Linear Algebra Exercises-n-Answers.pdf

120 Linear Algebra, by Hefferon
and so there are infinitely many solutions to the matrix equation.
( ) 1 0 c ∣∣
{
c, f ∈ R}
0 1 f
With the bases still fixed at E 2 , E 2 , for instance taking c = 2 and f = 3 gives a matrix representing
this map.
⎛
⎝ x ⎞
( )
y⎠ f 2,3 x + 2z
↦−→
y + 3z
z
The check that f 2,3 ◦ η is the identity map on R 2 is easy.
Three.IV.4.25 By Lemma 4.3 it cannot have infinitely many left inverses, because a matrix with both
left and right inverses has only one of each (and that one of each is one of both — the left and right
inverse matrices are equal).
Three.IV.4.26 The associativity of matrix multiplication gives on the one hand H −1 (HG) = H −1 Z =
Z, and on the other that H −1 (HG) = (H −1 H)G = IG = G.
Three.IV.4.27 Multiply both sides of the first equation by H.
Three.IV.4.28 Checking that when I − T is multiplied on both sides by that expression (assuming
that T 4 is the zero matrix) then the result is the identity matrix is easy. The obvious generalization
is that if T n is the zero matrix then (I − T ) −1 = I + T + T 2 + · · · + T n−1 ; the check again is easy.
Three.IV.4.29 The powers of the matrix are formed by taking the powers of the diagonal entries.
That is, D 2 is all zeros except for diagonal entries of d 2 1,1 , d 2 2,2 , etc. This suggests defining D 0 to be
the identity matrix.
Three.IV.4.30 Assume that B is row equivalent to A and that A is invertible. Because they are
row-equivalent, there is a sequence of row steps to reduce one to the other. That reduction can be
done with matrices, for instance, A can be changed by row operations to B as B = R n · · · R 1 A. This
equation gives B as a product of invertible matrices and by Lemma 4.5 then, B is also invertible.
Three.IV.4.31 (a) See the answer to Exercise 28.
(b) We will show that both conditions are equivalent to the condition that the two matrices be
nonsingular.
As T and S are square and their product is defined, they are equal-sized, say n×n. Consider
the T S = I half. By the prior item the rank of I is less than or equal to the minimum of the rank
of T and the rank of S. But the rank of I is n, so the rank of T and the rank of S must each be n.
Hence each is nonsingular.
The same argument shows that ST = I implies that each is nonsingular.
Three.IV.4.32
Exercise 31.
Inverses are unique, so we need only show that it works. The check appears above as
Three.IV.4.33 (a) See the answer for Exercise 25.
(b) See the answer for Exercise 25.
(c) Apply the first part to I = AA −1 to get I = I trans = (AA −1 ) trans = (A −1 ) trans A trans .
(d) Apply the prior item with A trans = A, as A is symmetric.
Three.IV.4.34 For the answer to the items making up the first half, see Exercise 30. For the proof
in the second half, assume that A is a zero divisor so there is a nonzero matrix B with AB = Z (or
else BA = Z; this case is similar), If A is invertible then A −1 (AB) = (A −1 A)B = IB = B but also
A −1 (AB) = A −1 Z = Z, contradicting that B is nonzero.
Three.IV.4.35
Three.IV.4.36
No, there are at least four.
( ) ±1 0
0 ±1
It is not reflexive since, for instance,
( )
1 0
H =
0 2
is not a two-sided inverse of itself. The same example shows that it is not transitive. That matrix has
this two-sided inverse
( )
1 0
G =
0 1/2