Matrix multiplication order: A B AB A B B B B A A A 3x3 determinant sign errors: The cofactor expansion alternates signs + + + − - − + + + − - − Singular matrix checks: Before finding an inverse, always verify det ( A ) ≠ 0 \det(A) \neq 0 det ( A ) = 0 A x = b A\mathbf{x} = \mathbf{b} A x = b Eigenvectors are not unique: Any non-zero scalar multiple of an eigenvector is also an
eigenvector. When diagonalising, ensure consistency: the columns of P P P D D D Repeated eigenvalues: A repeated eigenvalue does not necessarily give two independent
eigenvectors. Check by attempting to solve ( A − λ I ) v = 0 (A - \lambda I)\mathbf{v} = \mathbf{0} ( A − λ I ) v = 0 9. Additional Exam-Style Questions Question 5 The matrix A = ( 3 1 − 1 1 ) A = \begin{pmatrix} 3 & 1 \\ -1 & 1 \end{pmatrix} A = ( 3 − 1 1 1 )
(a) Find the eigenvalues and eigenvectors of A A A
(b) Write down a matrix P P P D D D P − 1 A P = D P^{-1}AP = D P − 1 A P = D
(c) Hence find A 5 A^5 A 5
Solution (a) Characteristic equation: ( 3 − λ ) ( 1 − λ ) + 1 = 0 (3 - \lambda)(1 - \lambda) + 1 = 0 ( 3 − λ ) ( 1 − λ ) + 1 = 0
λ 2 − 4 λ + 4 = 0 ⟹ ( λ − 2 ) 2 = 0 \lambda^2 - 4\lambda + 4 = 0 \implies (\lambda - 2)^2 = 0 λ 2 − 4 λ + 4 = 0 ⟹ ( λ − 2 ) 2 = 0
λ = 2 \lambda = 2 λ = 2
( 1 1 − 1 − 1 ) ( x y ) = 0 ⟹ x + y = 0 \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{0} \implies x + y = 0 ( 1 − 1 1 − 1 ) ( x y ) = 0 ⟹ x + y = 0
Eigenvector: ( 1 − 1 ) \begin{pmatrix} 1 \\ -1 \end{pmatrix} ( 1 − 1 )
Only one independent eigenvector, so A A A
(b) Since A A A P P P D D D
(c) For A n A^n A n 2 × 2 2 \times 2 2 × 2 λ \lambda λ
A n = λ n I + n λ n − 1 ( A − λ I ) A^n = \lambda^n I + n\lambda^{n-1}(A - \lambda I) A n = λ n I + n λ n − 1 ( A − λ I )
A − 2 I = ( 1 1 − 1 − 1 ) A - 2I = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix} A − 2 I = ( 1 − 1 1 − 1 )
A 5 = 2 5 I + 5 ⋅ 2 4 ( 1 1 − 1 − 1 ) = ( 32 0 0 32 ) + ( 80 80 − 80 − 80 ) A^5 = 2^5 I + 5 \cdot 2^4 \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix} = \begin{pmatrix} 32 & 0 \\ 0 & 32 \end{pmatrix} + \begin{pmatrix} 80 & 80 \\ -80 & -80 \end{pmatrix} A 5 = 2 5 I + 5 ⋅ 2 4 ( 1 − 1 1 − 1 ) = ( 32 0 0 32 ) + ( 80 − 80 80 − 80 )
= ( 112 80 − 80 − 48 ) = \begin{pmatrix} 112 & 80 \\ -80 & -48 \end{pmatrix} = ( 112 − 80 80 − 48 )
Question 6 (a) Find the 3 × 3 3 \times 3 3 × 3 M M M 90 ∘ 90^\circ 9 0 ∘ x x x
(b) Verify that det ( M ) = 1 \det(M) = 1 det ( M ) = 1
(c) The point ( 1 , 1 , 0 ) (1, 1, 0) ( 1 , 1 , 0 ) M M M
Solution (a) A rotation of θ \theta θ x x x x x x y y y z z z
M = ( 1 0 0 0 cos 90 ∘ − sin 90 ∘ 0 sin 90 ∘ cos 90 ∘ ) = ( 1 0 0 0 0 − 1 0 1 0 ) M = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos 90^\circ & -\sin 90^\circ \\ 0 & \sin 90^\circ & \cos 90^\circ \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix} M = 1 0 0 0 cos 9 0 ∘ sin 9 0 ∘ 0 − sin 9 0 ∘ cos 9 0 ∘ = 1 0 0 0 0 1 0 − 1 0
(b) Expanding along the first row:
det M = 1 ∣ 0 − 1 1 0 ∣ − 0 + 0 = 0 − ( − 1 ) = 1 \det M = 1\begin{vmatrix} 0 & -1 \\ 1 & 0 \end{vmatrix} - 0 + 0 = 0 - (-1) = 1 det M = 1 0 1 − 1 0 − 0 + 0 = 0 − ( − 1 ) = 1
(c)
( 1 0 0 0 0 − 1 0 1 0 ) ( 1 1 0 ) = ( 1 0 1 ) \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} 1 0 0 0 0 1 0 − 1 0 1 1 0 = 1 0 1
The image is ( 1 , 0 , 1 ) (1, 0, 1) ( 1 , 0 , 1 )
Question 7 The transformation T T T A = ( 2 3 0 2 ) A = \begin{pmatrix} 2 & 3 \\ 0 & 2 \end{pmatrix} A = ( 2 0 3 2 )
(a) Find the invariant points of T T T
(b) Show that the line y = 0 y = 0 y = 0 T T T
(c) Find another invariant line of T T T
Solution (a) Invariant points satisfy A x = x A\mathbf{x} = \mathbf{x} A x = x
( 2 3 0 2 ) ( x y ) = ( x y ) \begin{pmatrix} 2 & 3 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix} ( 2 0 3 2 ) ( x y ) = ( x y )
2 x + 3 y = x ⟹ x + 3 y = 0 2x + 3y = x \implies x + 3y = 0 2 x + 3 y = x ⟹ x + 3 y = 0 2 y = y ⟹ y = 0 2y = y \implies y = 0 2 y = y ⟹ y = 0
So x = 0 x = 0 x = 0 y = 0 y = 0 y = 0
(b) Points on y = 0 y = 0 y = 0 ( x , 0 ) (x, 0) ( x , 0 )
( 2 3 0 2 ) ( x 0 ) = ( 2 x 0 ) \begin{pmatrix} 2 & 3 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} x \\ 0 \end{pmatrix} = \begin{pmatrix} 2x \\ 0 \end{pmatrix} ( 2 0 3 2 ) ( x 0 ) = ( 2 x 0 )
The image ( 2 x , 0 ) (2x, 0) ( 2 x , 0 ) y = 0 y = 0 y = 0 y = 0 y = 0 y = 0
(c) For an invariant line y = m x y = mx y = m x A ( 1 m ) = λ ( 1 m ) A\begin{pmatrix} 1 \\ m \end{pmatrix} = \lambda\begin{pmatrix} 1 \\ m \end{pmatrix} A ( 1 m ) = λ ( 1 m )
2 + 3 m = λ 2 + 3m = \lambda 2 + 3 m = λ 2 m = λ m 2m = \lambda m 2 m = λm
From the second equation: m ( 2 − λ ) = 0 m(2 - \lambda) = 0 m ( 2 − λ ) = 0
If m = 0 m = 0 m = 0 y = 0 y = 0 y = 0
If λ = 2 \lambda = 2 λ = 2 2 + 3 m = 2 ⟹ m = 0 2 + 3m = 2 \implies m = 0 2 + 3 m = 2 ⟹ m = 0
For a line not through the origin, try y = m x + c y = mx + c y = m x + c c ≠ 0 c \neq 0 c = 0
