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Practice Exam 2 Solutions - Linear Algebra | MATH 333, Exams of Linear Algebra

Material Type: Exam; Class: Linear Algebra; Subject: Mathematics; University: SUNY at Geneseo; Term: Unknown 1989;

Typology: Exams

Pre 2010

Uploaded on 08/16/2009

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Math 333 - Practice Exam 2 with Some Solutions
(Note that the exam will NOT be this long.)
1 Definitions
1. (0 points) Let T:VWbe a transformation. Let Abe a square matrix.
(a) Define Tis linear”.
(b) Define the null space of T, null(T).
(c) Define the image of T, image(T).
(d) Define Tis one-to-one”.
(e) Define Tis onto”.
(f) Define Tis invertible”.
(g) Define Tis an isomorphism”.
(h) Define rank(T) and nullity(T).
(i) Define Ais invertible”.
Solution: See your notes or textbook.
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Math 333 - Practice Exam 2 with Some Solutions

(Note that the exam will NOT be this long.)

1 Definitions

  1. (0 points) Let T : V → W be a transformation. Let A be a square matrix. (a) Define “T is linear”.

(b) Define the null space of T , null(T ).

(c) Define the image of T , image(T ).

(d) Define “T is one-to-one”.

(e) Define “T is onto”.

(f) Define “T is invertible”.

(g) Define “T is an isomorphism”.

(h) Define rank(T ) and nullity(T ).

(i) Define “A is invertible”.

Solution: See your notes or textbook.

2 Linear Transformations, Null Spaces, and Images

  1. (0 points) Let T : P 2 (R) → P 2 (R) be given by T (f (x)) = f (x) − xf ′(x). (a) Show T is linear.

Solution: Let a ∈ R and f (x), g(x) ∈ P 2 (R). Then

T (af (x) + g(x)) = [af (x) + g(x)] − x[af (x) + g(x)]′ = af (x) + g(x) − axf ′(x) − xg′(x) = a[f (x) − xf ′(x)] + [g(x) − xg′(x)] = aT (f (x)) + T (g(x)).

(b) Find a basis for the image of T.

Solution: We know that a generating set for the image of T is the image of the standard basis of P 2 (R). Thus

image(T ) = span({T (1), T (x), T (x^2 )})

= span({ 1 , x − x, x^2 − 2 x^2 }) = span({ 1 , −x^2 }).

The vectors { 1 , −x^2 } are clearly linearly independent, so it will also be a basis.

(c) Is T one-to-one? Is T onto? Justify your answer.

Solution: Since rank(T ) = 2 and dim(P 2 (R)) = 3, T is clearly not onto. Fur- thermore, the Dimension Theorem says the nullity(T ) = 1, so T is not one-to-one either.

  1. (0 points) Let T : V −→ W be a linear transformation. Prove the following theorems. (a) Theorem 2.1: The sets null(T ) and image(T ) are subspaces of V and W , respec- tively.

(b) Theorem 2.2: Let β be a basis of V. Then the set {T (β)} is a generating set for image(T ).

(c) Theorem 2.4: T is one-to-one if and only if null(T ) = { 0 }.

Solution: See your notes or textbook.

3 Matrix Representations and Change of Basis

  1. (0 points) Consider the vector space V = P 1 (R). (a) Explain why you know that the set β = {1 + x, 1 − 2 x} is a basis of V.

Solution: Since neither vector is a multiple of the other, β is linearly independent. Since the dimension of V is 2, β is a basis.

(b) Determine [p(x)]β , where p(x) = 2x − 3 ∈ V.

Solution: Notice that p(x) = 2x − 3 = (− 4 /3)(1 + x) + (− 5 /3)(1 − 2 x). Therefore [p(x)]β =

  1. (0 points) Let T : P 2 (R) −→ R^2 be given by T (f (x)) = (f (0), f ′(1)). (a) Show that T is linear.

Solution: Let f (x), g(x) ∈ P 2 (R) and c ∈ R. Then

T (cf (x) + g(x)) = (cf (0) + g(0), cf ′(1) + g′(1))

= c(f (0), f ′(1)) + (g(0), g′(1)) = cT (f (x)) + T (g(x)).

(b) Determine the matrix of T with respect to the standard bases of P 2 (R) and R^2.

Solution: First we recall that the standard basis of P 2 (R) is β = { 1 , x, x^2 } and that the standard basis of R^2 is γ = {(1, 0), (0, 1)}. Now we look at the image of each element of the basis β under T. T (1) = (1, 0), T (x) = (0, 1), and T (x^2 ) = (0, 2). Since we are using the standard basis of R^2 the columns of our matrix are the vectors we have just written. So our matrix is

[T ]γβ =

  1. (0 points) Let β and γ be the following standard ordered bases of M 2 × 2 (R) and P 2 (R), respectively:

β =

and γ = { 1 , x, x^2 }.

Compute [T ]βγ if we define the linear transformation T : P 2 (R) −→ M 2 × 2 (R) by

T (f (x)) =

f ′(0) 2 f (1) 0 f ′′(3)

Solution: First we see that T (1) =

. So the first column of [T ]βγ is the coordi-

nate vector [T (1)]β = (0, 2 , 0 , 0). Next T (x) =

. So the second column of [T ]βγ

is the coordinate vector [T (x)]β = (1, 2 , 0 , 0). Finally T (x^2 ) =

. So the third

column of [T ]βγ is the coordinate vector [T (x^2 )]β = (0, 2 , 0 , 2). So in total we get

[T ]βγ =

Solution: (a) Plugging basis β into T and writing as a linear combination of the elements of γ, we

get [T ]γβ =

(b) Plugging basis α into T and writing as a linear combination of the elements of γ, we

get [T ]γα =

(c) To get the change of basis matrix, we must find the coordinate vectors of the elements of β with respect to α:

[(1, 0 , 0)]α =

, [(0, 1 , 0)]α =

, and [(0, 0 , 1)]α =

Therefore the change of basis matrix is Q =

(d) [T ]γα · Q =

= [T ]γβ.

(e) x = (1, 5 , 7) =⇒ [x]β =

=⇒ [x]α = Q · [x]β =