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Homework 8 Solutions
Math 110
Section 5.
- (a) The characteristic polynomial of A is (1−t)(2−t)−6 = t^2 − 3 t−4 = (t−4)(t+1); therefore, the eigenvalues of A are 4 and −1. For λ = 4, we have E 4 = N S
[
]
= span{(2, 3)}.
Similarly, for λ = −1, we have
E 1 = N S
[
]
= span{(− 1 , 1)}.
Thus, {(2, 3), (− 1 , 1)} is a basis for R^2 consisting of eigenvectors; we thus get Q−^1 AQ = D for Q =
[
]
, D =
[
]
- (h) Let ε = {E 11 , E 12 , E 21 , E 22 } be the standard basis for M 2 × 2 (R). Then we see T E 11 = E 22 , T E 12 = E 12 , T E 21 = E 21 , and T E 22 = E 11. Thus,
[T ]ε =
Now the characteristic polynomial of T is
det
−t 0 0 1 0 1 − t 0 0 0 0 1 − t 0 1 0 0 −t
Expanding by minors along the second and third columns, this is equal to (1 − t)^2 [(−t)^2 − 12 ] = (t − 1)^2 (t^2 − 1) = (t − 1)^3 (t + 1). Therefore, the eigenvalues of T are 1 and −1. Now the eigenspace of 1 corresponds to
N S
= span{(0,^1 ,^0 ,^ 0),^ (0,^0 ,^1 ,^ 0),^ (1,^0 ,^0 ,^ 1)}.
In M 2 × 2 (R), this corresponds to span{E 12 , E 21 , E 11 + E 22 }. On the other hand, the eigenspace of −1 corresponds to
N S
= span{(−^1 ,^0 ,^0 ,^ 1)}.
In M 2 × 2 (R), this corresponds to span{−E 11 + E 22 }. Therefore, β =
{[
]
[
]
[
]
[
]}
diagonalizes T.
- (a) We see that zero is an eigenvalue of T if and only if T x = 0 for some nonzero vector x ∈ V , or in other words if and only if T is not one-to-one. Thus, T is one-to-one if and only if zero is not an eigenvalue of T ; however, since V is finite-dimensional, T is invertible if and only if T is one-to-one. (b) (⇒) Suppose λ is an eigenvalue of T , and find a corresponding eigenvector x ∈ V. Then since T x = λx, we also have T (λ−^1 x) = x, so T −^1 x = λ−^1 x. Therefore, λ−^1 is an eigenvalue of T −^1. (⇐) If λ−^1 is an eigenvalue of T −^1 , then by the previous part, λ = (λ−^1 )−^1 is an eigenvalue of T = (T −^1 )−^1. (c) If A ∈ Mn×n(F ), then A is invertible if and only if 0 is not an eigenvalue of A. Also, if A is invertible, then λ is an eigenvalue of A if and only if λ−^1 is an eigenvalue of A−^1. (These results follow from the previous parts by letting T = LA.)
- Since (A − tI)t^ = At^ − tI, we have
det(A − tI) = det[(A − tI)t] = det(At^ − tI).
- (a) We prove this by induction on m; for m = 1, the statement is exactly what we are assuming. Now suppose x is an eigenvector of T m^ with eigenvalue λm; then we calculate
T m+1x = T (T mx) = T (λmx) = λm(T x) = λm(λx) = λm+1x.
Therefore, x is also an eigenvector of T m+1^ with eigenvalue λm+1, completing the induction. (b) Let A ∈ Mn×n(F ), and let x ∈ F n^ be an eigenvector of A with eigenvalue λ. Then for any positive integer m, x is an eigenvector of Am^ with eigenvalue λm. (This follows from the previous part by letting T = LA.)
- (a) The ij-entry of A − tI is Aij − tδij. Therefore, we get
f (t) = det(A − tI) =
σ∈Sn
sgn(σ)(Aσ(1), 1 − tδσ(1), 1 )(Aσ(2), 2 − tδσ(2), 2 ) · · · (Aσ(n),n − tδσ(n),n).
If σ is the identity permutation, this term gives exactly (A 11 − t)(A 22 − t) · · · (Ann − t). On the other hand, if σ is any other permutation, then σ(i) 6 = i for at least two values of i. (This is because if σ(i) = j 6 = i for some i, then σ(j) also cannot be equal to j.) Therefore, any other term of the sum gives a polynomial of degree at most n − 2, so the sum of all the other terms is also a polynomial q(t) of degree at most n − 2. (b) From the previous part, the coefficient of tn−^1 in f (t) is equal to the coefficient of tn−^1 in (A 11 − t)(A 22 − t) · · · (Ann − t), which is (−1)n−^1 (A 11 + A 22 + · · · + Ann). Therefore, an− 1 = (−1)n−^1 (tr A), so tr A = (−1)n−^1 an− 1.
with y 1 ∈ span(β 1 ),... , yk− 1 ∈ span(βk− 1 ). This, x = y 1 + · · · + yk− 1 + z ∈ span(β 1 ) + · · · + span(βk− 1 ) + span(βk). Also, suppose that y 1 + · · · + yk− 1 + yk = 0 with y 1 ∈ span(β 1 ),... , yk ∈ span(βk). Then yk = −y 1 − · · · − yk− 1 ∈ W 1 ∩ W 2 , so yk = 0. This implies that y 1 + · · · + yk− 1 = 0, so y 1 = · · · = yk− 1 = 0 also since W 1 = span(β 1 ) ⊕ · · · ⊕ span(βk− 1 ).