( 2 3 0 2 ) ( x m x + c ) = ( ( 2 + 3 m ) x + 3 c 2 m x + 2 c ) \begin{pmatrix} 2 & 3 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} x \\ mx + c \end{pmatrix} = \begin{pmatrix} (2 + 3m)x + 3c \\ 2mx + 2c \end{pmatrix} ( 2 0 3 2 ) ( x m x + c ) = ( ( 2 + 3 m ) x + 3 c 2 m x + 2 c )
For this to lie on y = m x + c y = mx + c y = m x + c 2 m x + 2 c = m ( 2 + 3 m ) x + 3 m c + c 2mx + 2c = m(2 + 3m)x + 3mc + c 2 m x + 2 c = m ( 2 + 3 m ) x + 3 m c + c
Comparing coefficients: 2 m = m ( 2 + 3 m ) ⟹ 3 m 2 = 0 ⟹ m = 0 2m = m(2 + 3m) \implies 3m^2 = 0 \implies m = 0 2 m = m ( 2 + 3 m ) ⟹ 3 m 2 = 0 ⟹ m = 0
Then: 2 c = c ⟹ c = 0 2c = c \implies c = 0 2 c = c ⟹ c = 0
The only invariant line is y = 0 y = 0 y = 0
10. Advanced Worked Examples Example 10.1: 3x3 eigenvalues and eigenvectors Problem. Find the eigenvalues and eigenvectors of
A = ( 2 1 0 1 3 1 0 1 2 ) A = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 2 \end{pmatrix} A = 2 1 0 1 3 1 0 1 2
Solution. Characteristic equation:
det ( A − λ I ) = ∣ 2 − λ 1 0 1 3 − λ 1 0 1 2 − λ ∣ = 0 \det(A - \lambda I) = \begin{vmatrix} 2-\lambda & 1 & 0 \\ 1 & 3-\lambda & 1 \\ 0 & 1 & 2-\lambda \end{vmatrix} = 0 det ( A − λ I ) = 2 − λ 1 0 1 3 − λ 1 0 1 2 − λ = 0
Expanding along the first row:
( 2 − λ ) ∣ 3 − λ 1 1 2 − λ ∣ − 1 ∣ 1 1 0 2 − λ ∣ + 0 (2-\lambda)\begin{vmatrix} 3-\lambda & 1 \\ 1 & 2-\lambda \end{vmatrix} - 1\begin{vmatrix} 1 & 1 \\ 0 & 2-\lambda \end{vmatrix} + 0 ( 2 − λ ) 3 − λ 1 1 2 − λ − 1 1 0 1 2 − λ + 0
= ( 2 − λ ) [ ( 3 − λ ) ( 2 − λ ) − 1 ] − ( 2 − λ ) = (2-\lambda)[(3-\lambda)(2-\lambda)-1] - (2-\lambda) = ( 2 − λ ) [( 3 − λ ) ( 2 − λ ) − 1 ] − ( 2 − λ )
= ( 2 − λ ) [ ( 3 − λ ) ( 2 − λ ) − 2 ] = ( 2 − λ ) [ λ 2 − 5 λ + 4 ] = ( 2 − λ ) ( λ − 1 ) ( λ − 4 ) = (2-\lambda)[(3-\lambda)(2-\lambda) - 2] = (2-\lambda)[\lambda^2 - 5\lambda + 4] = (2-\lambda)(\lambda-1)(\lambda-4) = ( 2 − λ ) [( 3 − λ ) ( 2 − λ ) − 2 ] = ( 2 − λ ) [ λ 2 − 5 λ + 4 ] = ( 2 − λ ) ( λ − 1 ) ( λ − 4 )
Eigenvalues: λ 1 = 1 \lambda_1 = 1 λ 1 = 1 λ 2 = 2 \lambda_2 = 2 λ 2 = 2 λ 3 = 4 \lambda_3 = 4 λ 3 = 4
For λ 1 = 1 \lambda_1 = 1 λ 1 = 1
( 1 1 0 1 2 1 0 1 1 ) ( x y z ) = 0 \begin{pmatrix} 1 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{0} 1 1 0 1 2 1 0 1 1 x y z = 0
x + y = 0 x + y = 0 x + y = 0 x + 2 y + z = 0 x + 2y + z = 0 x + 2 y + z = 0 y + z = 0 y + z = 0 y + z = 0 x = − y x = -y x = − y z = − y z = -y z = − y ( 1 − 1 1 ) \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} 1 − 1 1
For λ 2 = 2 \lambda_2 = 2 λ 2 = 2
( 0 1 0 1 1 1 0 1 0 ) ( x y z ) = 0 \begin{pmatrix} 0 & 1 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{0} 0 1 0 1 1 1 0 1 0 x y z = 0
y = 0 y = 0 y = 0 x + z = 0 x + z = 0 x + z = 0 ( 1 0 − 1 ) \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} 1 0 − 1
For λ 3 = 4 \lambda_3 = 4 λ 3 = 4
( − 2 1 0 1 − 1 1 0 1 − 2 ) ( x y z ) = 0 \begin{pmatrix} -2 & 1 & 0 \\ 1 & -1 & 1 \\ 0 & 1 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{0} − 2 1 0 1 − 1 1 0 1 − 2 x y z = 0
− 2 x + y = 0 ⟹ y = 2 x -2x + y = 0 \implies y = 2x − 2 x + y = 0 ⟹ y = 2 x y − 2 z = 0 ⟹ z = x y - 2z = 0 \implies z = x y − 2 z = 0 ⟹ z = x ( 1 2 1 ) \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} 1 2 1
Example 10.2: Diagonalisation of a 3x3 matrix Problem. Using the eigenvalues and eigenvectors from Example 10.1, diagonalise A A A A 4 A^4 A 4
Solution. P = ( 1 1 1 − 1 0 2 1 − 1 1 ) P = \begin{pmatrix} 1 & 1 & 1 \\ -1 & 0 & 2 \\ 1 & -1 & 1 \end{pmatrix} P = 1 − 1 1 1 0 − 1 1 2 1 D = ( 1 0 0 0 2 0 0 0 4 ) D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{pmatrix} D = 1 0 0 0 2 0 0 0 4
det P = 1 ( 0 − ( − 2 ) ) − 1 ( ( − 1 ) − 2 ) + 1 ( 1 − 0 ) = 2 + 3 + 1 = 6 \det P = 1(0-(-2)) - 1((-1)-2) + 1(1-0) = 2 + 3 + 1 = 6 det P = 1 ( 0 − ( − 2 )) − 1 (( − 1 ) − 2 ) + 1 ( 1 − 0 ) = 2 + 3 + 1 = 6
P − 1 = 1 6 ( 2 0 2 3 0 − 3 1 2 1 ) P^{-1} = \frac{1}{6}\begin{pmatrix} 2 & 0 & 2 \\ 3 & 0 & -3 \\ 1 & 2 & 1 \end{pmatrix} P − 1 = 6 1 2 3 1 0 0 2 2 − 3 1
A 4 = P D 4 P − 1 A^4 = PD^4P^{-1} A 4 = P D 4 P − 1
D 4 = ( 1 0 0 0 16 0 0 0 256 ) D^4 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 256 \end{pmatrix} D 4 = 1 0 0 0 16 0 0 0 256
P D 4 = ( 1 16 256 − 1 0 512 1 − 16 256 ) PD^4 = \begin{pmatrix} 1 & 16 & 256 \\ -1 & 0 & 512 \\ 1 & -16 & 256 \end{pmatrix} P D 4 = 1 − 1 1 16 0 − 16 256 512 256
A 4 = 1 6 ( 1 16 256 − 1 0 512 1 − 16 256 ) ( 2 0 2 3 0 − 3 1 2 1 ) A^4 = \frac{1}{6}\begin{pmatrix} 1 & 16 & 256 \\ -1 & 0 & 512 \\ 1 & -16 & 256 \end{pmatrix}\begin{pmatrix} 2 & 0 & 2 \\ 3 & 0 & -3 \\ 1 & 2 & 1 \end{pmatrix} A 4 = 6 1 1 − 1 1 16 0 − 16 256 512 256 2 3 1 0 0 2 2 − 3 1
= 1 6 ( 2 + 48 + 256 512 − 768 + 512 2 − 48 + 256 − 2 + 512 1024 − 2 − 768 + 512 2 − 48 + 256 512 + 256 2 + 48 + 256 ) = \frac{1}{6}\begin{pmatrix} 2+48+256 & 512-768+512 & 2-48+256 \\ -2+512 & 1024 & -2-768+512 \\ 2-48+256 & 512+256 & 2+48+256 \end{pmatrix} = 6 1 2 + 48 + 256 − 2 + 512 2 − 48 + 256 512 − 768 + 512 1024 512 + 256 2 − 48 + 256 − 2 − 768 + 512 2 + 48 + 256
