Math. 311 (Fulling) 24 March 1999

Test B – Solutions

1. (18 pts.) Determine whether each set is linearly independent. If it is not, find an independent set with the same span.

(a) \{( \begin{matrix}1 \\ 2 \\ 1 \end{matrix} ) , ( \begin{matrix}2 \\ −1 \\ 4 \end{matrix} )\}

YES — obviously these two vectors are not multiples of each other. More formally,

( \begin{matrix}1 & 2 & 1 \\ 2 & −1 & 4 \end{matrix} ) → ( \begin{matrix}1 & 2 & 1 \\ 0 & −5 & 2 \end{matrix} ) → ( \begin{matrix}1 & 0 & \frac{9}{5} \\ 0 & 1 & − \frac{2}{5} \end{matrix} ) ,

and we see that the number of nonzero rows has not decreased as we put the matrix into row echelon form.

(b) \{( \begin{matrix}1 \\ 2 \\ 3 \end{matrix} ) , ( \begin{matrix}2 \\ −1 \\ 4 \end{matrix} ) , ( \begin{matrix}1 \\ 7 \\ 5 \end{matrix} )\}

Let’s put the vectors into a matrix as rows and reduce:

( \begin{matrix}1 & 2 & 3 \\ 2 & −1 & 4 \\ 1 & 7 & 5 \end{matrix} ) → ( \begin{matrix}1 & 2 & 3 \\ 0 & −5 & −2 \\ 0 & 5 & 2 \end{matrix} ) → ( \begin{matrix}1 & 2 & 3 \\ 0 & 1 & \frac{2}{5} \\ 0 & 0 & 0 \end{matrix} ) → ( \begin{matrix}1 & 0 & \frac{11}{5} \\ 0 & 1 & \frac{2}{5} \\ 0 & 0 & 0 \end{matrix} ) .

So the answer is NO. A basis for the span is

\{( \begin{matrix}1 \\ 2 \\ 3 \end{matrix} ) , ( 0 1 \frac{2}{5} )\} ,  or  \{( 1 0 \frac{11}{5} ) , ( 0 1 \frac{2}{5} )\} ,

or any two of the original three vectors (since in this case none of the vectors is a multiple of any of the others).

2. (12 pts.) Two bases for the solution space of y′′ = 9y are

A = \{a⃗1 = e3t, a⃗2 = e−3t\} and B = \{b⃗1 = cosh(3t), b⃗2 = \frac{1}{3} sinh(3t)\}.

Find the matrix that expresses the coordinates of an arbitrary vector with respect to the A basis in terms of its coordinates with respect to the B basis.

From the definition of the hyperbolic functions, we have

\begin{aligned}b⃗_{1} & = \frac{1}{2} a⃗_{1} + \frac{1}{2} a⃗_{2} , \\ b⃗_{2} & = \frac{1}{6} a⃗_{1} − \frac{1}{6} a⃗_{2} . \end{aligned}

From here there are many equivalent ways to proceed:

311B-S99 Page 2

Method 1: Therefore,

( \frac{1}{6} − \frac{1}{6}^{\frac{1}{2}  \frac{1}{2}} )^{t} = ( \frac{1}{2} − \frac{1}{6}^{\frac{1}{2}  \frac{1}{6}} )

maps the B -coordinates into the A -coordinates, as demanded.

Method 2: Therefore,

(b⃗_{1}, b⃗_{2}) = (a⃗_{1}, a⃗_{2}) ( \frac{1}{2} − \frac{1}{6}^{\frac{1}{2} \frac{1}{6}} ) ,

and the matrix appearing in this equation is the desired one.

Method 3: Therefore, if y = x_{1}a⃗_{1} + x_{2}a⃗_{2} and also y = y_{1}b⃗_{1} + y_{2}b⃗_{2} , then

y = y_{1} (\frac{1}{2} a⃗_{1} + \frac{1}{2} a⃗_{2}) + y_{2} (\frac{1}{6} a⃗_{1} − \frac{1}{6} a⃗_{2}) = (\frac{1}{2} y_{1} + \frac{1}{6} y_{2}) a⃗_{1} + (\frac{1}{2} y_{1} − \frac{1}{6} y_{2}) a⃗_{2} .

Therefore,

( \begin{matrix}x_{1} \\ x_{2} \end{matrix} ) = ( \frac{1}{2} − \frac{1}{6}^{\frac{1}{2} \frac{1}{6}} ) ( \begin{matrix}y_{1} \\ y_{2} \end{matrix} ) .

Remark: The factor \frac{1}{3} was not just thrown in to make the problem harder. b⃗_{2} is a “natural” solution to use, because it satisfies the initial conditions y(0) = 0 , y′(0) = 1 .

3. (32 pts.) A linear mapping L: P2 → P2 is defined by

[L(p)](t) = (t2 − 4)p′′(t) + tp′(t) − 4p(t) (p′ ≡ \frac{dp}{dt} , etc.) .

Note that L(t2) = −8 (more precisely: if p(t) = t2 , then L(p)(t) = −8 for all t ). (This is free information! You don’t have to rederive it.)

(a) Find the matrix that represents L with respect to the standard basis {t2, t, 1} for P2 .

\begin{aligned}L(t^{2}) & = −8, \\ L(t) & = 0 + t − 4t = −3t, \\ L(1) & = −4. \end{aligned} Therefore, the matrix is ( \begin{matrix}0 & 0 & 0 \\ 0 & −3 & 0 \\ −8 & 0 & −4 \end{matrix} ) .

