# 21-241 (Fall 15) Solutions for Review Session (Nov 18, 2015)

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21-241 (Fall 15) Solutions for Review Session (Nov 18, 2015)
```21-241 (Fall 15)
Solutions for Review Session (Nov 18, 2015)
Page 1 of 6
1. In this problem, x, y, z, 0 (zero vector) are n × 1 vectors, Q is an n × n matrix, c is a scalar.
Circle the following expressions that should never appear on any of your workings.
(a) projx (y) =
y·x
x·x
·x
(b) x · y = y · x
(c) (x · y) · z = x · (y · z)
(d) (x · y)z = (x · z)(y · z)
(e) cx · y = x · cy
(f) Qx · y = x · Qy
(g) Qx · Qy = Q(x · y)
(h) Qx · y = x · QT y
(i) x · y = xT y
(j) x ⊥ y =⇒ x · y = 0
(k) (x · y)z = (x · z)y
(l) cx = x[ c ]
(m) (x · y)z = (xT y)z
(n) (x · y)z = (zxT )y
(o) (x · y)z = (zxT )y = (xz T )T y = (xz T ) · y
(p) (x + y) · z = x · z + y · z
(q) x · 0 = 0
(r)
y·x
z·x
=
y
z
(s) 0 = 0
Solution.
Wrong because of invalid operations/LHS-RHS mismatch: (a), (c), (d), (g), (m), (o), (r).
Wrong because of difference between zero vector and scalar zero: (j), (q), (s).
Also wrong: (f), (k).
Notation: (i). We often write x · y = xT y. However we have to be careful that x · y is a
scalar whereas xT y is a 1 × 1 matrix.
2. True or false?
(a) R(A) ⊥ N (A)
(b) C(A) ⊥ N (AT )
(c) Let V be a vector space and U ⊆ V be a subspace of V . Then U ∪ U ⊥ = V .
(d) From a basis B = {v 1 , v 2 , . . . , v n } of vector space V , it is always possible to generate an
orthogonal basis of V .
(e) Given a linearly dependent set of vectors {v 1 , v 2 , . . . , v k }, we can express v 1 as a linear
combination of v 2 , . . . , v k .
(f) dim R(A) = dim C(A).
(g) rank AT + nul AT = number of rows of A.
21-241 (Fall 15)
Solutions for Review Session (Nov 18, 2015)
Page 2 of 6
(h) For all vectors v ∈ Rm , there are vectors x ∈ C(A) and y ∈ N (AT ) such that v = x + y.
Solution.
(a) True
(b) True
(c) False
(d) True
(e) False
(f) True
(g) True
(h) True

1 2
 2 0
3. Let A = 
 5 0
1 −2

0
1 
. State the dimension and find a basis for
2 
1
(a) R(A)
(b) C(A)
(c) N (A)
Solution.
(a) 3. B = {e1 , e2 , e3 }.
(b) 3. B = columns of A.
(c) 0. B = ∅.
Note. There are other choices for the bases.

1 2
 2 0
4. Let A = 
 5 2
1 −2
(a) R(A)
(b) C(A)
(c) N (A)

0
1 
. State the dimension and find a basis for
2 
1
21-241 (Fall 15)
Solutions for Review Session (Nov 18, 2015)
Page 3 of 6
Solution.
(a) 2. B = {(1 2 0), (2 0 1)}.
(b) 2. B = {(1 2 5 1)T , (2 0 2 − 2)T }.
(c) 1. B = {(2 − 1 − 4)T }.
Note. There are other choices for the bases.
5. Let P be a hyperplane in R3 with equation x − 2y + 3z = 0.
(a) Find a vector perpendicular to P .
(b) What matrix has the plane P as its nullspace?
(c) What matrix has P as its row space?


 −1 
(d) What is the orthogonal complement of span  1  in N (P )?


1
Solution.
(a) (1 − 2
3)T
(b) (1 − 2 3)
−3 0 1
(c)
2 1 0
6. Let {v 1 , v 2 , . . . , v k } be a set of k orthogonal non-zero vectors, i.e. v i · v j = 0 for all 1 ≤ i <
j ≤ k. Prove that the set is linearly independent.
Solution. Let
k
P
i=1
that v i ·
k
P
ci v i = 0. So v i ·
k
P
ci v i = v i · 0 = 0 for all i = 1 . . . k. We have
i=1
ci v i = v i · ci v i = ci kv i k2 . So 0 = ci kv i k2 also. Since v i 6= 0, kv i k > 0. So
i=1
ci = 0. Since this applies for all i = 1 . . . k, we have by definition that {v 1 , v 2 , . . . , v k } is
a linearly independent set.
7. Let B = {v 1 , . . . , v k , . . . , v n } be an orthogonal basis for Rn . Find (with proof) the intersection
of span{v 1 , . . . , v k } and span{v k , . . . , v n }.
