Mathematics 244 – Linear Algebra Solutions – Makeup Exam 2 April 2, 2004

I. See class notes or text.

II. A) Let (a_{1}, a_{2}, a_{3}, a_{4}), (b_{1}, b_{2}, b_{3}, b_{4}) ∈ R^{4} and c ∈ R. Then

T (c(a_{1}, a_{2}, a_{3}, a_{4}) + (b_{1}, b_{2}, b_{3}, b_{4}))
= T (ca_{1} + b_{1}, ca_{2} + b_{2}, ca_{3} + b_{3}, ca_{4} + b_{4})
= ((ca_{1} + b_{1}) + (ca_{2} + b_{2}) − (ca_{4} + b_{4}) + ((ca_{2} + b_{2}) − 3(ca_{3} + b_{3}))x
+ (−(ca_{1} + b_{1}) − 2(ca_{2} + b_{2}) + 3(ca_{3} + b_{3}) + (ca_{4} + b_{4}))x^{2}
= c ((a_{1} + a_{2} − a_{4}) + (a_{2} − 3a_{3})x + (−a_{1} − 2a_{2} + 3a_{3} + a_{4})x^{2})
+ (b_{1} + b_{2} − b_{4}) + (b_{2} − 3b_{3})x + (−b_{1} − 2b_{2} + 3b_{3} + b_{4})x^{2}
= cT (a_{1}, a_{2}, a_{3}, a_{4}) + T (b_{1}, b_{2}, b_{3}, b_{4})

Hence T is linear.

B) We have

\begin{aligned}T (1, 0, 0, 0) & = 1 − x^{2} = 0(1 + x) + 0(x) + (−1)(x^{2} − 1) \\ T (0, 1, 0, 0) & = 1 + x − 2x^{2} = (−1)(1 + x) + 2x + (−2)(x^{2} − 1) \\ T (0, 0, 1, 0) & = −3x + 3x^{2} = 3(1 + x) + (−6)x + 3(x^{2} − 1) \\ T (0, 0, 0, 1) & = −1 + x^{2} = 0(1 + x) + 0(x) + 1(x^{2} − 1) \end{aligned}

These expansions are found by solving systems of linear equations. For instance: If

−3x + 3x^{2} = a_{1}(1 + x) + a_{2}x + a_{3}(x^{2} − 1) = (a_{1} − a_{3}) + (a_{1} + a_{2})x + a_{3}x^{2},

then equating coefficients of the powers of x on both sides of the equations: a_{1} − a_{3} = 0, a_{1} + a_{2} = −3, a_{3} = 3. So a_{1} = 3, and a_{2} = −6, which shows how T (0, 0, 1, 0) is derived. Therefore the matrix is

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

C) From the formula given for T above (the matrix could also be used for this, but I’ll do it this way): Ker(T) is the set of all (a_{1}, a_{2}, a_{3}, a_{4}) ∈ R^{4} such that T (a_{1}, a_{2}, a_{3}, a_{4}) = the zero polynomial, or equivalently

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

The equivalent echelon form system is

\begin{aligned}a_{1} & & + & 3a_{3} & − & a_{4} & = 0 \\ & a_{2} & − & 3a_{3} & & & = 0 \end{aligned}

(one equation cancels out completely). This system has 2 free variables (a_{3}, a_{4}), so dim Ker(T) = 2. A basis for the kernel is

{(−3, 3, 1, 0), (1, 0, 0, 1)}

(the first comes from a_{3} = 1, a_{4} = 0, the second from a_{3} = 0, a_{4} = 1).

D) The Dimension Theorem says 4 = dim Ker(T)+dim Im(T). By part C, dim Ker(T) = 2, so dim Im(T) = 2. Since dim P_{2}(R) = 3, this says that T is not surjective.

III. A) False. The only way this statement would have to be true would be if dimKer(T) ≤ 2.

But there are linear mappings for this V, W for which Ker(T) is 3 or larger, so a set of three vectors in the kernel could be linearly independent. For instance, take 3 linearly independent vectors in the kernel of the zero mapping (kernel is all of V).

B) False. Composing the rotations gives a rotation through the sum of the two rotation angles. 2π/3 + 4π/3 = 2π. So ST = R_{2π} = I_{R^{2}}. Using the standard basis β, we would have

[ST]_{β}^{β} = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) .

C) True. If dim Ker(T) = 0, then the only element of the kernel is the zero vector in V. Then if T (v_{1}) = T (v_{2}) for some vectors v_{1} and v_{2} in V, because T is linear, T (v_{1} − v_{2}) = 0. This says v_{1} − v_{2} ∈ Ker(T) so v_{1} − v_{2} = 0, which implies v_{1} = v_{2}. Hence T is injective.