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Practice Final Exam
UCLA: Math 115A
Instructor: Jens Eberhardt
- This exam has 6 questions, for a total of 60 points.
- Please print your working and answers neatly.
- Write your solutions in the space provided showing working.
- Indicate your final answer clearly.
- You may write on the reverse of a page or on the blank pages found at the back of the booklet however these will not be graded unless very clearly indicated.
- Non programmable and non graphing calculators are allowed.
Name:
ID number:
Question Points Score
1 10 2 10 3 10
4 10 5 10 6 10
Total: 60
- Consider the vector space V = P 2 (R) with its standard ordered basis
β =
1 , x, x^2
and the linear maps T : P 2 (R) → P 2 (R), T (f ) = f (1) + f (−1)x + f (0)x^2 S : P 2 (R) → P 2 (R), S(ax^2 + bx + c) = cx^2 + bx + a.
(a) (3 points) What is [T ]β and [S]β? Show that
[T S]β =
(b) (6 points) Compute [(T S)−^1 ]β. (c) (1 point) What is (T S)−^1 (x^2 + x + 1)?
Solution: (a) We have
T (1) = 1 + x + x^2 T (x) = 1 − x T (x^2 ) = 1 + x
Hence
[T ]β =
Also clearly
[S]β =
Using [T S]β = [T ]β [S]β we get
[T S]β =
(b) We have [(T + S)−^1 ]β = [T + S]− β 1. We hence have to invert [T + S]β
Substract 2nd from 1st (^)
Norm 2nd (^)
- Consider the matrix
A =
∈ M 3 , 3 (R).
(a) (2 points) Compute the characteristic polynomial of A and determine the eigenvalues and their algebraic multiplicity. (b) (6 points) Is A is diagonalizable? If yes, compute a basis β of eigenvectors of A. (c) (2 points) Compute [LA]β , where the LA is the linear transformation given by
LA : R^3 → R^3 , v 7 → Av.
Solution: (a) We compute
det(A − tI 3 ) = det
−t 1 − 2 1 −t − 2 0 0 − 1 − t
(^) = (t^2 − 1)(− 1 − t) = −(t + 1)^2 (t − 1)
(b) The eigenvalues are the roots of the characteristic polynomial and hence λ = 1, −1 with multi- plicity 1 and 2 respectively.
(c) We compute N(A − 1 I 3 ) = Span((1, 1 , 0)t) N(A − (−1)I 3 ) = Span((− 1 , 1 , 0)t, (2, 0 , 1)t) with our favorite algorithm (Wolfram Alpha). Hence
β = {(1, 1 , 0)t, (− 1 , 1 , 0)t, (2, 0 , 1)t}
is a basis of eigenvectors.
(d) A diagonal matrix with entries 1, − 1 , −1 in an appropriate order.
- Consider the vector space V = R^4 with its standard inner product. Consider the linearly independent subset S = {w 1 = (1, 0 , 1 , 0), w 2 = (1, 1 , 1 , 1), w 3 = (2, 2 , 0 , 2)}. (a) (6 points) Apply the Gram-Schmidt orthogonalization algorithm to S to compute an orthogonal basis β′^ of Span(S). (b) (2 points) Use your result to compute an orthonormal basis β of Span(S). (c) (2 points) Let x = (1, 2 , 3 , 2) ∈ Span(S). Compute the coordinate vector [x]β.
Solution: (a)
v 1 = w 1 = (1, 0 , 1 , 0)
v 2 = w 2 − 〈w 2 , v 1 〉 〈v 1 , v 1 〉
v 1 = (1, 1 , 1 , 1) −
v 3 = w 3 −
〈w 3 , v 1 〉 〈v 1 , v 1 〉
v 1 −
〈w 3 , v 2 〉 〈v 2 , v 2 〉
v 2 = (2, 2 , 0 , 2) −
(b)
u 1 =
||v 1 ||
v 1 =
u 2 =
||v 2 || v 2 =
u 3 =
||v 3 ||
v 3 =
(c)
〈x, v 1 〉 =
〈x, v 2 〉 =
〈x, v 3 〉 =
Hence [x]β = √^12 (4, 4 , −2).
