Linear Algebra Exercises-n-Answers.pdf

190 Linear Algebra, by Hefferon
to get this eigenspace.
⎛ ⎞
b 1 { ⎜ 0
⎟
⎝ 0 ⎠
0
B
∣ b 1 ∈ C} = {b 1 + 0 · x + 0 · x 2 + 0 · x 3 ∣ ∣ b 1 ∈ C} = {b 1
∣ ∣ b 1 ∈ C}
Five.II.3.28 The determinant of the triangular matrix T − xI is the product down the diagonal, and
so it factors into the product of the terms t i,i − x.
Five.II.3.29 Just expand the determinant of T − xI.
∣ a − x c
b d − x∣ = (a − x)(d − x) − bc = x2 + (−a − d) · x + (ad − bc)
Five.II.3.30 Any two representations of that transformation are similar, and similar matrices have the
same characteristic polynomial.
Five.II.3.31 (a) Yes, use λ = 1 and the identity map.
(b) Yes, use the transformation that multiplies by λ.
Five.II.3.32
If t(⃗v) = λ · ⃗v then ⃗v ↦→ ⃗0 under the map t − λ · id.
Five.II.3.33 The characteristic equation
0 =
∣ a − x b
c d − x∣ = (a − x)(d − x) − bc
simplifies to x 2 + (−a − d) · x + (ad − bc). Checking that the values x = a + b and x = a − c satisfy the
equation (under the a + b = c + d condition) is routine.
Five.II.3.34 Consider an eigenspace V λ . Any ⃗w ∈ V λ is the image ⃗w = λ · ⃗v of some ⃗v ∈ V λ (namely,
⃗v = (1/λ)· ⃗w). Thus, on V λ (which is a nontrivial subspace) the action of t −1 is t −1 ( ⃗w) = ⃗v = (1/λ)· ⃗w,
and so 1/λ is an eigenvalue of t −1 .
Five.II.3.35 (a) We have (cT + dI)⃗v = cT⃗v + dI⃗v = cλ⃗v + d⃗v = (cλ + d) · ⃗v.
(b) Suppose that S = P T P −1 is diagonal. Then P (cT + dI)P −1 = P (cT )P −1 + P (dI)P −1 =
cP T P −1 + dI = cS + dI is also diagonal.
Five.II.3.36 The scalar λ is an eigenvalue if and only if the transformation t − λid is singular. A
transformation is singular if and only if it is not an isomorphism (that is, a transformation is an
isomorphism if and only if it is nonsingular).
Five.II.3.37 (a) Where the eigenvalue λ is associated with the eigenvector ⃗x then A k ⃗x = A · · · A⃗x =
A k−1 λ⃗x = λA k−1 ⃗x = · · · = λ k ⃗x. (The full details can be put in by doing induction on k.)
(b) The eigenvector associated wih λ might not be an eigenvector associated with µ.
Five.II.3.38
Five.II.3.39
No. These are two same-sized,
(
equal
)
rank,
(
matrices
)
with different eigenvalues.
1 0 1 0
0 1 0 2
The characteristic polynomial has an odd power and so has at least one real root.
Five.II.3.40 The characteristic polynomial x 3 − 5x 2 + 6x has distinct roots λ 1 = 0, λ 2 = −2, and
λ 3 = −3. Thus the matrix can be diagonalized
⎛
into this form.
⎝ 0 0 0
⎞
0 −2 0 ⎠
0 0 −3
Five.II.3.41 We must show that it is one-to-one and onto, and that it respects the operations of matrix
addition and scalar multiplication.
To show that it is one-to-one, suppose that t P (T ) = t P (S), that is, suppose that P T P −1 = P SP −1 ,
and note that multiplying both sides on the left by P −1 and on the right by P gives that T = S. To
show that it is onto, consider S ∈ M n×n and observe that S = t P (P −1 SP ).
The map t P preserves matrix addition since t P (T + S) = P (T + S)P −1 = (P T + P S)P −1 =
P T P −1 + P SP −1 = t P (T + S) follows from properties of matrix multiplication and addition that we
have seen. Scalar multiplication is similar: t P (cT ) = P (c · T )P −1 = c · (P T P −1 ) = c · t P (T ).
Five.II.3.42 This is how the answer was given in the cited source. If the argument of the characteristic
function of A is set equal to c, adding the first (n − 1) rows (columns) to the nth row (column) yields
a determinant whose nth row (column) is zero. Thus c is a characteristic root of A.