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MATH 425 Written Homework 7 Solution Radford 04/21/
- (20 points) A =
. Therefore cA(x) = (9 − x)^3 (8 − x) from which
we deduce Dim N ((A − 9I 4 )^3 ) = 3, Dim N (A − 8I 4 ) = 1. Note that v 1 =
forms
a basis for the space of eigenvectors for A belonging to λ − 8, hence forms a basis for N (A − 8I 4 ). Now N ((A − 9I 4 )^3 ) = R(A − 8I 4 ) has basis {e 4 , e 3 , e 1 + 9e 2 }. Under multiplication
by A − 9I 4 =
observe that e 4 −→ 0, e 3 −→ 10 e 4 −→ 0, and e 1 +
9 e 2 −→ 111 e 4 −→ 0. Therefore v 2 = e 1 + 9e 2 − 11110 e 3 , v 3 = 10e 4 form an independent set of eigenvectors for A belonging to λ = 9. Set v 4 = e 3. Let Let S = (v 1 · · · v 4 ) =
( 12 ) and J =
Comment: I was not particular about the order of the blocks.
- (20 points) A =
. Therefore cA(x) = (9 − x)^2 x^2 from which we
deduce Dim N ((A − 9I 4 )^2 ) = 2 = Dim N (A^2 ). By inspection {e 2 , e 3 } form a basis for the space of eigenvectors for A belonging to λ = 0, hence form a basis for N (A^2 ). Let v 1 = e 2 , v 2 = e 3. Now R(A^2 ) = N ((A − 9I 4 )^2 ) has basis { 9 e 1 + 9e 4 , e 4 }. Let v 4 = 9e 1 + 9e 4 and v 3 = (A−9I 4 )v 4 = 81e 4. Then {v 3 , v 4 } is a basis for N ((A−I 3 )^2 ). Let S = (v 1 · · · v 4 ) =
( 12 ) and J =
. ( 8 )
Comment: Note that {e 1 , e 4 } is a basis for N ((A−9I 4 )^2 ) as well. Thus v 4 = e 1 and v 3 = 9e 4 work also. I was not particular about the order of the blocks.
- (20 points) We are to find cA(x) and mA(x), where A is the matrix of the reflection
R : R^2 −→ R^2 through S = span(
( 3 7
) ).
Now R is a reflection through a line in R^2. Therefore A has eigenvalues 1, −1. Thus (x−1)(x+1) divides cA(x), since the roots of cA(x) are the eigenvalues of A, and (x−1)(x+1) divides mA(x) by part d) of Proposition 7.6.1. ( 10 ) Since mA(x) divides cA(x) by part c) of the same, both have highest coefficient 1, and the degree of cA(x) is 2, cA(x) = (x − 1)(x + 1) = mA(x). ( 10 )
Comment: The matrix of R is A =
. The solution can be based on this.
- (20 points) Let m ≥ 0. Then D(1) = 0 and D(xm) = mxm−^1 for all m ≥ 1. Let B = { 1 , x,... , xn} be the natural basis for P n. Then [D]B is an (n + 1)×(n + 1) upper triangular matric with zeros on the diagonal. Therefore cD(x) = c[D]B (x) = (−x)n+1. ( 8 ) To find mD(x) one can verify Exercise 8.1.5 and use Proposition 7.6.1 to show that mD(x) = xn+1(= ±cD(x)). A more direct way is to first note that Dn+1^ = 0 and Dn^6 = 0. In particular {I,... , Dn+1} is dependent. We will show {I,... , Dn} is independent. Since Dn+1^ = 0 by definition mD(x) = xn+1. Suppose that {I,... , Dn} is dependent. Then a 0 I + a 1 D + · · · + an− 1 Dn^ = 0 for some a 0 ,... , an ∈ R not all of which are zero. Thus amDm^ + · · · + anDn^ = 0 where 0 ≤ m ≤ n and am 6 = 0. Applying Dn−m^ to both sides of the preceding equation amDn^ = 0. Since Dn^6 = 0 necessarily am = 0, a contradiction. Thus {I,... , Dn} is independent after all. ( 12 )
- (20 points) Let v ∈ V , λ ∈ R and suppose that T (v) = λv. Then T m(v) = λmv for all m ≥ 0 by induction on m. Since T 0 (v) = v = λ^0 v the assertion follows for m = 0. ( 8 ) Suppose that m ≥ 0 and T m(v) = λmv. The calculation
T m+1(v) = T (T m(v)) = T (λmv) = λmT (v) = λm(λv) = λm+1v
shows that the assertion holds for m + 1. Thus the assertion holds for all m ≥ 0 by induction. ( 8 ) Next let p(x) = a 0 + a 1 x + · · · + anxn^ ∈ R[x]. Then the calculation
p(T )(v) = (a 0 I + a 1 T + · · · + anT n)(v) = a 0 I(v) + a 1 T (v) + · · · + anT (v) = a 0 v + a 1 (λv) + · · · + an(λnv) = (a 0 + a 1 λ + · · · + anλn)v = p(λ)v
shows that p(T )(v) = p(λ)v. ( 8 )
