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Material Type: Exam; Class: Linear Algebra; Subject: Mathematics; University: SUNY at Geneseo; Term: Unknown 1989;
Typology: Exams
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(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
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.
(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
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 =
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 =
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 =
