Linear Algebra Exercises-n-Answers.pdf

66 Linear Algebra, by Hefferon
But there is an upper bound (other than the size of the matrices). In general, rank(A + B) ≤
rank(A) + rank(B).
To prove this, note that Gaussian elimination can be performed on A + B in either of two ways:
we can first add A to B and then apply the appropriate sequence of reduction steps
(A + B) step 1
−→ · · · step k
−→ echelon form
or we can get the same results by performing step 1 through step k separately on A and B, and then
adding. The largest rank that we can end with in the second case is clearly the sum of the ranks.
(The matrices above give examples of both possibilities, rank(A + B) < rank(A) + rank(B) and
rank(A + B) = rank(A) + rank(B), happening.)
Subsection Two.III.4: Combining Subspaces
Two.III.4.20 With each of these we can apply Lemma 4.15.
(a) Yes. The plane is the sum of this W 1 and W 2 because for any scalars a and b
( ( ) (
a a − b b
= +
b)
0 b)
shows that the general vector is a sum of vectors from the two parts. And, these two subspaces are
(different) lines through the origin, and so have a trivial intersection.
(b) Yes. To see that any vector in the plane is a combination of vectors from these parts, consider
this relationship. ( ( ) ( )
a 1 1
= c
b)
1 + c
1 2
1.1
We could now simply note that the set
( ( )
1 1
{ , }
1)
1.1
is a basis for the space (because it is clearly linearly independent, and has size two in R 2 ), and thus
ther is one and only one solution to the above equation, implying that all decompositions are unique.
Alternatively, we can solve
c 1 + c 2 = a −ρ 1+ρ 2 c −→ 1 + c 2 = a
c 1 + 1.1c 2 = b
0.1c 2 = −a + b
to get that c 2 = 10(−a + b) and c 1 = 11a − 10b, and so we have
( ( ) ( )
a 11a − 10b −10a + 10b
=
+
b)
11a − 10b 1.1 · (−10a + 10b)
as required. As with the prior answer, each of the two subspaces is a line through the origin, and
their intersection is trivial.
(c) Yes. Each vector in the plane is a sum in this way
( ( (
x x 0
= +
y)
y)
0)
and the intersection of the two subspaces is trivial.
(d) No. The intersection is not trivial.
(e) No. These are not subspaces.
Two.III.4.21 With each of these we can use Lemma 4.15.
(a) Any vector in R 3 can be decomposed as this sum.
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
x x 0
⎝y⎠ = ⎝y⎠ + ⎝0⎠
z 0 z
And, the intersection of the xy-plane and the z-axis is the trivial subspace.
(b) Any vector in R 3 can be decomposed as
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎝ x y⎠ = ⎝ x − z
y − z⎠ + ⎝ z z⎠
z 0 z
and the intersection of the two spaces is trivial.