= 1 6 ( 306 256 210 510 1024 − 258 210 768 306 ) = ( 51 128 / 3 35 85 512 / 3 − 43 35 128 51 ) = \frac{1}{6}\begin{pmatrix} 306 & 256 & 210 \\ 510 & 1024 & -258 \\ 210 & 768 & 306 \end{pmatrix} = \begin{pmatrix} 51 & 128/3 & 35 \\ 85 & 512/3 & -43 \\ 35 & 128 & 51 \end{pmatrix} = 6 1 306 510 210 256 1024 768 210 − 258 306 = 51 85 35 128/3 512/3 128 35 − 43 51
Example 10.3: Reflection in an arbitrary line Problem. Find the 2 × 2 2 \times 2 2 × 2 y = m x y = mx y = m x
Solution. The line y = m x y = mx y = m x θ = arctan m \theta = \arctan m θ = arctan m x x x
Rotate by − θ -\theta − θ x x x
Reflect in the x x x
Rotate back by θ \theta θ
R − θ = ( cos θ sin θ sin θ − cos θ ) R_{-\theta} = \begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix} R − θ = ( cos θ sin θ sin θ − cos θ )
Wait -- the reflection matrix in a line at angle θ \theta θ x x x
M = ( cos 2 θ sin 2 θ sin 2 θ − cos 2 θ ) M = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix} M = ( cos 2 θ sin 2 θ sin 2 θ − cos 2 θ )
This can be derived as R θ ( 1 0 0 − 1 ) R − θ R_\theta \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} R_{-\theta} R θ ( 1 0 0 − 1 ) R − θ
For m = 1 m = 1 m = 1 θ = π / 4 \theta = \pi/4 θ = π /4 M = ( 0 1 1 0 ) M = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} M = ( 0 1 1 0 ) y = x y = x y = x
Problem. The transformation T T T A = ( 3 − 4 1 − 1 ) A = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} A = ( 3 1 − 4 − 1 ) T T T
Solution. Characteristic equation: ( 3 − λ ) ( − 1 − λ ) + 4 = 0 (3-\lambda)(-1-\lambda) + 4 = 0 ( 3 − λ ) ( − 1 − λ ) + 4 = 0
− 3 − 3 λ + λ + λ 2 + 4 = 0 ⟹ λ 2 − 2 λ + 1 = 0 ⟹ ( λ − 1 ) 2 = 0 -3 - 3\lambda + \lambda + \lambda^2 + 4 = 0 \implies \lambda^2 - 2\lambda + 1 = 0 \implies (\lambda-1)^2 = 0 − 3 − 3 λ + λ + λ 2 + 4 = 0 ⟹ λ 2 − 2 λ + 1 = 0 ⟹ ( λ − 1 ) 2 = 0
λ = 1 \lambda = 1 λ = 1 ( A − I ) v = 0 (A-I)\mathbf{v} = \mathbf{0} ( A − I ) v = 0
( 2 − 4 1 − 2 ) ( x y ) = 0 ⟹ x − 2 y = 0 \begin{pmatrix} 2 & -4 \\ 1 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{0} \implies x - 2y = 0 ( 2 1 − 4 − 2 ) ( x y ) = 0 ⟹ x − 2 y = 0
Eigenvector: ( 2 1 ) \begin{pmatrix} 2 \\ 1 \end{pmatrix} ( 2 1 ) y = x / 2 y = x/2 y = x /2
For non-trivial invariant lines not through the origin, try y = m x + c y = mx + c y = m x + c c ≠ 0 c \neq 0 c = 0
( 3 − 4 1 − 1 ) ( x m x + c ) = ( ( 3 − 4 m ) x − 4 c ( 1 − m ) x − c ) \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ mx+c \end{pmatrix} = \begin{pmatrix} (3-4m)x - 4c \\ (1-m)x - c \end{pmatrix} ( 3 1 − 4 − 1 ) ( x m x + c ) = ( ( 3 − 4 m ) x − 4 c ( 1 − m ) x − c )
For this to lie on y = m x + c y = mx + c y = m x + c ( 1 − m ) x − c = m ( 3 − 4 m ) x − 4 m c + c (1-m)x - c = m(3-4m)x - 4mc + c ( 1 − m ) x − c = m ( 3 − 4 m ) x − 4 m c + c
Comparing coefficients of x x x 1 − m = m ( 3 − 4 m ) = 3 m − 4 m 2 1 - m = m(3 - 4m) = 3m - 4m^2 1 − m = m ( 3 − 4 m ) = 3 m − 4 m 2
4 m 2 − 4 m + 1 = 0 ⟹ ( 2 m − 1 ) 2 = 0 ⟹ m = 1 / 2 4m^2 - 4m + 1 = 0 \implies (2m - 1)^2 = 0 \implies m = 1/2 4 m 2 − 4 m + 1 = 0 ⟹ ( 2 m − 1 ) 2 = 0 ⟹ m = 1/2
Comparing constants: − c = − 4 m c + c ⟹ 4 m c = 2 c ⟹ c ( 2 m − 1 ) = 0 -c = -4mc + c \implies 4mc = 2c \implies c(2m - 1) = 0 − c = − 4 m c + c ⟹ 4 m c = 2 c ⟹ c ( 2 m − 1 ) = 0
Since m = 1 / 2 m = 1/2 m = 1/2 c ( 0 ) = 0 c(0) = 0 c ( 0 ) = 0 c c c
Therefore every line of the form y = x / 2 + c y = x/2 + c y = x /2 + c T T T
Example 10.5: Matrix representation of rotation about an arbitrary point Problem. Find the 3 × 3 3 \times 3 3 × 3 θ \theta θ ( a , b ) (a, b) ( a , b )
Solution. Using the homogeneous coordinate system where a point ( x , y ) (x, y) ( x , y ) ( x y 1 ) \begin{pmatrix}x\\y\\1\end{pmatrix} x y 1
Translate by ( − a , − b ) (-a, -b) ( − a , − b )
Rotate by θ \theta θ
Translate back by ( a , b ) (a, b) ( a , b )
M = ( 1 0 a 0 1 b 0 0 1 ) ( cos θ − sin θ 0 sin θ cos θ 0 0 0 1 ) ( 1 0 − a 0 1 − b 0 0 1 ) M = \begin{pmatrix} 1 & 0 & a \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & -a \\ 0 & 1 & -b \\ 0 & 0 & 1 \end{pmatrix} M = 1 0 0 0 1 0 a b 1 cos θ sin θ 0 − sin θ cos θ 0 0 0 1 1 0 0 0 1 0 − a − b 1
= ( cos θ − sin θ a ( 1 − cos θ ) + b sin θ sin θ cos θ b ( 1 − cos θ ) − a sin θ 0 0 1 ) = \begin{pmatrix} \cos\theta & -\sin\theta & a(1-\cos\theta)+b\sin\theta \\ \sin\theta & \cos\theta & b(1-\cos\theta)-a\sin\theta \\ 0 & 0 & 1 \end{pmatrix} = cos θ sin θ 0 − sin θ cos θ 0 a ( 1 − cos θ ) + b sin θ b ( 1 − cos θ ) − a sin θ 1
Example 10.6: Determinant and area of a triangle Problem. The vertices of a triangle are A ( 1 , 2 ) A(1, 2) A ( 1 , 2 ) B ( 4 , 6 ) B(4, 6) B ( 4 , 6 ) C ( 3 , − 1 ) C(3, -1) C ( 3 , − 1 )
Solution.
