Worked examples · Sheet 3

Diagonalization with the diagonalizing matrix

Theory used: Diagonalization procedure Diagonalizability and bases of eigenvectors Diagonalizability criterion Distinct real eigenvalues Computing an eigenspace

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

Determine whether the matrix

A = (2123)

is diagonalizable and, if so, find an invertible $P$ and a diagonal $D$ such that $P^{- 1} A P = D$ .

p_{A} (λ) = λ^{2} - 5 λ + 4 = (λ - 1) (λ - 4)

A - I_{2} = (1122) ⟹ x + 2 y = 0 ⟹ V_{1} = ⟨(- 2, 1)⟩

A - 4 I_{2} = (- 2 1 2 - 1) ⟹ x - y = 0 ⟹ V_{4} = ⟨(1, 1)⟩

P = (- 2 1 11), D = (1004), P^{- 1} = \frac{1}{3} (- 1 1 12)

A P = (- 2 1 44) = P D

The trace and the determinant are $5$ and $4$ , therefore the characteristic polynomial is $λ^{2} - 5 λ + 4$ , with roots $1$ and $4$ . The matrix has two distinct real eigenvalues, so it is diagonalizable with no further checks. The columns of $P$ are one generator for each eigenspace, and the diagonal of $D$ contains the eigenvalues in the same order: the first column $(- 2, 1)$ is associated with $1$ , the second $(1, 1)$ is associated with $4$ . The matrix $P$ is invertible, since $det P = - 2 - 1 = - 3 \neq = 0$ , and its inverse is obtained by swapping the entries of the main diagonal, changing the sign of the other two and dividing by the determinant. The check $A P = P D$ confirms the result without having to compute the product $P^{- 1} A P$ .

Exercise 2

Diagonalize the matrix

A = 10 - 2 214104 .

p_{A} (λ) = (1 - λ) 1 - λ - 2 1 4 - λ = (1 - λ) [(1 - λ) (4 - λ) + 2] = (1 - λ) (λ^{2} - 5 λ + 6) = (1 - λ) (λ - 2) (λ - 3)

A - I_{3} = 00 - 2 204103 ⟹ {2 y + z = 0 - 2 x + 4 y + 3 z = 0 ⟹ {z = - 2 y x = - y ⟹ V_{1} = ⟨(1, - 1, 2)⟩

A - 2 I_{3} = - 1 0 - 2 2 - 1 4 102 ⟹ {y = 0 x = z ⟹ V_{2} = ⟨(1, 0, 1)⟩

A - 3 I_{3} = - 2 0 - 2 2 - 2 4 101 ⟹ {y = 0 z = 2 x ⟹ V_{3} = ⟨(1, 0, 2)⟩

P = 1 - 1 2 101102, D = 100020003

A P = 1 - 1 2 202306 = P D

The second row of $A - λ I_{3}$ contains only the entry $1 - λ$ , therefore the Laplace expansion along that row immediately gives the factor $1 - λ$ multiplied by a determinant of order $2$ . The trinomial $λ^{2} - 5 λ + 6$ has roots $2$ and $3$ , so the eigenvalues are $1$ , $2$ , $3$ , distinct, and the matrix is diagonalizable. For $λ = 1$ the first equation gives $z = - 2 y$ ; substituting into the second one obtains $- 2 x + 4 y - 6 y = 0$ , that is, $x = - y$ , and with $y = - 1$ one finds $(1, - 1, 2)$ . For $λ = 2$ and $λ = 3$ the second row gives $y = 0$ , and the first row becomes $- x + z = 0$ and $- 2 x + z = 0$ respectively. The columns of $P$ are the three generators in the order of the eigenvalues $1, 2, 3$ ; the product $A P$ has as columns the same vectors multiplied by $1$ , $2$ and $3$ , therefore it coincides with $P D$ .

Exercise 3

Determine whether the matrix

A = 301211 - 2 00

is diagonalizable and, if so, find a diagonalizing matrix.

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

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

A - I_{3} = 201201 - 2 0 - 1, rk (A - I_{3}) = 1, m_{g} (1) = 3 - 1 = 2 = m_{a} (1)

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

A - 2 I_{3} = 101 2 - 1 1 - 2 0 - 2 ⟹ {y = 0 x = 2 z ⟹ V_{2} = ⟨(2, 0, 1)⟩

P = 101011201, D = 100010002, det P = - 1

The expansion along the second row gives the factor $1 - λ$ , and the trinomial $λ^{2} - 3 λ + 2$ again has the root $1$ , in addition to $2$ : the eigenvalue $1$ is therefore double. By the diagonalizability criterion one must check that $m_{g} (1) = 2$ . The matrix $A - I_{3}$ has its first row equal to twice the third and a zero second row, so it has rank $1$ and the eigenspace $V_{1}$ has dimension $2$ , equal to the algebraic multiplicity; the eigenvalue $2$ is simple and requires no checks. The matrix is therefore diagonalizable. A basis of $V_{1}$ is obtained from the equation $z = x + y$ by setting first $x = 1, y = 0$ and then $x = 0, y = 1$ . For $λ = 2$ the second row gives $y = 0$ and the first becomes $x - 2 z = 0$ . Since $det P = - 1 \neq = 0$ , the three columns form a basis of $R^{3}$ made of eigenvectors, as guaranteed by the criterion.