Linear Algebra Exercises-n-Answers.pdf

150 Linear Algebra, by Hefferon
4 Represent it with respect to the standard bases E 1 , E 1 , then the only entry in the resulting 1×1 matrix
is the scalar k.
5 We can show this by induction on the number of components in the vector. In the n = 1 base case
the only permutation is the trivial one, and the map
(
x1
)
↦→
(
x1
)
is indeed expressible as a composition of swaps — as zero swaps. For the inductive step we assume
that the map induced by any permutation of fewer than n numbers can be expressed with swaps only,
and we consider the map induced by a permutation p of n numbers.
⎛ ⎞ ⎛ ⎞
x 1 x p(1)
x 2
⎜
⎝
⎟
. ⎠ ↦→ x p(2)
⎜
⎝
⎟
. ⎠
x n x p(n)
Consider the number i such that p(i) = n. The map
⎛ ⎞ ⎛
x 1
x 2
.
ˆp
x i
↦−→
⎜ ⎟ ⎜
⎝
. ⎠ ⎝
x n
x p(1)
x p(2)
will, when followed by the swap of the i-th and n-th components, give the map p. Now, the inductive
hypothesis gives that ˆp is achievable as a composition of swaps.
6 (a) A line is a subset of R n of the form {⃗v = ⃗u + t · ⃗w ∣ ∣ t ∈ R}. The image of a point on that line
is h(⃗v) = h(⃗u + t · ⃗w) = h(⃗u) + t · h( ⃗w), and the set of such vectors, as t ranges over the reals, is a
line (albeit, degenerate if h( ⃗w) = ⃗0).
(b) This is an obvious extension of the prior argument.
(c) If the point B is between the points A and C then the line from A to C has B in it. That is,
there is a t ∈ (0 .. 1) such that ⃗ b = ⃗a + t · (⃗c − ⃗a) (where B is the endpoint of ⃗ b, etc.). Now, as in the
argument of the first item, linearity shows that h( ⃗ b) = h(⃗a) + t · h(⃗c − ⃗a).
7 The two are inverse. For instance, for a fixed x ∈ R, if f ′ (x) = k (with k ≠ 0) then (f −1 ) ′ (x) = 1/k.
.
x p(n)
.
x n
⎞
⎟
⎠
f(x)
x
f −1 (f(x))
Topic: Markov Chains
1 (a) With this file coin.m
# Octave function for Markov coin game. p is chance of going down.
function w = coin(p,v)
q = 1-p;
A=[1,p,0,0,0,0;
0,0,p,0,0,0;
0,q,0,p,0,0;
0,0,q,0,p,0;
0,0,0,q,0,0;
0,0,0,0,q,1];
w = A * v;
endfunction