Area = 1 2 ∣ det ( 1 2 1 4 6 1 3 − 1 1 ) ∣ \text{Area} = \frac{1}{2}\left|\det\begin{pmatrix} 1 & 2 & 1 \\ 4 & 6 & 1 \\ 3 & -1 & 1 \end{pmatrix}\right| Area = 2 1 det 1 4 3 2 6 − 1 1 1 1
Expanding along the third column:
= 1 2 ∣ 1 ⋅ ∣ 4 6 3 − 1 ∣ − 1 ⋅ ∣ 1 2 3 − 1 ∣ + 1 ⋅ ∣ 1 2 4 6 ∣ ∣ = \dfrac{1}{2}\left|1\cdot\begin{vmatrix} 4 & 6 \\ 3 & -1 \end{vmatrix} - 1\cdot\begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} + 1\cdot\begin{vmatrix} 1 & 2 \\ 4 & 6 \end{vmatrix}\right| = 2 1 1 ⋅ 4 3 6 − 1 − 1 ⋅ 1 3 2 − 1 + 1 ⋅ 1 4 2 6
= 1 2 ∣ ( − 4 − 18 ) − ( − 1 − 6 ) + ( 6 − 8 ) ∣ = 1 2 ∣ − 22 + 7 − 2 ∣ = 17 2 = \dfrac{1}{2}|(-4-18) - (-1-6) + (6-8)| = \dfrac{1}{2}|-22 + 7 - 2| = \dfrac{17}{2} = 2 1 ∣ ( − 4 − 18 ) − ( − 1 − 6 ) + ( 6 − 8 ) ∣ = 2 1 ∣ − 22 + 7 − 2∣ = 2 17
Example 10.7: Solving a system using the inverse Problem. Solve the system x + 2 y + z = 4 x + 2y + z = 4 x + 2 y + z = 4 2 x + y + z = 3 2x + y + z = 3 2 x + y + z = 3 x + y + 2 z = 5 x + y + 2z = 5 x + y + 2 z = 5
Solution. The system is A x = b A\mathbf{x} = \mathbf{b} A x = b
A = ( 1 2 1 2 1 1 1 1 2 ) , b = ( 4 3 5 ) A = \begin{pmatrix} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 4 \\ 3 \\ 5 \end{pmatrix} A = 1 2 1 2 1 1 1 1 2 , b = 4 3 5
det A = 1 ( 2 − 1 ) − 2 ( 4 − 1 ) + 1 ( 2 − 1 ) = 1 − 6 + 1 = − 4 \det A = 1(2-1) - 2(4-1) + 1(2-1) = 1 - 6 + 1 = -4 det A = 1 ( 2 − 1 ) − 2 ( 4 − 1 ) + 1 ( 2 − 1 ) = 1 − 6 + 1 = − 4
Using Cramer's rule:
x = ◆ L B ◆ det ( 4 2 1 3 1 1 5 1 2 ) ◆ R B ◆◆ L B ◆ − 4 ◆ R B ◆ = 4 ( 2 − 1 ) − 2 ( 6 − 5 ) + 1 ( 3 − 5 ) − 4 = 4 − 2 − 2 − 4 = 0 x = \frac◆LB◆\det\begin{pmatrix} 4 & 2 & 1 \\ 3 & 1 & 1 \\ 5 & 1 & 2 \end{pmatrix}◆RB◆◆LB◆-4◆RB◆ = \frac{4(2-1) - 2(6-5) + 1(3-5)}{-4} = \frac{4 - 2 - 2}{-4} = 0 x = L ◆ B ◆ det 4 3 5 2 1 1 1 1 2 ◆ R B ◆◆ L B ◆ − 4◆ R B ◆ = − 4 4 ( 2 − 1 ) − 2 ( 6 − 5 ) + 1 ( 3 − 5 ) = − 4 4 − 2 − 2 = 0
y = ◆ L B ◆ det ( 1 4 1 2 3 1 1 5 2 ) ◆ R B ◆◆ L B ◆ − 4 ◆ R B ◆ = 1 ( 6 − 5 ) − 4 ( 4 − 1 ) + 1 ( 10 − 3 ) − 4 = 1 − 12 + 7 − 4 = 1 y = \frac◆LB◆\det\begin{pmatrix} 1 & 4 & 1 \\ 2 & 3 & 1 \\ 1 & 5 & 2 \end{pmatrix}◆RB◆◆LB◆-4◆RB◆ = \frac{1(6-5) - 4(4-1) + 1(10-3)}{-4} = \frac{1 - 12 + 7}{-4} = 1 y = L ◆ B ◆ det 1 2 1 4 3 5 1 1 2 ◆ R B ◆◆ L B ◆ − 4◆ R B ◆ = − 4 1 ( 6 − 5 ) − 4 ( 4 − 1 ) + 1 ( 10 − 3 ) = − 4 1 − 12 + 7 = 1
z = ◆ L B ◆ det ( 1 2 4 2 1 3 1 1 5 ) ◆ R B ◆◆ L B ◆ − 4 ◆ R B ◆ = 1 ( 5 − 3 ) − 2 ( 10 − 3 ) + 4 ( 2 − 1 ) − 4 = 2 − 14 + 4 − 4 = 2 z = \frac◆LB◆\det\begin{pmatrix} 1 & 2 & 4 \\ 2 & 1 & 3 \\ 1 & 1 & 5 \end{pmatrix}◆RB◆◆LB◆-4◆RB◆ = \frac{1(5-3) - 2(10-3) + 4(2-1)}{-4} = \frac{2 - 14 + 4}{-4} = 2 z = L ◆ B ◆ det 1 2 1 2 1 1 4 3 5 ◆ R B ◆◆ L B ◆ − 4◆ R B ◆ = − 4 1 ( 5 − 3 ) − 2 ( 10 − 3 ) + 4 ( 2 − 1 ) = − 4 2 − 14 + 4 = 2
Solution: x = 0 x = 0 x = 0 y = 1 y = 1 y = 1 z = 2 z = 2 z = 2
11. Connections to Other Topics 11.1 Matrices and complex numbers Complex numbers a + b i a + bi a + bi ( a − b b a ) \begin{pmatrix}a & -b\\b & a\end{pmatrix} ( a b − b a ) ∣ z ∣ 2 = det |z|^2 = \det ∣ z ∣ 2 = det Complex Numbers .
11.2 Matrices and vectors The cross product a × b \mathbf{a}\times\mathbf{b} a × b i , j , k \mathbf{i}, \mathbf{j}, \mathbf{k} i , j , k Vectors in 3D .
11.3 Eigenvalues and differential equations Diagonalisation is used to solve systems of coupled linear differential equations. The eigenvalues
determine the form of the solution. See
Differential Equations .
