MATH-225 Final Exam — 20.05.2021 — Solutions
N.B. Correct answers without sufficient correct mathematical explanations will not get full credit.
Q 1: Let A be an n × n matrix, λ1 an eigenvalue of A, and let In denote the identity matrix of
size n × n. Recall that the multiplicity of λ1 is the largest integer k such that (λ − λ1 )k is a factor of
the characteristic polynomial |λIn − A|.
(a) (5 pts) Show by an example that the dimension of Null(λ1 In − A) can be different from the
multiplicity of λ1 .
(b) (5 pts) If A and B are two similar square matrices, show that they have the same characteristic
polynomial.
(c) (5 pts) Let A be an 8 × 8 matrix, let {v1 , v2 , . . . , v8 } be a basis of R8 such that {v1 , v2 , v3 , v4 }
is a basis of Null(I8 − A), and let S be the 8 × 8 matrix with columns v1 , v2 , . . . , v8 in this order.
Calculate the leftmost 4 columns of the matrix S −1 AS.
(d) (5 pts) What is the minimal multiplicity of the eigenvalue 1 that the matrix A in (c) may
have?
1 1
Solution. (a) Take A = and observe that the eigenspace corresponding to 1 has dimension
0 1
2
1 while pA (λ) = (λ − 1) , hence the multiplicity is 2.
(b) If A = S −1 BS for some invertible matrix S, then
pA (λ) = det(λIn − A) = det(λIn − S −1 BS) = det(S −1 (λIn − B)S)
= det(S −1 ) det(λIn − B) det(S) = det(S)−1 det(λIn − B) det(S)
= det(λIn − B) = pB (λ).
(c) Let v1 , v2 , v3 , v4 be a basis of the eigenspace Null(In − A), and let v5 , . . . , v8 be vectors in R8
such that {v1 , . . . , v8 } is a basis of R8 . Let S be the matrix with columns the vectors v1 , . . . , v8 , in
this order. Then
AS = A[v1 v2 v3 v4 v5 v6 v7 v8 ]
= [Av1 Av2 Av3 Av4 Av5 Av6 Av7 Av8 ] = [v1 v2 v3 v4 Av5 Av6 Av7 Av8 ]
I4 B1,2
=S = SB,
0 B2,2
where B1,2 and B2,2 are 4 × 4 matrices and B is the partitioned matrix
I4 B1,2
B= .
0 B2,2
This shows that the leftmost four columns of the matrix B = S −1 AS are the first vectors of the
standard basis of R8 .
(d) By the result at items (b) and (c), it follows that
(λ − 1)I4 −B1,2
pA (λ) = pB (λ) = det(λI8 − B) = det( = (λ − 1)4 pB2,2 (λ),
0 λI4 − B2,2
where, in order to get the last equality, we have used the column cofactor expansions. This shows
that the multiplicity of 1 is at least 4.
We can have exactly the multiplicity 4 if we choose, for example, A to be the matrix
I 0
A= 4 .
0 2I4
Q 2: (a) (8 pnts) Show that the equation y 2 dx + (2xy − y 2 )dy = 0 is exact and then find the
general solution.
We have learnt four different methods to solve first order differential equations of the types: sep-
arable, linear, Bernoulli, and homogeneous.
, dy
(b) (6 pnts) and (c) (6 pnts): Solve the initial value problem (2x − y) dx + y = 0, y(1) = 1, by
using any two of these different methods.
Solution. (a) Exact Differential Equation with M (x, y) = y 2 , N (x, y) = 2xy − y 2 then
∂M ∂N
− = 2y − 2y = 0.
∂y ∂x
Hence there exists a function F such that
∂F ∂F
= M (x, y) = y 2 , = N (x, y) = 2xy − y 2 .
∂x ∂y
Solving the first one we find F (x, y) = y 2 x + h(y) where h is any fuction of y. Using this result in
the second equation we find h0 = −y 2 . Then the function F becomes F (x, y) = xy 2 − 13 y 3 . Hence the
general solution of the differential equation is
1
xy 2 − y 3 = C,
3
where C is any constant.
(b) Linear Equation. If we take the x variable as dependent variable the the equation takes the
form
dx 2x
+ =1
dy y
Integrating factor ρ(y) = y 2 . Multiplying the equation by the integrating factor we get
d
(xy 2 ) = y 2
dy
Hence we get xy 2 − 13 y 3 = C. To satisfy the initial condition C = 2/3
(c) This equation is homogeneous. We can write it as
dy y
+ =0
dx 2x − y
y
Let u = x then we get a separable equation
(2 − u)du dx
+ =0
3u − u2 x
which has the solution
u2 (3 − u) = C/x3
inserting u = y/x we get
y 2 (3x − y) = C
Using the initial condtion we obtain C = 2.
4 −5 1
Q 3: (a) (12 pnts) Find all eigenvalues and associated eigenvectors of A = 1 0 −1.
0 1 −1
dx
(b) (4 pnts) Find 3 linearly independent solutions to the system of differential equations = Ax,
T dt
where x(t) = x1 (t) x2 (t) x3 (t) .
(c) (4 pnts) Check that your solutions in part (b) are actually linearly independent using the
Wronskian.