(b) Is L surjective (onto P2 )? If not, what is its range?

NO. The range is P_{1} , the first-degree polynomials. This is clear either from the matrix — the span of the columns being the same as the span of

\{( \begin{matrix}0 \\ 1 \\ 0 \end{matrix} ) , ( \begin{matrix}0 \\ 0 \\ 1 \end{matrix} )\}

— or from the action of L on the basis polynomials, which yields the constants and the multiples of t but no t^{2} terms.

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

NO. From (b) and the dimension theorem, the kernel must have dimension 3 − 2 = 1 . To see what the kernel is, reduce the matrix:

( \begin{matrix}0 & 0 & 0 & 0 \\ 0 & −3 & 0 & 0 \\ −8 & 0 & −4 & 0 \end{matrix} ) → ( \begin{matrix}1 & 0 & \frac{1}{2} & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} ) .

Thus p(t) = at^{2} + bt + c is in the kernel if

a + \frac{1}{2} c = 0, b = 0.

That is, the kernel consists of the multiples of the polynomial

− \frac{1}{2} t^{2} + 1.

(d) Use “superposition principles” to find all solutions in P2 of the differential equation

(t2 − 4)p′′(t) + tp′(t) − 4p(t) = −16.

Obviously one solution is p_{p}(t) = 2t^{2} . The general solution, therefore, is p = p_{p} + p_{c} , where p_{c} is the general element of the kernel (general solution of the corresponding homogeneous equation, L(p) = 0 ). Thus, all the solutions in P_{2} are

p(t) = 2t^{2} + (− \frac{1}{2} t^{2} + 1) c for arbitrary numbers c .

(Of course, there are other solutions of the differential equation that are not quadratic polynomials.)

4. (22 pts.) The matrix M = ( \begin{matrix}1 & 2 & 1 \\ 2 & −1 & 4 \end{matrix} ) represents a linear operator L: R3 → R2 with

respect to the natural bases.

(a) What are the rank and the nullity of M ? (The nullity is the dimension of the kernel of L .)

The rank is 2 : Method 1: It can’t be greater than 2, because the dimension of the whole column space is just 2. It can’t be less than 2, because the three columns are not all proportional. Method 2: The (column) rank is equal to the row rank, and we already saw in Qu. 1(a) that the row rank of this matrix is 2 .

Therefore, by the dimension theorem, the nullity is 3 − 2 = 1 . This can also be checked directly, by reducing the augmented matrix to (compare Qu. 1(a))

( \begin{matrix}1 & 0 & \frac{9}{5} & 0 \\ 0 & 1 & − \frac{2}{5} & 0 \end{matrix} ) ,

from which it is clear that the third coordinate of the solution is arbitrary and the first and second coordinates are then determined.

(b) What matrix represents L if we switch to the basis \{( \begin{matrix}1 \\ 1 \end{matrix} ) , ( \begin{matrix}−1 \\ 1 \end{matrix} )\} for R2 ? (The basis for the domain is still the natural one.)

Call the new basis vectors b⃗_{1} and b⃗_{2} . In terms of the natural basis vectors, we have

\begin{aligned}b⃗_{1} & = ê_{1} + ê_{2} , \\ b⃗_{2} & = −ê_{1} + ê_{2} . \end{aligned}

Therefore (reasoning as in Qu. 2),

C = ( \begin{matrix}1 & −1 \\ 1 & 1 \end{matrix} )

maps b⃗ -coordinates into natural coordinates. (More quickly, this matrix is obtained simply by “stacking the new basis vectors together”.) Therefore,

C^{−1} = \frac{1}{2} ( \begin{matrix}1 & 1 \\ −1 & 1 \end{matrix} )

maps natural coordinates to b⃗ -coordinates. To solve our problem we need to postprocess the naturalbasis calculation of L with C^{−1} . Therefore, the desired matrix is

C^{−1}M = \frac{1}{2} ( \begin{matrix}1 & 1 \\ −1 & 1 \end{matrix} ) ( \begin{matrix}1 & 2 & 1 \\ 2 & −1 & 4 \end{matrix} ) = \frac{1}{2} ( \begin{matrix}3 & 1 & 5 \\ 1 & −3 & 3 \end{matrix} ) .

5. (16 pts.)

(a) Explain what a “subspace” is.

A subspace of a vector space V is a subset of V that is closed under addition and scalar multiplication. The meaning of “closed” is made clear by the proofs below.

(b) Prove that one of these is a subspace (of a vector space V ). [Do either (A) or (B) – your choice.] (A) The kernel of a linear function L: V → V .

Suppose that we have two elements of the kernel: L(v⃗_{1}) = 0 = L(v⃗_{2}) . Then

L(rv⃗_{1} + v⃗_{2}) = rL(v⃗_{1}) + L(v⃗_{2}).

That is, rv⃗_{1} + v⃗_{2} is also in the kernel. Thus the kernel is closed under the vector operations.

(B) The span of a list of vectors in V , {v⃗1, v⃗2, …, v⃗N } .

Suppose that we have two elements of the span:

u⃗_{1} = ∑_{j=1}^{N} c_{j}v⃗_{j} and u⃗_{2} = ∑_{j=1}^{N} d_{j}v⃗_{j} .

Then

ru⃗_{1} + u⃗_{2} = r ∑_{j=1}^{N} c_{j}v⃗_{j} + ∑_{j=1}^{N} d_{j}v⃗_{j} = ∑_{j=1}^{N}(rc_{j} + d_{j})v⃗_{j} ,

which is also an element of the span, as was to be proved.