12. Additional Exam-Style Questions Question 8 The matrix B = ( 1 0 2 0 2 0 2 0 1 ) B = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 2 & 0 \\ 2 & 0 & 1 \end{pmatrix} B = 1 0 2 0 2 0 2 0 1
(a) Find the eigenvalues and eigenvectors of B B B
(b) Verify that B B B P P P D D D
Solution (a)
det ( B − λ I ) = ∣ 1 − λ 0 2 0 2 − λ 0 2 0 1 − λ ∣ = ( 2 − λ ) [ ( 1 − λ ) 2 − 4 ] \det(B - \lambda I) = \begin{vmatrix} 1-\lambda & 0 & 2 \\ 0 & 2-\lambda & 0 \\ 2 & 0 & 1-\lambda \end{vmatrix} = (2-\lambda)[(1-\lambda)^2 - 4] det ( B − λ I ) = 1 − λ 0 2 0 2 − λ 0 2 0 1 − λ = ( 2 − λ ) [( 1 − λ ) 2 − 4 ]
= ( 2 − λ ) ( λ 2 − 2 λ − 3 ) = ( 2 − λ ) ( λ − 3 ) ( λ + 1 ) = (2-\lambda)(\lambda^2 - 2\lambda - 3) = (2-\lambda)(\lambda-3)(\lambda+1) = ( 2 − λ ) ( λ 2 − 2 λ − 3 ) = ( 2 − λ ) ( λ − 3 ) ( λ + 1 )
Eigenvalues: λ 1 = − 1 \lambda_1 = -1 λ 1 = − 1 λ 2 = 2 \lambda_2 = 2 λ 2 = 2 λ 3 = 3 \lambda_3 = 3 λ 3 = 3
λ = − 1 \lambda = -1 λ = − 1 ( 2 0 2 0 3 0 2 0 2 ) v = 0 ⟹ x + z = 0 , y = 0 \begin{pmatrix} 2 & 0 & 2 \\ 0 & 3 & 0 \\ 2 & 0 & 2 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies x + z = 0, y = 0 2 0 2 0 3 0 2 0 2 v = 0 ⟹ x + z = 0 , y = 0 ( 1 0 − 1 ) \begin{pmatrix}1\\0\\-1\end{pmatrix} 1 0 − 1
λ = 2 \lambda = 2 λ = 2 ( − 1 0 2 0 0 0 2 0 − 1 ) v = 0 ⟹ x = 2 z \begin{pmatrix} -1 & 0 & 2 \\ 0 & 0 & 0 \\ 2 & 0 & -1 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies x = 2z − 1 0 2 0 0 0 2 0 − 1 v = 0 ⟹ x = 2 z ( 2 1 1 ) \begin{pmatrix}2\\1\\1\end{pmatrix} 2 1 1 y y y
Actually: − x + 2 z = 0 ⟹ x = 2 z -x + 2z = 0 \implies x = 2z − x + 2 z = 0 ⟹ x = 2 z y y y ( 0 1 0 ) \begin{pmatrix}0\\1\\0\end{pmatrix} 0 1 0
λ = 3 \lambda = 3 λ = 3 ( − 2 0 2 0 − 1 0 2 0 − 2 ) v = 0 ⟹ x = z , y = 0 \begin{pmatrix} -2 & 0 & 2 \\ 0 & -1 & 0 \\ 2 & 0 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies x = z, y = 0 − 2 0 2 0 − 1 0 2 0 − 2 v = 0 ⟹ x = z , y = 0 ( 1 0 1 ) \begin{pmatrix}1\\0\\1\end{pmatrix} 1 0 1
(b) Three independent eigenvectors, so B B B
P = ( 1 0 1 0 1 0 − 1 0 1 ) , D = ( − 1 0 0 0 2 0 0 0 3 ) P = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} P = 1 0 − 1 0 1 0 1 0 1 , D = − 1 0 0 0 2 0 0 0 3
Question 9 Find the matrix representing an enlargement of scale factor 3 3 3 ( 1 , 2 ) (1, 2) ( 1 , 2 )
Solution In homogeneous coordinates, this is the composite of translate by ( − 1 , − 2 ) (-1, -2) ( − 1 , − 2 ) 3 3 3 ( 1 , 2 ) (1, 2) ( 1 , 2 )
M = ( 1 0 1 0 1 2 0 0 1 ) ( 3 0 0 0 3 0 0 0 1 ) ( 1 0 − 1 0 1 − 2 0 0 1 ) = ( 3 0 − 2 0 3 − 4 0 0 1 ) M = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 0 & -2 \\ 0 & 3 & -4 \\ 0 & 0 & 1 \end{pmatrix} M = 1 0 0 0 1 0 1 2 1 3 0 0 0 3 0 0 0 1 1 0 0 0 1 0 − 1 − 2 1 = 3 0 0 0 3 0 − 2 − 4 1
Question 10 Prove that if A A A λ 1 , λ 2 \lambda_1, \lambda_2 λ 1 , λ 2 λ 1 ≠ λ 2 \lambda_1 \neq \lambda_2 λ 1 = λ 2 A A A
Solution Since λ 1 ≠ λ 2 \lambda_1 \neq \lambda_2 λ 1 = λ 2 v 1 \mathbf{v}_1 v 1 v 2 \mathbf{v}_2 v 2 ( A − λ 1 I ) v 1 = 0 (A - \lambda_1 I)\mathbf{v}_1 = \mathbf{0} ( A − λ 1 I ) v 1 = 0 ( A − λ 2 I ) v 2 = 0 (A - \lambda_2 I)\mathbf{v}_2 = \mathbf{0} ( A − λ 2 I ) v 2 = 0
Suppose v 1 \mathbf{v}_1 v 1 v 2 \mathbf{v}_2 v 2 v 2 = c v 1 \mathbf{v}_2 = c\mathbf{v}_1 v 2 = c v 1 c c c
Then ( A − λ 2 I ) v 1 = 0 (A - \lambda_2 I)\mathbf{v}_1 = \mathbf{0} ( A − λ 2 I ) v 1 = 0 c c c λ 1 \lambda_1 λ 1 λ 2 \lambda_2 λ 2 v 1 \mathbf{v}_1 v 1 ( A − λ 1 I ) v 1 = 0 (A - \lambda_1 I)\mathbf{v}_1 = \mathbf{0} ( A − λ 1 I ) v 1 = 0 ( A − λ 2 I ) v 1 = 0 (A - \lambda_2 I)\mathbf{v}_1 = \mathbf{0} ( A − λ 2 I ) v 1 = 0 ( λ 1 − λ 2 ) v 1 = 0 (\lambda_1 - \lambda_2)\mathbf{v}_1 = \mathbf{0} ( λ 1 − λ 2 ) v 1 = 0 λ 1 ≠ λ 2 \lambda_1 \neq \lambda_2 λ 1 = λ 2 v 1 ≠ 0 \mathbf{v}_1 \neq \mathbf{0} v 1 = 0
Therefore v 1 \mathbf{v}_1 v 1 v 2 \mathbf{v}_2 v 2 P P P A = P D P − 1 A = PDP^{-1} A = P D P − 1 ■ \blacksquare ■
