its second row equal to the first multiplied by $\frac{3}{2}$ , so the system reduces to the equation $4 x - 2 y = 0$ , that is, $2 x - y = 0$ , and the generator with $x = 1$ is $(1, 2)$ . In the same way the matrix $A - 3 I_{2}$ gives the equation $3 x - 2 y = 0$ , satisfied by $(2, 3)$ . The checks show that $A (1, 2) = 2 (1, 2)$ and $A (2, 3) = 3 (2, 3)$ , therefore $V_{2} = ⟨(1, 2)⟩$ and $V_{3} = ⟨(2, 3)⟩$ .

Exercise 3

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

A = (3 - 1 11) .

p_{A} (λ) = (3 - λ) (1 - λ) + 1 = λ^{2} - 4 λ + 4 = (λ - 2)^{2}

λ = 2, m_{a} (2) = 2

A - 2 I_{2} = (1 - 1 1 - 1) R_{2} \to R_{2} + R_{1} (1010), rk (A - 2 I_{2}) = 1

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

The characteristic polynomial is a perfect square, therefore the matrix has the single eigenvalue $2$ , a double root: $m_{a} (2) = 2$ . The geometric multiplicity is the dimension of the eigenspace, that is, $2 - rk (A - 2 I_{2})$ . The matrix $A - 2 I_{2}$ is not zero and has opposite rows, so its rank is $1$ and $m_{g} (2) = 1$ . The eigenspace is the line $x + y = 0$ , spanned by $(1, - 1)$ ; indeed $A (1, - 1) = (2, - 2) = 2 (1, - 1)$ . In this case $m_{g} (2) = 1 < 2 = m_{a} (2)$ , therefore the eigenvectors of $A$ all lie on a single line and there is no basis of $R^{2}$ made of eigenvectors of $A$ .