Linear Algebra Exercises-n-Answers.pdf

204 Linear Algebra, by Hefferon
Five.IV.2.20 For each, because many choices of basis are possible, many other answers are possible.
Of course, the calculation to check if an answer gives that P T P −1 is in Jordan form is the arbiter of
what’s correct.
(a) Here is the arrow diagram.
C 3 t
w.r.t. E 3
−−−−→ C 3 w.r.t. E T 3
⏐
⏐
id↓P
id↓P
C 3 t
w.r.t. B −−−−→ C 3 w.r.t. B
J
The matrix to move from the lower left to the upper left is this.
⎛
P −1 = ( Rep ) −1
E3,B(id) = RepB,E3 (id) = ⎝ 1 −2 0
⎞
1 0 1⎠
−2 0 0
The matrix P to move from the upper right to the lower right is the inverse of P −1 .
(b) We want this matrix and its inverse.
⎛
P −1 = ⎝ 1 0 3
⎞
0 1 4⎠
0 −2 0
(c) The concatenation of these bases for the generalized null spaces will do for the basis for the entire
space.
⎛ ⎞ ⎛ ⎞
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
−1 −1
1 0 −1
0
B −1 = 〈
⎜ 0
⎟
⎝ 1 ⎠ , 0
⎜−1
⎟
⎝ 0 ⎠ 〉 B 1
3 = 〈
⎜−1
⎟
⎝ 0 ⎠ , 0
⎜ 0
⎟
⎝−2⎠ , −1
⎜ 1
⎟
⎝ 2 ⎠ 〉
0 1
0 2 0
The change of basis matrices are this one
⎛
and its inverse.
⎞
−1 −1 1 0 −1
0 0 1 0 −1
P −1 =
⎜ 0 −1 −1 0 1
⎟
⎝ 1 0 0 −2 2 ⎠
0 1 0 2 0
Five.IV.2.21 The general procedure is to factor the characteristic polynomial c(x) = (x − λ 1 ) p 1
(x −
λ 2 ) p2 · · · to get the eigenvalues λ 1 , λ 2 , etc. Then, for each λ i we find a string basis for the action of the
transformation t − λ i when restricted to N ∞ (t − λ i ), by computing the powers of the matrix T − λ i I
and finding the associated null spaces, until these null spaces settle down (do not change), at which
point we have the generalized null space. The dimensions of those null spaces (the nullities) tell us
the action of t − λ i on a string basis for the generalized null space, and so we can write the pattern of
subdiagonal ones to have N λi . From this matrix, the Jordan block J λi associated with λ i is immediate
J λi = N λi + λ i I. Finally, after we have done this for each eigenvalue, we put them together into the
canonical form.
(a) The characteristic polynomial of this matrix is c(x) = (−10 − x)(10 − x) + 100 = x 2 , so it has
only the single eigenvalue λ = 0.
power p (T + 0 · I) p N ((t − 0) p ) nullity
( ) ( )
−10 4 2y/5 ∣∣
1
{ y ∈ C} 1
−25
(
10
)
y
0 0
2
C 2 2
0 0
(Thus, this transformation is nilpotent: N ∞ (t − 0) is the entire space). From the nullities we know
that t’s action on a string basis is β ⃗ 1 ↦→ β ⃗ 2 ↦→ ⃗0. This is the canonical form matrix for the action of
t − 0 on N ∞ (t − 0) = C 2 ( )
0 0
N 0 =
1 0
and this is the Jordan form of the matrix.
( )
0 0
J 0 = N 0 + 0 · I =
1 0