13. Advanced Worked Examples Example 13.1: Finding eigenvectors of a symmetric matrix Problem. Find the eigenvalues and a set of orthonormal eigenvectors of
A = ( 4 2 2 1 ) A = \begin{pmatrix}4&2\\2&1\end{pmatrix} A = ( 4 2 2 1 )
Solution. det ( A − λ I ) = ( 4 − λ ) ( 1 − λ ) − 4 = λ 2 − 5 λ = 0 \det(A-\lambda I) = (4-\lambda)(1-\lambda)-4 = \lambda^2-5\lambda = 0 det ( A − λ I ) = ( 4 − λ ) ( 1 − λ ) − 4 = λ 2 − 5 λ = 0 λ = 0 , 5 \lambda = 0, 5 λ = 0 , 5
λ = 0 \lambda = 0 λ = 0 ( 4 2 2 1 ) v = 0 ⟹ v 1 = − v 2 / 2 \begin{pmatrix}4&2\\2&1\end{pmatrix}\mathbf{v}=\mathbf{0} \implies v_1 = -v_2/2 ( 4 2 2 1 ) v = 0 ⟹ v 1 = − v 2 /2 ( 1 , − 2 ) (1,-2) ( 1 , − 2 ) ◆ L B ◆ 1 ◆ R B ◆◆ L B ◆ 5 ◆ R B ◆ ( 1 , − 2 ) \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{5}◆RB◆(1,-2) L ◆ B ◆1◆ R B ◆◆ L B ◆ 5 ◆ R B ◆ ( 1 , − 2 )
λ = 5 \lambda = 5 λ = 5 ( − 1 2 2 − 4 ) v = 0 ⟹ v 1 = 2 v 2 \begin{pmatrix}-1&2\\2&-4\end{pmatrix}\mathbf{v}=\mathbf{0} \implies v_1 = 2v_2 ( − 1 2 2 − 4 ) v = 0 ⟹ v 1 = 2 v 2 ( 2 , 1 ) (2,1) ( 2 , 1 ) ◆ L B ◆ 1 ◆ R B ◆◆ L B ◆ 5 ◆ R B ◆ ( 2 , 1 ) \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{5}◆RB◆(2,1) L ◆ B ◆1◆ R B ◆◆ L B ◆ 5 ◆ R B ◆ ( 2 , 1 )
Orthogonality check: ( 1 ) ( 2 ) + ( − 2 ) ( 1 ) = 0 (1)(2)+(-2)(1) = 0 ( 1 ) ( 2 ) + ( − 2 ) ( 1 ) = 0
Example 13.2: Using diagonalisation to compute a matrix power Problem. Given A = ( 3 1 0 2 ) A = \begin{pmatrix}3&1\\0&2\end{pmatrix} A = ( 3 0 1 2 ) A 10 A^{10} A 10
Solution. Eigenvalues: ( 3 − λ ) ( 2 − λ ) = 0 ⟹ λ = 2 , 3 (3-\lambda)(2-\lambda) = 0 \implies \lambda = 2, 3 ( 3 − λ ) ( 2 − λ ) = 0 ⟹ λ = 2 , 3
λ = 3 \lambda = 3 λ = 3 ( 0 1 0 − 1 ) v = 0 ⟹ v = ( 1 , 0 ) \begin{pmatrix}0&1\\0&-1\end{pmatrix}\mathbf{v}=\mathbf{0} \implies \mathbf{v}=(1,0) ( 0 0 1 − 1 ) v = 0 ⟹ v = ( 1 , 0 ) λ = 2 \lambda = 2 λ = 2 ( 1 1 0 0 ) v = 0 ⟹ v = ( 1 , − 1 ) \begin{pmatrix}1&1\\0&0\end{pmatrix}\mathbf{v}=\mathbf{0} \implies \mathbf{v}=(1,-1) ( 1 0 1 0 ) v = 0 ⟹ v = ( 1 , − 1 )
P = ( 1 1 0 − 1 ) P = \begin{pmatrix}1&1\\0&-1\end{pmatrix} P = ( 1 0 1 − 1 ) D = ( 3 0 0 2 ) D = \begin{pmatrix}3&0\\0&2\end{pmatrix} D = ( 3 0 0 2 ) P − 1 = ( 1 1 0 − 1 ) P^{-1} = \begin{pmatrix}1&1\\0&-1\end{pmatrix} P − 1 = ( 1 0 1 − 1 )
A 10 = P D 10 P − 1 = ( 1 1 0 − 1 ) ( 3 10 0 0 2 10 ) ( 1 1 0 − 1 ) A^{10} = PD^{10}P^{-1} = \begin{pmatrix}1&1\\0&-1\end{pmatrix}\begin{pmatrix}3^{10}&0\\0&2^{10}\end{pmatrix}\begin{pmatrix}1&1\\0&-1\end{pmatrix} A 10 = P D 10 P − 1 = ( 1 0 1 − 1 ) ( 3 10 0 0 2 10 ) ( 1 0 1 − 1 )
= ( 1 1 0 − 1 ) ( 59049 59049 0 − 1024 ) = ( 59049 58025 0 1024 ) = \begin{pmatrix}1&1\\0&-1\end{pmatrix}\begin{pmatrix}59049&59049\\0&-1024\end{pmatrix} = \boxed{\begin{pmatrix}59049&58025\\0&1024\end{pmatrix}} = ( 1 0 1 − 1 ) ( 59049 0 59049 − 1024 ) = ( 59049 0 58025 1024 )
Example 13.3: Rotation matrix properties Problem. Show that
R θ = ( cos θ − sin θ sin θ cos θ ) R_\theta = \begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix} R θ = ( cos θ sin θ − sin θ cos θ ) R θ R ϕ = R θ + ϕ R_\theta R_\phi = R_{\theta+\phi} R θ R ϕ = R θ + ϕ R θ − 1 = R − θ R_\theta^{-1} = R_{-\theta} R θ − 1 = R − θ
Solution.
R θ R ϕ = ( cos θ cos ϕ − sin θ sin ϕ − cos θ sin ϕ − sin θ cos ϕ sin θ cos ϕ + cos θ sin ϕ − sin θ sin ϕ + cos θ cos ϕ ) R_\theta R_\phi = \begin{pmatrix}\cos\theta\cos\phi-\sin\theta\sin\phi&-\cos\theta\sin\phi-\sin\theta\cos\phi\\\sin\theta\cos\phi+\cos\theta\sin\phi&-\sin\theta\sin\phi+\cos\theta\cos\phi\end{pmatrix} R θ R ϕ = ( cos θ cos ϕ − sin θ sin ϕ sin θ cos ϕ + cos θ sin ϕ − cos θ sin ϕ − sin θ cos ϕ − sin θ sin ϕ + cos θ cos ϕ )
= ( cos ( θ + ϕ ) − sin ( θ + ϕ ) sin ( θ + ϕ ) cos ( θ + ϕ ) ) = R θ + ϕ = \begin{pmatrix}\cos(\theta+\phi)&-\sin(\theta+\phi)\\\sin(\theta+\phi)&\cos(\theta+\phi)\end{pmatrix} = R_{\theta+\phi} = ( cos ( θ + ϕ ) sin ( θ + ϕ ) − sin ( θ + ϕ ) cos ( θ + ϕ ) ) = R θ + ϕ