2
N.B. Correct answers without sufficient correct mathematical explanations will not get full credit.
Q 1: Let A be an n × n matrix, λ1 an eigenvalue of A, and let In denote the identity matrix of
size n × n. Recall that the multiplicity of λ1 is the largest integer k such that (λ − λ1 )k is a factor of
the characteristic polynomial |λIn − A|.
(a) (5 pts) Show by an example that the dimension of Null(λ1 In − A) can be different from the
multiplicity of λ1 .
(b) (5 pts) If A and B are two similar square matrices, show that they have the same characteristic
polynomial.
(c) (5 pts) Let A be an 8 × 8 matrix, let {v1 , v2 , . . . , v8 } be a basis of R8 such that {v1 , v2 , v3 , v4 }
is a basis of Null(I8 − A), and let S be the 8 × 8 matrix with columns v1 , v2 , . . . , v8 in this order.
Calculate the leftmost 4 columns of the matrix S −1 AS.
(d) (5 pts) What is the minimal multiplicity of the eigenvalue 1 that the matrix A in (c) may
have?
1 1
Solution. (a) Take A = and observe that the eigenspace corresponding to 1 has dimension
0 1
2
1 while pA (λ) = (λ − 1) , hence the multiplicity is 2.
(b) If A = S −1 BS for some invertible matrix S, then
pA (λ) = det(λIn − A) = det(λIn − S −1 BS) = det(S −1 (λIn − B)S)
= det(S −1 ) det(λIn − B) det(S) = det(S)−1 det(λIn − B) det(S)
= det(λIn − B) = pB (λ).
(c) Let v1 , v2 , v3 , v4 be a basis of the eigenspace Null(In − A), and let v5 , . . . , v8 be vectors in R8
such that {v1 , . . . , v8 } is a basis of R8 . Let S be the matrix with columns the vectors v1 , . . . , v8 , in
this order. Then
AS = A[v1 v2 v3 v4 v5 v6 v7 v8 ]
= [Av1 Av2 Av3 Av4 Av5 Av6 Av7 Av8 ] = [v1 v2 v3 v4 Av5 Av6 Av7 Av8 ]
I4 B1,2
=S = SB,
0 B2,2
where B1,2 and B2,2 are 4 × 4 matrices and B is the partitioned matrix
I4 B1,2
B= .
0 B2,2
This shows that the leftmost four columns of the matrix B = S −1 AS are the first vectors of the
standard basis of R8 .
(d) By the result at items (b) and (c), it follows that
(λ − 1)I4 −B1,2
pA (λ) = pB (λ) = det(λI8 − B) = det( = (λ − 1)4 pB2,2 (λ),
0 λI4 − B2,2
where, in order to get the last equality, we have used the column cofactor expansions. This shows
that the multiplicity of 1 is at least 4.
We can have exactly the multiplicity 4 if we choose, for example, A to be the matrix
I 0
A= 4 .
0 2I4
Q 2: (a) (8 pnts) Show that the equation y 2 dx + (2xy − y 2 )dy = 0 is exact and then find the
general solution.
We have learnt four different methods to solve first order differential equations of the types: sep-
arable, linear, Bernoulli, and homogeneous.
, dy
(b) (6 pnts) and (c) (6 pnts): Solve the initial value problem (2x − y) dx + y = 0, y(1) = 1, by
using any two of these different methods.
Solution. (a) Exact Differential Equation with M (x, y) = y 2 , N (x, y) = 2xy − y 2 then
∂M ∂N
− = 2y − 2y = 0.
∂y ∂x
Hence there exists a function F such that
∂F ∂F
= M (x, y) = y 2 , = N (x, y) = 2xy − y 2 .
∂x ∂y
Solving the first one we find F (x, y) = y 2 x + h(y) where h is any fuction of y. Using this result in
the second equation we find h0 = −y 2 . Then the function F becomes F (x, y) = xy 2 − 13 y 3 . Hence the
general solution of the differential equation is
1
xy 2 − y 3 = C,
3
where C is any constant.
(b) Linear Equation. If we take the x variable as dependent variable the the equation takes the
form
dx 2x
+ =1
dy y
Integrating factor ρ(y) = y 2 . Multiplying the equation by the integrating factor we get
d
(xy 2 ) = y 2
dy
Hence we get xy 2 − 13 y 3 = C. To satisfy the initial condition C = 2/3
(c) This equation is homogeneous. We can write it as
dy y
+ =0
dx 2x − y
y
Let u = x then we get a separable equation
(2 − u)du dx
+ =0
3u − u2 x
which has the solution
u2 (3 − u) = C/x3
inserting u = y/x we get
y 2 (3x − y) = C
Using the initial condtion we obtain C = 2.
4 −5 1
Q 3: (a) (12 pnts) Find all eigenvalues and associated eigenvectors of A = 1 0 −1.
0 1 −1
dx
(b) (4 pnts) Find 3 linearly independent solutions to the system of differential equations = Ax,
T dt
where x(t) = x1 (t) x2 (t) x3 (t) .
(c) (4 pnts) Check that your solutions in part (b) are actually linearly independent using the
Wronskian.
2
