Math. 311 (Fulling) 26 October 1998

Test B – Solutions

1. (18 pts.) For each of these sets, tell whether or not it is a basis for R3 ; and if not, explain briefly why not.

(a) {(1, 1, 1), (0, 1, 1), (0, 0, 2)}

Yes — From the matrix ( \begin{matrix}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 2 \end{matrix} ) it is obvious that these vectors are independent. Alternatively, the determinant equals 2 ≠ 0 .

(b) {(3, 5, 9), (0, 6, −20)}

No — There are not enough vectors to span a 3-dimensional space.

(c) {(1, 0, 1), (0, 1, 1), (1, 2, 3)}

No — Form a matrix and reduce:

( \begin{matrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 2 & 3 \end{matrix} ) → ( \begin{matrix}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{matrix} ) .

We see that the vectors are not independent (and don’t span) because the third vector equals the first plus twice the second. Or, you can calculate the determinant and get 0 .

2. (10 pts.) Let V be the vector space spanned by the functions {cos t, cos 2t, sin t, sin 2t} . Let S be the space of solutions of the differential equation y′′ + y = 0 . You may rest assured that S is a subset of V . Is S a subspace of V ? Explain!

Yes — Use any one of these 5 (really only 2) arguments:

1. S is the span of {cos t, sin t} , and the span of a set is always a subspace of the larger vector space involved.

1′. The solutions are the functions c_{1} cos t + c_{2} sin t , and we can show directly that this set is closed

under addition and scalar multiplication:

r(c_{1} cos t + c_{2} sin t) + (d_{1} cos t + d_{2} sin t) = (rc_{1} + d_{1}) cos t + (rc_{2} + d_{2}) sin t.

2. The solution space of a homogeneous linear equation is always a subspace. 2′. S is the kernel of the linear operator L ≡ \frac{d^{2}}{dt^{2}} + 1 (that is, L(y) ≡ y′′ + y ), and the kernel of

a linear operator is always a subspace. 2′′. We can show by brute force that S is closed under addition and scalar multiplication: If y′′_{1} +

y_{1} = 0 and y′′_{2} + y_{2} = 0 , then

(ry_{1} + y_{2})′′ + (ry_{1} + y_{2}) = r(y′′_{1} + y_{1}) + (y′′_{2} + y_{2}) = 0.

311B-F98 Page 2

3. (12 pts.) Find the matrix M that changes coordinates from the natural basis in R2 to the basis

B = \{( \begin{matrix}0 \\ 2 \end{matrix} ) , ( \begin{matrix}1 \\ 1 \end{matrix} )\} ≡ \{b⃗1, b⃗2\}.

Let

B = (\begin{matrix}⊤ & ⊤ \\ b⃗_{1} & b⃗_{2} \\ ⊥ & ⊥ \end{matrix} ) = ( \begin{matrix}0 & 1 \\ 2 & 1 \end{matrix} ) .

B maps coordinates from B to the natural basis. Therefore, what we want is

M = B^{−1} = \frac{1}{(−2)} ( \begin{matrix}1 & −1 \\ −2 & 0 \end{matrix} ) = ( \begin{matrix}− \frac{1}{2} & \frac{1}{2} \\ 1 & 0 \end{matrix} ) .

Alternative argument: We have

\begin{matrix}b⃗_{1} = 0ê_{1} + 2ê_{2} , \\ b⃗_{2} = 1ê_{1} + 1ê_{2} . \end{matrix}

That is, ( \begin{matrix}0 & 2 \\ 1 & 1 \end{matrix} ) is the matrix that expresses the B basis vectors in terms of the natural basis.

Therefore, its transpose (which is what we called B above) is the matrix that expresses the natural coordinates in terms of the coordinates with respect to basis B . Again, the desired M is the inverse of that.

4. (25 pts.) Suppose that the linear function L: R2 → R3 is represented by A = ( \begin{matrix}1 & 2 \\ 3 & 4 \end{matrix} )

with respect to the natural basis.

(a) What is the rank of A ?  \underline{2}  (since there are two independent columns).

(b) Find the matrix representing f if the basis in the domain is changed to B (see Question 3), the basis for the codomain remaining unchanged.

First map the coordinates from B to the natural coordinates, then apply A :

AM^{−1} = AB = ( \begin{matrix}1 & 2 \\ 3 & 4 \end{matrix} ) ( \begin{matrix}0 & 1 \\ 2 & 1 \end{matrix} ) = ( \begin{matrix}4 & 3 \\ 8 & 7 \end{matrix} ) .

(c) Find the matrix representing f if the basis B is used for both domain and codomain.

Same as (b), except that at the end we need to map the coordinates of the output vector from natural to B :

MAM^{−1} = B^{−1}AB = ( \begin{matrix}− \frac{1}{2} & \frac{1}{2} \\ 1 & 0 \end{matrix} ) ( \begin{matrix}4 & 3 \\ 8 & 7 \end{matrix} ) = ( \begin{matrix}2 & 2 \\ 4 & 3 \end{matrix} ) .

5. (Essay – 15 pts.) Explain how the “superposition principles” are used in solving a nonho-

mogeneous differential equation such as \frac{d^{2}y}{dt^{2}} + 4y = cos(3t) .

Best student response received:

1st apply superposltion principle #1 which states that the linear combination of solution to a homogeneous problem is a solution to that problem. From this we have L(y_{1}) = 0 and L(y_{2}) = 0 , where y_{1} and y_{2} are 2 solutions and when added make up a solution to the homogeneous equation. So in general y = c_{1}y_{1} + c_{2}y_{2} . Now we apply the superposition principle #2, which states that the sum of a particular solution of a nonhomogeneous and the solution of its corresponding homogeneous solution, give a solution of the original nonhomogeneous problem. So if L(y_{p}) = f , then a solution to the nonhomogeneous differential equation to problem 5 is

y = y_{p} + c_{1}y_{1} + c_{2}y_{2} .

To this I would add only that the solutions y_{1} and y_{2} must be chosen to be linearly independent, and that the 3rd superposition principle (difference of two solutions of the same nonhomogeneous problem is a solution of the homogeneous one) proves that the final formula gives all the solutions.

6. (20 pts.) Let V be the span of the functions {v⃗1 = e−t, v⃗2 = te−t, v⃗3 = t2e−t} (that is, each element of V is e−t times an element of P2 ). Let L: V → V be the differentiation operator with V as both domain and codomain (that is, L: V → V and Lf ≡ f′ ).

(a) Find the matrix representing L with respect to the basis {v⃗1, v⃗2, v⃗3} .

Calculate

L(v⃗_{1}) = −e^{−t} = −v⃗_{1} , L(v⃗_{2}) = e^{−t} − te^{−t} = v⃗_{1} − v⃗_{2} , L(v⃗_{3}) = 2te^{−t} − t^{2}e^{−t} = 2v⃗_{2} − v⃗_{3} .

Put the coefficients in these equations into the columns of the matrix in the appropriate order:

( \begin{matrix}−1 & 1 & 0 \\ 0 & −1 & 2 \\ 0 & 0 & −1 \end{matrix} ) .

(b) Is L surjective (onto)? If not, what is its range?

Yes — Columns span the whole space.

(c) Is L injective (one-to-one)? If not, what is its kernel?

Yes — Matrix is nonsingular, kernel is just {0} . (Note also that (c) follows from (b) or vice versa by the theorem (rank) + (nullity) = (dimension of domain).)