R θ − 1 = ◆ L B ◆ 1 ◆ R B ◆◆ L B ◆ cos 2 θ + sin 2 θ ◆ R B ◆ ( cos θ sin θ − sin θ cos θ ) = ( cos θ sin θ − sin θ cos θ ) = R − θ R_\theta^{-1} = \dfrac◆LB◆1◆RB◆◆LB◆\cos^2\theta+\sin^2\theta◆RB◆\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix} = \begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix} = R_{-\theta} R θ − 1 = L ◆ B ◆1◆ R B ◆◆ L B ◆ cos 2 θ + sin 2 θ ◆ R B ◆ ( cos θ − sin θ sin θ cos θ ) = ( cos θ − sin θ sin θ cos θ ) = R − θ
Example 13.4: Determinant and area scaling Problem. The triangle with vertices ( 1 , 0 ) (1,0) ( 1 , 0 ) ( 0 , 2 ) (0,2) ( 0 , 2 ) ( 3 , 4 ) (3,4) ( 3 , 4 ) T = ( 2 − 1 1 3 ) T = \begin{pmatrix}2&-1\\1&3\end{pmatrix} T = ( 2 1 − 1 3 )
Solution. Original area:
1 2 ∣ det ( 0 − 1 3 − 1 2 − 0 4 − 0 ) ∣ = 1 2 ∣ − 2 + 2 ∣ = 0 \dfrac{1}{2}\left|\det\begin{pmatrix}0-1&3-1\\2-0&4-0\end{pmatrix}\right| = \dfrac{1}{2}|-2+2| = 0 2 1 det ( 0 − 1 2 − 0 3 − 1 4 − 0 ) = 2 1 ∣ − 2 + 2∣ = 0
Wait, the points are collinear? Let me use ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 1 , 0 ) (1,0) ( 1 , 0 ) ( 0 , 1 ) (0,1) ( 0 , 1 ) = 1 2 = \dfrac{1}{2} = 2 1
det ( T ) = 6 + 1 = 7 \det(T) = 6+1 = 7 det ( T ) = 6 + 1 = 7 = 7 × 1 2 = 3.5 = 7 \times \dfrac{1}{2} = \boxed{3.5} = 7 × 2 1 = 3.5
Problem. The matrix S = ( 1 k 0 1 ) S = \begin{pmatrix}1&k\\0&1\end{pmatrix} S = ( 1 0 k 1 )
Solution. det ( S − λ I ) = ( 1 − λ ) 2 = 0 \det(S-\lambda I) = (1-\lambda)^2 = 0 det ( S − λ I ) = ( 1 − λ ) 2 = 0 λ = 1 \lambda = 1 λ = 1
( S − I ) v = ( 0 k 0 0 ) v = 0 ⟹ v 2 = 0 (S-I)\mathbf{v} = \begin{pmatrix}0&k\\0&0\end{pmatrix}\mathbf{v} = \mathbf{0} \implies v_2 = 0 ( S − I ) v = ( 0 0 k 0 ) v = 0 ⟹ v 2 = 0 ( 1 , 0 ) (1,0) ( 1 , 0 )
The x x x y = 0 y=0 y = 0 y = c y = c y = c c c c y = 0 y=0 y = 0
Example 13.6: Matrix equation A X = B AX = B A X = B Problem. Solve A X = B AX = B A X = B A = ( 1 2 3 5 ) A = \begin{pmatrix}1&2\\3&5\end{pmatrix} A = ( 1 3 2 5 ) B = ( 4 7 7 12 ) B = \begin{pmatrix}4&7\\7&12\end{pmatrix} B = ( 4 7 7 12 )
Solution. X = A − 1 B X = A^{-1}B X = A − 1 B det ( A ) = 5 − 6 = − 1 \det(A) = 5-6 = -1 det ( A ) = 5 − 6 = − 1
A − 1 = ( − 5 2 3 − 1 ) A^{-1} = \begin{pmatrix}-5&2\\3&-1\end{pmatrix} A − 1 = ( − 5 3 2 − 1 )
X = ( − 5 2 3 − 1 ) ( 4 7 7 12 ) = ( − 20 + 14 − 35 + 24 12 − 7 21 − 12 ) = ( − 6 − 11 5 9 ) X = \begin{pmatrix}-5&2\\3&-1\end{pmatrix}\begin{pmatrix}4&7\\7&12\end{pmatrix} = \begin{pmatrix}-20+14&-35+24\\12-7&21-12\end{pmatrix} = \boxed{\begin{pmatrix}-6&-11\\5&9\end{pmatrix}} X = ( − 5 3 2 − 1 ) ( 4 7 7 12 ) = ( − 20 + 14 12 − 7 − 35 + 24 21 − 12 ) = ( − 6 5 − 11 9 )
14. Additional Exam-Style Questions Question 11 The matrix M = ( 2 − 1 4 − 3 ) M = \begin{pmatrix}2&-1\\4&-3\end{pmatrix} M = ( 2 4 − 1 − 3 ) λ 1 = 1 \lambda_1 = 1 λ 1 = 1 λ 2 = − 2 \lambda_2 = -2 λ 2 = − 2 M 4 + 3 M M^4 + 3M M 4 + 3 M
Solution By Cayley--Hamilton: M 2 + M − 2 I = O ⟹ M 2 = − M + 2 I M^2 + M - 2I = O \implies M^2 = -M + 2I M 2 + M − 2 I = O ⟹ M 2 = − M + 2 I
M 3 = M ( − M + 2 I ) = − M 2 + 2 M = − ( − M + 2 I ) + 2 M = 3 M − 2 I M^3 = M(-M+2I) = -M^2+2M = -(-M+2I)+2M = 3M-2I M 3 = M ( − M + 2 I ) = − M 2 + 2 M = − ( − M + 2 I ) + 2 M = 3 M − 2 I
M 4 = M ( 3 M − 2 I ) = 3 M 2 − 2 M = 3 ( − M + 2 I ) − 2 M = − 5 M + 6 I M^4 = M(3M-2I) = 3M^2-2M = 3(-M+2I)-2M = -5M+6I M 4 = M ( 3 M − 2 I ) = 3 M 2 − 2 M = 3 ( − M + 2 I ) − 2 M = − 5 M + 6 I
M 4 + 3 M = − 5 M + 6 I + 3 M = − 2 M + 6 I = − 2 ( 2 − 1 4 − 3 ) + 6 ( 1 0 0 1 ) = ( − 4 + 6 2 − 8 6 + 6 ) = ( 2 2 − 8 12 ) M^4+3M = -5M+6I+3M = -2M+6I = -2\begin{pmatrix}2&-1\\4&-3\end{pmatrix}+6\begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}-4+6&2\\-8&6+6\end{pmatrix} = \begin{pmatrix}2&2\\-8&12\end{pmatrix} M 4 + 3 M = − 5 M + 6 I + 3 M = − 2 M + 6 I = − 2 ( 2 4 − 1 − 3 ) + 6 ( 1 0 0 1 ) = ( − 4 + 6 − 8 2 6 + 6 ) = ( 2 − 8 2 12 )
Question 12 Prove that similar matrices have the same trace and determinant.