X1. (a) Let xk denote the kth vector; we then see that Axk = (ωk^ +ω^6 k, 1+ω^2 k, ωk^ +ω^3 k,... , ω^4 k^ + ω^6 k, 1 + ω^5 k). However, since ω^7 = 0, we have ω^6 k^ = ω−k^ and 1 = ω^7 k, so Axk = (ωk^ + ω−k)xk. Therefore, xk is an eigenvector of A, with eigenvalue ωk^ + ω−k. However, since ωk^ = cos( 2 πk 7 ) + i sin( 2 πk 7 ) and ω−k^ = cos( 2 πk 7 ) − i sin( 2 πk 7 ), this eigenvalue is equal to 2 cos( 2 πk 7 ). (b) For k = 1, 2 , 3, the eigenspace of 2 cos( 2 πk 7 ) has two linearly independent vectors xk, x−k. Therefore, {x− 3 , x− 2 ,... , x 3 } is linearly independent by 5.8, so it is a basis of C^7 consisting of eigenvectors of A. (In fact, A is diagonalizable over R as well: all the eigenvalues are real, and for k = 1 , 2 , 3, 12 (xk + x−k) = (1, cos( 2 πk 7 ),... , cos( 127 πk )) and (^21) i (xk − x−k) = (0, sin( 2 πk 7 ),... , sin( 127 πk )) are linearly independent real eigenvectors corresponding to 2 cos( 2 πk 7 ). Since x 0 = (1, 1 ,... , 1), we thus get a basis of R^7 consisting of real eigenvectors of A.) Alternately, A is symmetric, and we will see later that any symmetric matrix is diagonal- izable. (c) If we move the top row of A to the bottom, this does not change the sign of the determi- nant, while the result is almost upper triangular. We now reduce A as follows:
None of these steps changes the determinant; thus, det A = 2. (d) From part (b), the eigenvectors of A are 2 with multiplicity 1 and 2 cos( 27 π ), 2 cos( 47 π ), 2 cos( 67 π ) each with multiplicity 2. Since the product of the eigenvectors is equal to det A, we thus get 27 cos^2
2 π 7
cos^2
4 π 7
cos^2
6 π 7
Therefore, cos( 27 π ) cos( 47 π ) cos( 67 π ) = ± 18. However, cos( 27 π ) is positive, while cos( 47 π ) and cos( 67 π ) are negative, so we conclude
cos
2 π 7
cos
4 π 7
cos
6 π 7
Section 5.
- (a) If f ∈ W , then f has degree at most 2, so f ′^ has degree at most 1 ≤ 2, which implies f ′^ ∈ W also. Therefore, W is T -invariant.
(e) In this case, W is not T -invariant; for example, A =
[
]
∈ W , but T (A) =
[
]
∈ / W.
- (a) We calculate T (e 1 ) = (1, 0 , 1 , 1); T 2 (e 1 ) = (1, − 1 , 2 , 2); and T 3 (e 1 ) = (0, − 3 , 3 , 3) = 3 T 2 (e 1 ) − 6 T (e 1 ) + 3e 1. Therefore, the T -cyclic subspace generated by e 1 is span{(1, 0 , 0 , 0), (1, 0 , 1 , 1), (1, − 1 , 2 , 2)}. This subspace can also be identified as {(a, b, c, d) : c = d}.
- We can rewrite any linear combination c 0 In + c 1 A + c 2 A^2 + · · · + ckAk^ as p(A), where p(t) = cktk^ + · · · + c 1 t + c 0. We now divide p(t) by the characteristic polynomial f (t) of A to get p(t) = q(t)f (t)+r(t), where the degree of r(t) is less than n. This implies that p(A) = q(A)f (A)+r(A); however, by the Cayley-Hamilton theorem, f (A) = 0. Therefore, p(A) = r(A) is a linear combination of In, A, A^2 ,... , An−^1. We have thus shown that span{In, A, A^2 ,.. .} ⊆ span{In, A, A^2 ,... , An−^1 }; since the right hand side clearly has dimension at most n, this implies the desired result. Alternately, consider the linear transformation T : Mn×n(F ) → Mn×n(F ) such that T (B) = AB. Then the T -invariant subspace generated by In is exactly span{In, A, A^2 ,.. .}; but since the Cayley-Hamilton therom implies that An^ is a linear combination of In, A, A^2 ,... , An−^1 , our general results on T -cyclic subspaces imply that span{In, A, A^2 ,.. .} has dimension at most n.
- Since A has rank 1, the null space of A has dimension n − 1, which implies that A has eigenvalue 0 with multiplicity at least n − 1. On the other hand, A also has eigenvalue n with corresponding eigenvector x = (1, 1 ,... , 1). Therefore, the characteristic polynomial of A must be (−t)n−^1 (n − t).
