Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 121
and while H is a two-sided inverse of G and G is a two-sided inverse of H, we know that H is not
a two-sided inverse of H. However, the relation is symmetric: if G is a two-sided inverse of H then
GH = I = HG and therefore H is also a two-sided inverse of G.
Three.IV.4.37 This is how the answer was given in the cited source. Let A be m×m, non-singular,
with the stated property. Let B be its inverse. Then for n ≤ m,
m∑ m∑ m∑
m∑ m∑
m∑
1 = δ nr = b ns a sr = b ns a sr = k
(A is singular if k = 0).
r=1
r=1 s=1
s=1 r=1
s=1
b ns
Subsection Three.V.1: Changing Representations of Vectors
Three.V.1.6 For the matrix to change bases from D to E 2 we need that Rep E2
(id( ⃗ δ 1 )) = Rep E2
( ⃗ δ 1 )
and that Rep E2
(id( ⃗ δ 2 )) = Rep E2
( ⃗ δ 2 ). Of course, the representation of a vector in R 2 with respect to
the standard basis is easy.
( ( )
Rep E2
( ⃗ 2
δ 1 ) = Rep
1)
E2
( ⃗ −2
δ 2 ) =
4
Concatenating those two together to make the columns of the change of basis matrix gives this.
( )
2 −2
Rep D,E2 (id) =
1 4
The change of basis matrix in the other direction can be gotten by calculating Rep D (id(⃗e 1 )) = Rep D (⃗e 1 )
and Rep D (id(⃗e 2 )) = Rep D (⃗e 2 ) (this job is routine) or it can be found by taking the inverse of the above
matrix. Because of the formula for the inverse of a 2×2 matrix, this is easy.
Rep E2,D(id) = 1
10 ·
(
4 2
−1 2
)
=
(
4/10 2/10
−1/10 2/10
Three.V.1.7 In each case, the columns Rep D (id( β ⃗ 1 )) = Rep D ( β ⃗ 1 ) and Rep D (id( β ⃗ 2 )) = Rep D ( β ⃗ 2 ) are
concatenated to make the change of basis matrix Rep B,D (id).
( ) ( ) ( ) ( )
0 1
2 −1/2 1 1
1 −1
(a)
(b)
(c)
(d)
1 0
−1 1/2
2 4
−1 2
Three.V.1.8 One way to go is to find Rep B ( ⃗ δ 1 ) and Rep B ( ⃗ δ 2 ), and then concatenate them into the
columns of the desired change of basis matrix. Another way is to find the inverse of the matrices that
answer( Exercise ) 7. ( ) ( ) ( )
0 1
1 1
2 −1/2
2 1
(a)
(b)
(c)
(d)
1 0
2 4 −1 1/2
1 1
Three.V.1.9 The columns vector representations Rep D (id( β ⃗ 1 )) = Rep D ( β ⃗ 1 ), and Rep D (id( β ⃗ 2 )) =
Rep D ( β ⃗ 2 ), and Rep D (id( β ⃗ 3 )) = Rep D ( β ⃗ 3 ) make the change of basis matrix Rep B,D (id).
⎛ ⎞ ⎛
⎞ ⎛
⎞
0 0 1
1 −1 0
1 −1 1/2
(a) ⎝1 0 0⎠
(b) ⎝0 1 −1⎠
(c) ⎝1 1 −1/2⎠
0 1 0
0 0 1
0 2 0
E.g., for the first column of the first matrix, 1 = 0 · x 2 + 1 · 1 + 0 · x.
Three.V.1.10 A matrix changes bases if and only if it is nonsingular.
(a) This matrix is nonsingular and so changes bases. Finding to what basis E 2 is changed means
finding D such that
( ) 5 0
Rep E2 ,D(id) =
0 4
and by the definition of how a matrix represents a linear map, we have this.
( ( 5 0
Rep D (id(⃗e 1 )) = Rep D (⃗e 1 ) = Rep
0)
D (id(⃗e 2 )) = Rep D (⃗e 2 ) =
4)
)
Where
( ) ( )
x1 x2
D = 〈 , 〉
y 1 y 2