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Review Problems, Solutions
- Find all possible solutions (a, b, c) of the following system of equations.
a + 3b + 7c = 2 3 a + 4b + c = 1 4 a + 10b + 20c = 3
Solution: We must row reduce the augmented matrix:
If there were a solution of this system of equations, the last row in this reduced matrix would imply 0 = −3, which is a contradiction. Thus there are no solutions of the system.
- Let
A =
Find all the eigenvalues of A and give a basis for each of their corresponding eigenspaces. Solution: First we find the eigenvalues by solving the characteristic polynomial. The charac- teristic polynomial for this matrix is
det
1 − λ − 2 − 3 3 8 − λ 9 − 2 − 4 − 4 − λ
(^) = (1 − λ)(8 − λ)(− 4 − λ) + 72 + 36(1 − λ) + 6(− 4 − λ) − 6(8 − λ)
= (1 − λ)((8 − λ)(− 4 − λ) + 36) = (1 − λ)(λ^2 − 4 λ + 4) = (1 − λ)(λ − 2)^2. Therefore the roots of the characteristic polynomial, which are the eigenvalues, are λ = 1, 2. Note that λ = 1 has algebraic multiplicity 1, and λ = 2 has algebraic multiplicity 2. Now we find the eigenspaces, E 1 and E 2 , corresponding to each eigenvalue. First I’ll find E 1. We know that the geometric multiplicity for λ = 1 must be one, so if we find one eigenvector for λ = 1, then E 1 will be the span of this vector. Thus we seek to find any nonzero solution of
x 1 x 2 x 3
One way to find such a vector is to assume x 3 = 2. Then from the top row we obtain x 2 = −3, and plugging these into either of the other equations gives x 1 = 1. Thus
E 1 = span
Now we find E 2. The geometric multiplicity of λ = 2 could be either 2 or one 1, so we do not know in advance how many linearly independent eigenvectors we need to find. To figure this out we must find all solutions of
x 1 x 2 x 3
To find these solutions, we will row reduce (note we don’t need to consider the augmented matrix because the “augmented” part is just 0).
Therefore E 2 is the set of all vectors perpendicular to (− 1 , − 2 , −3)T^ , which is two dimensional. So to get a basis we need two linearly independent eigenvectors. We can find these by assuming alternately x 1 = 0 and x 2 = 0. One possible set of vectors so obtained is (^)
- Let
B =
Find bases for the range and null space of B. Solution The range of B will be the span of its columns, and the null space will be the set of all solutions to the corresponding homogeneous equation. Let’s find the null space first. As in number 2, we row reduce B to do this.
From this row reduced form we can see that in fact the only solution of Bx = 0 is x = 0, and so the null space of B is just { 0 }. By definition a basis for { 0 } is the empty set, ∅. If you did not know this, that is okay as long as you found that null(B) = { 0 }. Since the matrix is square this also implies that it is invertible, and so any three linearly independent vectors form a basis for its range. For example, you could answer (^)
as a basis for the range.
- Say whether each of the following statements is true or false. If the statement is true, give a brief explanation as to why. If it is false, provide a counter example.
Note that we can evaluate this integral immediately since it is an odd function inte- grated over a symmetric interval. If n 6 = 0, then we use integration by parts as follows ∫ (^) π
−π
x sin(nx) dx =
−x cos(nx) n
π
−π
∫ (^) π
−π
cos(nx) n
dx
− 2 π cos(nπ) n
sin(nx) n^2
π
−π = (−1)n+^ 2 π n
This problem is a partial example of calculating a Fourier series, which we will be studying later in class.
- Solve the following initial value problem
y′′^ + 2y = 0 , y(0) = 1 , y′(0) = 1.
Solution: We use the method from math 307. The characteristic polynomial of this equation is r^2 + 2 = 0, which has roots r = ±
2 i. Thus the general solution is
ygen(t) = c 1 sin(
2 t) + c 2 cos(
2 t).
Now we just have to use the initial conditions to solve for c 1 and c 2.
1 = y(0) = c 2.
1 = y′(0) =
2 c 1 ⇒
= c 1.
Thus the solution is y(t) =
sin(
2 t) √ 2
2 t).
Challenge Problem (Extra Credit):
(a) Let A be an n × m matrix with real entries. Show that the null space of A is the orthogonal complement of the range of AT^. In other words, show that the null space of A is exactly equal to the collection of all vectors that are orthogonal to the range of AT^. Solution: I will refer to the range of AT^ as R(AT^ ) and the null space of A as N (A). Also, the orthogonal complement of R(AT^ ), which is the set of all vectors perpendicular to R(AT^ ), will be written as R(AT^ )⊥. The object then is to show that N (A) = R(AT^ )⊥. First I will show that N (A) is contained in R(AT^ )⊥. To this end, take any vector x ∈ Rn^ in N (A) and vector v ∈ Rn^ in R(AT^ ). Then by the definition of R(AT^ ) there exists a vector y ∈ Rm^ such that AT^ y = v. Thus
vT^ x = (AT^ y)T^ x = yT^ Ax = 0.
This last equality holds since x ∈ N (A). Therefore v and x are perpendicular, and so since v was an arbitrary vector in R(AT^ ) this proves that x is perpendicular to R(AT^ ). This shows that N (A) is contained in R(A)⊥.
Now I will prove the opposite inclusion- that R(AT^ )⊥^ is contained in N (A). To this end, choose any vector x ∈ Rn^ in R(AT^ )⊥, and let y ∈ Rm^ be arbitrary. Then we essentially just use the previous calculation in reverse.
0 = (AT^ y)T^ x = yT^ Ax.
We can now choose y to be any one of the standard basis vectors for Rm, and thus prove that each component of Ax is zero and so Ax = 0. Therefore x ∈ N (A), which proves R(AT^ )⊥^ is contained in N (A). Having proved that R(AT^ )⊥^ is contained in N (A) and N (A) is contained in R(AT^ )⊥^ it must be the case that these two sets are equal. This fact is sometimes called the Fundamental Theorem of The Alternative.
(b) Use part (a) to show that the equation
Ax = b
has a solution x if and only if y · b = 0 for every y in the nullspace of AT^. Solution: This is really just a restatement of part (a). There is a solution of the equation if and only if b ∈ R(A). Since A = (AT^ )T^ , part (a) says that R(A)⊥^ = N (AT^ ). Since the orthogonal complement of the orthogonal complement of a subspace is the original subspace this implies that R(A) = N (AT^ )⊥. Thus b ∈ R(A) if and only if b ∈ N (AT^ )⊥, which means exactly that y · b = 0 for every y in the nullspace of AT