Solution If B = P − 1 A P B = P^{-1}AP B = P − 1 A P det ( B ) = det ( P − 1 A P ) = det ( P − 1 ) det ( A ) det ( P ) = det ( A ) \det(B) = \det(P^{-1}AP) = \det(P^{-1})\det(A)\det(P) = \det(A) det ( B ) = det ( P − 1 A P ) = det ( P − 1 ) det ( A ) det ( P ) = det ( A )
tr ( B ) = tr ( P − 1 A P ) \text{tr}(B) = \text{tr}(P^{-1}AP) tr ( B ) = tr ( P − 1 A P ) tr ( A B C ) = tr ( C A B ) \text{tr}(ABC) = \text{tr}(CAB) tr ( A B C ) = tr ( C A B )
tr ( P − 1 A P ) = tr ( A P P − 1 ) = tr ( A ) \text{tr}(P^{-1}AP) = \text{tr}(APP^{-1}) = \text{tr}(A) tr ( P − 1 A P ) = tr ( A P P − 1 ) = tr ( A ) ■ \blacksquare ■
Question 13 Find the reflection matrix in the line y = 2 x y = 2x y = 2 x
Solution The line y = 2 x y = 2x y = 2 x θ = arctan 2 \theta = \arctan 2 θ = arctan 2 x x x
R = ( cos 2 θ sin 2 θ sin 2 θ − cos 2 θ ) R = \begin{pmatrix}\cos 2\theta&\sin 2\theta\\\sin 2\theta&-\cos 2\theta\end{pmatrix} R = ( cos 2 θ sin 2 θ sin 2 θ − cos 2 θ )
cos θ = ◆ L B ◆ 1 ◆ R B ◆◆ L B ◆ 5 ◆ R B ◆ \cos\theta = \dfrac◆LB◆1◆RB◆◆LB◆\sqrt{5}◆RB◆ cos θ = L ◆ B ◆1◆ R B ◆◆ L B ◆ 5 ◆ R B ◆ sin θ = ◆ L B ◆ 2 ◆ R B ◆◆ L B ◆ 5 ◆ R B ◆ \sin\theta = \dfrac◆LB◆2◆RB◆◆LB◆\sqrt{5}◆RB◆ sin θ = L ◆ B ◆2◆ R B ◆◆ L B ◆ 5 ◆ R B ◆
cos 2 θ = cos 2 θ − sin 2 θ = 1 − 4 5 = − 3 5 \cos 2\theta = \cos^2\theta-\sin^2\theta = \dfrac{1-4}{5} = -\dfrac{3}{5} cos 2 θ = cos 2 θ − sin 2 θ = 5 1 − 4 = − 5 3
sin 2 θ = 2 sin θ cos θ = 4 5 \sin 2\theta = 2\sin\theta\cos\theta = \dfrac{4}{5} sin 2 θ = 2 sin θ cos θ = 5 4
R = ( − 3 5 4 5 4 5 3 5 ) R = \begin{pmatrix}-\frac{3}{5}&\frac{4}{5}\\\frac{4}{5}&\frac{3}{5}\end{pmatrix} R = ( − 5 3 5 4 5 4 5 3 )
16. Further Advanced Topics 16.1 Orthogonal matrices A matrix Q Q Q orthogonal if Q T Q = Q Q T = I Q^TQ = QQ^T = I Q T Q = Q Q T = I
Properties:
∣ det Q ∣ = 1 |\det Q| = 1 ∣ det Q ∣ = 1 Columns and rows form orthonormal bases
Q − 1 = Q T Q^{-1} = Q^T Q − 1 = Q T Orthogonal transformations preserve lengths and angles
Every 2 × 2 2\times 2 2 × 2 det = 1 \det = 1 det = 1 det = − 1 \det = -1 det = − 1
16.2 Diagonalisation revisited If A A A n n n A = P D P − 1 A = PDP^{-1} A = P D P − 1
P P P D D D Computing A k A^k A k A k = P D k P − 1 A^k = PD^kP^{-1} A k = P D k P − 1
Computing e A e^A e A e A = P e D P − 1 e^A = Pe^DP^{-1} e A = P e D P − 1 e D e^D e D e λ i e^{\lambda_i} e λ i
16.3 Cayley-Hamilton theorem Every square matrix satisfies its own characteristic equation.
If p ( λ ) = det ( λ I − A ) p(\lambda) = \det(\lambda I - A) p ( λ ) = det ( λ I − A ) p ( A ) = O p(A) = O p ( A ) = O
This can be used to express A n A^n A n n n n A A A
17. Further Exam-Style Questions Question 16 Prove that if A A A det A = ± 1 \det A = \pm 1 det A = ± 1
Solution det ( Q T Q ) = det ( Q T ) det ( Q ) = ( det Q ) 2 \det(Q^TQ) = \det(Q^T)\det(Q) = (\det Q)^2 det ( Q T Q ) = det ( Q T ) det ( Q ) = ( det Q ) 2
But Q T Q = I Q^TQ = I Q T Q = I det ( I ) = 1 \det(I) = 1 det ( I ) = 1
Therefore ( det Q ) 2 = 1 ⟹ det Q = ± 1 (\det Q)^2 = 1 \implies \det Q = \pm 1 ( det Q ) 2 = 1 ⟹ det Q = ± 1 ■ \blacksquare ■
Question 17 Find the eigenvalues and eigenvectors of A = ( 4 1 2 3 ) A = \begin{pmatrix}4&1\\2&3\end{pmatrix} A = ( 4 2 1 3 ) A 5 A^5 A 5
Solution det ( A − λ I ) = ( 4 − λ ) ( 3 − λ ) − 2 = λ 2 − 7 λ + 10 = ( λ − 5 ) ( λ − 2 ) \det(A-\lambda I) = (4-\lambda)(3-\lambda)-2 = \lambda^2-7\lambda+10 = (\lambda-5)(\lambda-2) det ( A − λ I ) = ( 4 − λ ) ( 3 − λ ) − 2 = λ 2 − 7 λ + 10 = ( λ − 5 ) ( λ − 2 )
Eigenvalues: λ 1 = 5 \lambda_1 = 5 λ 1 = 5 λ 2 = 2 \lambda_2 = 2 λ 2 = 2
λ = 5 \lambda=5 λ = 5 ( A − 5 I ) v = 0 ⟹ ( − 1 1 2 − 2 ) ( v 1 v 2 ) = 0 ⟹ v 1 = ( 1 1 ) (A-5I)\mathbf{v} = \mathbf{0} \implies \begin{pmatrix}-1&1\\2&-2\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix} = \mathbf{0} \implies \mathbf{v}_1 = \begin{pmatrix}1\\1\end{pmatrix} ( A − 5 I ) v = 0 ⟹ ( − 1 2 1 − 2 ) ( v 1 v 2 ) = 0 ⟹ v 1 = ( 1 1 )
λ = 2 \lambda=2 λ = 2 ( A − 2 I ) v = 0 ⟹ ( 2 1 2 1 ) ( v 1 v 2 ) = 0 ⟹ v 2 = ( 1 − 2 ) (A-2I)\mathbf{v} = \mathbf{0} \implies \begin{pmatrix}2&1\\2&1\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix} = \mathbf{0} \implies \mathbf{v}_2 = \begin{pmatrix}1\\-2\end{pmatrix} ( A − 2 I ) v = 0 ⟹ ( 2 2 1 1 ) ( v 1 v 2 ) = 0 ⟹ v 2 = ( 1 − 2 )
P = ( 1 1 1 − 2 ) P = \begin{pmatrix}1&1\\1&-2\end{pmatrix} P = ( 1 1 1 − 2 ) D = ( 5 0 0 2 ) D = \begin{pmatrix}5&0\\0&2\end{pmatrix} D = ( 5 0 0 2 ) P − 1 = 1 − 3 ( − 2 − 1 − 1 1 ) P^{-1} = \dfrac{1}{-3}\begin{pmatrix}-2&-1\\-1&1\end{pmatrix} P − 1 = − 3 1 ( − 2 − 1 − 1 1 )
A 5 = P D 5 P − 1 = 1 3 ( 1 1 1 − 2 ) ( 3125 0 0 32 ) ( 2 1 1 − 1 ) A^5 = PD^5P^{-1} = \dfrac{1}{3}\begin{pmatrix}1&1\\1&-2\end{pmatrix}\begin{pmatrix}3125&0\\0&32\end{pmatrix}\begin{pmatrix}2&1\\1&-1\end{pmatrix} A 5 = P D 5 P − 1 = 3 1 ( 1 1 1 − 2 ) ( 3125 0 0 32 ) ( 2 1 1 − 1 )
= 1 3 ( 3125 32 3125 − 64 ) ( 2 1 1 − 1 ) = 1 3 ( 6282 3093 6186 3189 ) = ( 2094 1031 2062 1063 ) = \dfrac{1}{3}\begin{pmatrix}3125&32\\3125&-64\end{pmatrix}\begin{pmatrix}2&1\\1&-1\end{pmatrix} = \dfrac{1}{3}\begin{pmatrix}6282&3093\\6186&3189\end{pmatrix} = \begin{pmatrix}2094&1031\\2062&1063\end{pmatrix} = 3 1 ( 3125 3125 32 − 64 ) ( 2 1 1 − 1 ) = 3 1 ( 6282 6186 3093 3189 ) = ( 2094 2062 1031 1063 )