Mathematics 244 – Linear Algebra Solutions – Exam 2 March 29, 2004

I. A) Let T : P_{3}(R) → R^{3} be the mapping defined by T (p) = (p(1), p′(1), p′′(1)). Then

for all p, q ∈ P_{3}(R) and all c ∈ R, we have by the sum and scalar multiple rules for derivatives:

T (cp + q) = ((cp + q)(1), (cp + q)′(1), (cp + q)′′(1))
= (cp(1) + q(1), cp′(1) + q′(1), cp′′(1) + q′′(1))
= c(p(1), p′(1), p′′(1)) + (q(1), q′(1), q′′(1))
= cT (p) + T (q)

So T is linear.

B) Let α = {1, x + x^{2}, x^{2} + x^{3}, x^{3} + 1}. Since we know dim P_{3}(R) = 4 and there are 4 vectors in α, to show α is a basis, it suffices to show that α is linearly independent. So if

0 = a_{1}(1) + a_{2}(x + x^{2}) + a_{3}(x^{2} + x^{3}) + a_{4}(x^{3} + 1)

then collecting like powers of x on both sides of the equation:

0 = (a_{1} + a_{4}) + a_{2}x + (a_{2} + a_{3})x^{2} + (a_{4} + a_{3})x^{3}

implies

a_{1} + a_{4} = 0, a_{2} = 0, a_{2} + a_{3} = 0, a_{4} + a_{3} = 0

It easily follows that all a_{i} = 0, so α is linearly independent.

C) To find the matrix of T with respect to α in the domain and the standard basis β = {e_{1}, e_{2}, e_{3}} in R^{3}, we follow the usual “recipe”:

\begin{aligned}T (1) & = (1, 0, 0) = 1e_{1} + 0e_{2} + 0e_{3} \\ T (x + x^{2}) & = (x + x^{2}, 1 + 2x, 2)|_{x=1} = (2, 3, 2) = 2e_{1} + 3e_{2} + 2e_{3} \\ T (x^{2} + x^{3}) & = (2, 5, 8) = 2e_{1} + 5e_{2} + 8e_{3} \\ T (x^{3} + 1) & = (2, 3, 6) = 2e_{1} + 3e_{2} + 6e_{3} \end{aligned}

The scalars from these linear combinations go in the columns of the matrix:

[T]_{α}^{β} = ( \begin{matrix}1 & 2 & 2 & 2 \\ 0 & 3 & 5 & 3 \\ 0 & 2 & 8 & 6 \end{matrix} )

D) Following the same procedure as in part C,

\begin{aligned}S(e_{1}) & = (3, 1, 1) = 3e_{1} + 1e_{2} + 1e_{3} \\ S(e_{2}) & = (−1, 1, 0) = −1e_{1} + 1e_{2} + 0e_{3} \\ S(e_{3}) & = (0, 1, 0) = 0e_{1} + 1e_{2} + 0e_{3} \end{aligned}

So

[S]_{β}^{β} = ( \begin{matrix}3 & −1 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} )

Hence by our general results about composition of linear mappings and matrix multiplication,

[ST]_{α}^{β} = [S]_{β}^{β}[T]_{α}^{β}

= ( \begin{matrix}3 & −1 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} ) ( \begin{matrix}1 & 2 & 2 & 2 \\ 0 & 3 & 5 & 3 \\ 0 & 2 & 8 & 6 \end{matrix} )

= ( \begin{matrix}3 & 3 & 1 & 3 \\ 1 & 7 & 15 & 11 \\ 1 & 2 & 2 & 2 \end{matrix} )

II. See class notes, notes from review session, or the text for the statement and the proof of the Dimension Theorem.

III. A) This is TRUE (i.e. true for all linear mappings T : R^{3} → R^{5}). Proof: If T : R^{3} → R^{5}

is any linear mapping, then the Dimension Theorem says

dim R^{3} = 3 = dim Ker(T) + dim Im(T)

Since dim Ker(T) ≥ 0, this says dim Im(T) ≤ 3. Hence Im(T) ≠ R^{5}, so there are some b ∈ R^{5} such that there are no solutions of the equation T (x) = b.

B) This is FALSE. We proved a statement saying the opposite, but only in the case that T : V → W with V, W finite-dimensional of the same dimension. Since P (R) is not finite-dimensional that result does not apply in this case. However just because our other result does not apply is not a sufficient reason to say the statement is false here. Instead, we need to give an explicit counterexample. First note that {1, x, x^{2}, x^{3}, … } = {x^{k} : k ≥ 0} is a basis of P (R). By one of our general results, to define a linear mapping from P (R) to itself, it suffices to say where the elements of this basis map. To get an example of a non-injective T that is surjective, we can say (for instance):

T (1) = 1
T (x) = 1
T (x^{2}) = x
⋮
T (x^{k}) = x^{k−1}

(all k ≥ 1) There is a unique linear mapping satisfying this. T is certainly not injective since T (1) = T (x). (Says that Ker(T) contains p(x) = x − 1.) On the other hand, T is surjective since a basis for P (R) is contained in Im(T). (Note: Another example of the same type is given by T : P (R) → P (R) where T (f) = f′ – the derivative mapping!)

C) This is FALSE. The three given vectors are linearly independent since if

a_{1}(1, 1, 1, 0, 0, 0, 0) + a_{2}(0, 0, 0, 1, 1, 1, 1) + a_{3}(1, 0, 1, 0, 1, 0, 1) = (0, 0, 0, 0, 0, 0, 0)

then components 1, 2, 4 give:

a_{1} + a_{3} = 0, a_{1} = 0, a_{2} = 0,

which implies a_{1} = a_{2} = a_{3} = 0. Hence if these vectors were in the image of T we would have dim Im(T) ≥ 3. (You can’t tell dim Im(T) ≥ 3 without checking this.) But the Dimension Theorem says

4 = dim R^{4} = dim Ker(T) + dim Im(T)

so if dim Ker(T) = 2, then dim Im(T) = 2.

Extra Credit

Let the system be

a_{11}x_{1} + ⋯ + a_{1n}x_{n} = 0
⋮  ...  ⋮  ⋮
a_{m1}x_{1} + ⋯ + a_{mn}x_{n} = 0

The m × n matrix A = (a_{ij}) defines the linear mapping T_{A} : R^{n} → R^{m} (where T_{A}(x) = Ax). The solutions of the homogeneous system are exactly the vectors in Ker(T) ⊂ R^{n}. The image of T_{A} is contained in R^{m} so has dimension ≤ m. Since m < n, the Dimension Theorem says

dim Ker(T) = n − dim Im(T) ≥ n − m > 0

Hence Ker(T) contains a nonzero vector – a nontrivial solution of the system of equations.