Solution. Let U1 = span{v 1 , . . . , v k } and U2 = span{v k , . . . , v n }. We claim that
U1 ∩ U2 = span{v k }
21-241 (Fall 15)
Solutions for Review Session (Nov 18, 2015)
Page 4 of 6
Step 1. Let v ∈ span{v k }. Then v = ck v k for some scalar ck . Then v = 0v 1 + . . . +
0v k−1 + ck v k ∈ U1 . Likewise v = ck v k + 0v k+1 + . . . + 0v n ∈ U2 . So v ∈ U1 ∩ U2 . We
have shown that span{v k } ⊆ U1 ∩ U2 .
Step 2. Let v ∈ U1 ∩ U2 . Then v = c1 v 1 + . . . + ck v k = dk v k + . . . + dn v n . For
i = 1 . . . k − 1,
vi · v = vi · v
=⇒ v i · (c1 v 1 + . . . + ck v k ) = v i · (dk v k + . . . + dn v n )
=⇒ v i · ci v i = 0
=⇒ ci kv i k2 = 0
=⇒ ci = 0
So v = 0v 1 + . . . + 0v k−1 + ck v k = ck v k . So v ∈ span{v k }. Thus we have shown that
U1 ∩ U2 ⊆ span{v k }.
Hence U1 ∩ U2 = span{v k }.
8. Prove (rigorously) that orthogonal projection onto a vector space V is linear.
(Hint: Write down some mathematical statements instead of an essay.)
Solution. We can certainly find an orthonormal basis for V .
{v 1 , v 2 , . . . , v k }. Consider x, y vectors, c scalar. We have
Let it be B =
projV (x + cy)
=
k
X
((x + cy) · v i )v i
i=1
=
=
=
=
k
X
i=1
k
X
i=1
k
X
i=1
k
X
i=1
(x · v i + cy · v i )v i
(x · v i )v i + (cy · v i )v i
(x · v i )v i +
k
X
c(y · v i )v i
i=1
k
X
(x · v i )v i + c
(y · v i )v i
i=1
= projV (x) + c projV (y)
This proves the linearity of projV .
9. (a) Let A be a matrix. Show that any vector in R(A) can be expressed as xA for some
vector x.
21-241 (Fall 15)
Solutions for Review Session (Nov 18, 2015)
Page 5 of 6
(b) State a similar claim for any vector in C(A) and prove it.
(c) Let Q be an orthogomal matrix and a ∈ R with |a| < 1. Is Q + aI invertible?
Solution.

v1


(a) Let A =  ... . Let v ∈ R(A). Then we may write v = c1 v 1 + . . . + cm v m . Let
vm
x = (c1 · · · cm ). By the Column-Row Representation we have


v1


xA = (c1 · · · cm )  ...  = c1 v 1 + . . . + cm v m = v

vm
(b) Any vector in C(A) can be expressed as Ax for some vector x. Proof similar. Use
Column-Row Representation again.
(c) Assume that Q + aI is not invertible. Then by the Fundamental Theorem of Invertible Matrices, there exists v 6= 0 such that (Q + aI)v = 0. Then Qv = −av.
Then kvk2 = v · v = v T v = v T (QT Q)v = (v T QT )(Qv) = (Qv)T (Qv) = Qv · Qv =
−av · −av = a2 v · v = a2 kvk2 . So (1 − a2 )kvk2 = 0. Since |a| < 1, 1 − a2 6= 0. So
kvk2 = 0, so kvk = 0, so v = 0, a contradiction to our assumption. Thus Q + aI is
invertible.
10. Using the Gram-Schmidt Process, derive an orthonormal basis from
  

 
7
5

 2
B =  0 , 1 , 0 


−6
−1
0
Solution. v 1 = (1 0 0)T , v 2 = (0

0
 1
11. Let A = 
 1
0
1
2
0
0


1

1 
 and b = 


1
1
√1
2
−
√1 )T ,
2

−1
0 
.
1 
1
(a) Find the approximate solution(s) to Ax = b.
(b) Find the projection of b onto C(A).
Solution.
(a) Solve AT Ax = AT b.
v 3 = (0
√1
2
√1 )T .
2
21-241 (Fall 15)
Solutions for Review Session (Nov 18, 2015)
Page 6 of 6
12. (a) Does Ax = b always have a solution? Prove or disprove.
(b) Does AT Ax = AT b always have a solution? Prove or disprove.
Solution.
(a) No. Construct counterexample.
(b) Yes. Note that C(A) ⊥ N (AT ). Let v ∈ Rn . By the Orthogonal Decomposition
Theorem, we can express b as v = y + z where y ∈ C(A) and z ∈ N (AT ). We can
express y as Av for some v ∈ Rn by Problem 9(b). So b = Av + z. Bearing in mind
z ∈ N (AT ), we have
AT b = AT Av + AT z = AT Av + 0 = AT Av.
So v is a solution to AT Ax = AT b, thus proving the existence of a solution.
END OF REVIEW PAPER
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