- Let V be a finite dimensional vector space over a field F. Recall that
L(V, V ) = {T : V → V | T is a linear transformation}
denotes the vector space of linear transformations from V to V (also called linear operators on V ). Fix a vector v ∈ V and define Z = {T ∈ L(V, V ) | T (v) = 0}. One calls Z the annihilator of v in L(V, V ). (a) (4 points) Show that Z is a subspace of L(V, V ). (b) (2 points) Let λ ∈ F such that λ 6 = 0. Prove or disprove (by finding a counterexample) that
Z′^ = {T ∈ L(V, V ) | T (v) = λv}
is a subspace of L(V, V ). (c) (2 points) Assume that β = {v 1 ,... , vn} is an ordered basis of V , such that v 1 = v. Let T ∈ L(V, V ). Show that T ∈ Z if and only if the first column of A = [T ]β equals 0. (d) (2 points) Assuming v 6 = 0, what is dim(Z)?
Solution: (a) One easily checks φv : L(V, V ) → F, T 7 → T (v) is linear. Hence Z = N(φv ) is a subspace.
(b) Let V = R, v = 1, T 0 : R → R, x 7 → 0 be the zero map and λ = 1. Then T 0 ∈/ Z′, since T 0 (v) = 0 6 = 1 = λv. Hence Z′^ is not a subspace.
(c) We know that the first column in A is given by [T (v)]β. Assume that T ∈ Z, then T (v) = 0 and clearly [T (v)]β = 0. Assume that the first column of A equals zero, i.e [T (v)]β = 0. Then clearly T (v) = 0.
(d) Denote by X ⊂ Mn,n(F ) the subspace of all matrices whose first column is zero. Then clearly dim Zm = n^2 − n. By the last part
[−]β : Z → X, T 7 → [T ]β
is an isomorphism (since [−]β : L(V, V ) → Mn,n(F ) is). Hence dim Z = dim X = n^2 − n.
- Let V be a finite dimensional vector space over R with an inner product 〈x, y〉 ∈ R for x, y ∈ V.
(a) (3 points) Let λ ∈ R with λ > 0. Show that
〈x, y〉′^ = λ〈x, y〉, for x, y ∈ V,
defines an inner product on V. (b) (2 points) Let T : V → V be a linear operator, such that
〈T (x), T (y)〉 = 〈x, y〉, for all x, y ∈ V.
Show that T is one-to-one. (c) (2 points) Recall that the norm of a vector x ∈ V is defined by ||x|| =
〈x, x〉. Show that
〈x, y〉 =
(||x + y||^2 − ||x||^2 − ||y||^2 ), for all x, y ∈ V.
Hence, the inner product can be recovered from the norm. Hint: Rewrite 〈x + y, x + y〉 using the properties of inner products. Use that 〈x, y〉 ∈ R is a real number by assumption. (d) (3 points) Let β = {v 1 ,... , vn} be a basis of V. The Gram matrix G ∈ Mn,n(R) of the inner product 〈−, −〉 with respect to β is defined by
Gi,j = 〈vi, vj 〉.
Show that G is invertible.
Solution: (a) Show the 4 properties.
(b) Use non-degeneracy.
(c) Follow hint.
(d) Let x ∈ V with [x]β = (a 1 ,... , an)t. Then
G[x]β = (
∑^ n
j=
G 1 ,j aj ,... ,
∑^ n
j=
Gn,j aj )t
∑^ n
j=
〈v 1 , vj 〉aj ,... ,
∑^ n
j=
〈vn, vj 〉aj )t
= (〈v 1 ,
∑^ n
j=
aj vj 〉,... , 〈vn,
∑^ n
j=
aj vj 〉)t
= (〈v 1 , x〉,... , 〈vn, x〉)t
Hence G[x]β = 0 ⇔ x ∈ V ⊥^ = { 0 } ⇔ x = 0, and G is invertible.
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