Linear Algebra Exercises-n-Answers.pdf

110 Linear Algebra, by Hefferon
(a) Yes; we have both r · (g ◦ h) (⃗v) = r · g( h(⃗v) ) = (r · g) ◦ h (⃗v) and g ◦ (r · h) (⃗v) = g( r · h(⃗v) ) =
r · g(h(⃗v)) = r · (g ◦ h) (⃗v) (the second equality holds because of the linearity of g).
(b) Both answers are yes. First, f◦(rg+sh) and r·(f◦g)+s·(f◦h) both send ⃗v to r·f(g(⃗v))+s·f(h(⃗v));
the calculation is as in the prior item (using the linearity of f for the first one). For the other,
(rf + sg) ◦ h and r · (f ◦ h) + s · (g ◦ h) both send ⃗v to r · f(h(⃗v)) + s · g(h(⃗v)).
Three.IV.2.25 We have not seen a map interpretation of the transpose operation, so we will verify
these by considering the entries.
(a) The i, j entry of GH trans is the j, i entry of GH, which is the dot product of the j-th row of G
and the i-th column of H. The i, j entry of H trans G trans is the dot product of the i-th row of H trans
and the j-th column of G trans , which is the the dot product of the i-th column of H and the j-th
row of G. Dot product is commutative and so these two are equal.
(b) By the prior item each equals its transpose, e.g., (HH trans ) trans = H transtrans H trans = HH trans .
Three.IV.2.26 Consider r x , r y : R 3 → R 3 rotating all vectors π/2 radians counterclockwise about the
x and y axes (counterclockwise in the sense that a person whose head is at ⃗e 1 or ⃗e 2 and whose feet are
at the origin sees, when looking toward the origin, the rotation as counterclockwise).
Rotating r x first and then r y is different than rotating r y first and then r x . In particular, r x (⃗e 3 ) = −⃗e 2
so r y ◦ r x (⃗e 3 ) = −⃗e 2 , while r y (⃗e 3 ) = ⃗e 1 so r x ◦ r y (⃗e 3 ) = ⃗e 1 , and hence the maps do not commute.
Three.IV.2.27 It doesn’t matter (as long as the spaces have the appropriate dimensions).
For associativity, suppose that F is m × r, that G is r × n, and that H is n × k. We can take
any r dimensional space, any m dimensional space, any n dimensional space, and any k dimensional
space — for instance, R r , R m , R n , and R k will do. We can take any bases A, B, C, and D, for those
spaces. Then, with respect to C, D the matrix H represents a linear map h, with respect to B, C the
matrix G represents a g, and with respect to A, B the matrix F represents an f. We can use those
maps in the proof.
The second half is done similarly, except that G and H are added and so we must take them to
represent maps with the same domain and codomain.
Three.IV.2.28 (a) The product of rank n matrices can have rank less than or equal to n but not
greater than n.
To see that the rank can fall, consider the maps π x , π y : R 2 → R 2 projecting onto the axes. Each
is rank one but their composition π x ◦π y , which is the zero map, is rank zero. That can be translated
over to matrices representing those maps in this way.
( ) ( ) ( )
1 0 0 0 0 0
Rep E2,E 2
(π x ) · Rep E2,E 2
(π y ) =
=
0 0 0 1 0 0
To prove that the product of rank n matrices cannot have rank greater than n, we can apply the
map result that the image of a linearly dependent set is linearly dependent. That is, if h: V → W
and g : W → X both have rank n then a set in the range R(g ◦ h) of size larger than n is the image
under g of a set in W of size larger than n and so is linearly dependent (since the rank of h is n).
Now, the image of a linearly dependent set is dependent, so any set of size larger than n in the range
is dependent. (By the way, observe that the rank of g was not mentioned. See the next part.)
(b) Fix spaces and bases and consider the associated linear maps f and g. Recall that the dimension
of the image of a map (the map’s rank) is less than or equal to the dimension of the domain, and
consider the arrow diagram.
V
f
↦−→
R(f)
g
↦−→ R(g ◦ f)
First, the image of R(f) must have dimension less than or equal to the dimension of R(f), by the
prior sentence. On the other hand, R(f) is a subset of the domain of g, and thus its image has
dimension less than or equal the dimension of the domain of g. Combining those two, the rank of a
composition is less than or equal to the minimum of the two ranks.
The matrix fact follows immediately.