Download Solutions to Math 51 First Exam — April 21, 2011

Solutions to Math 51 First Exam — April 21, 2011
1. (12 points)
(a) Give the precise definition of a (linear) subspace V of Rn .
(4 points) A linear subspace V of Rn is a subset V ⊆ Rn which satisfies
• 0∈V.
• If x, y ∈ V then x + y ∈ V (“V is closed under addition”).
• If x ∈ V and c ∈ R then cx ∈ V (“V is closed under scalar multiplication”).
(b) Complete the following sentence: A set of vectors {v1 , . . . , vk } is defined to be linearly independent
if
(4 points)
... the equation c1 v1 + · · · + ck vk = 0 for c1 , . . . , ck ∈ R implies c1 = · · · = ck = 0.
OR
... no vector in {v1 , . . . , vk } can be written as linear combination of the others.
(c) Suppose {u, v, w} is a linearly independent set of vectors in R7 . Is {u + v + w, 3v − w, 2w}
linearly independent? Justify your answer completely.
(4 points) The vectors in {u + v + w, 3v − w, 2w} are linearly independent.
Proof: For c1 , c2 , c3 ∈ R, consider the equation
0 = c1 (u + v + w) + c2 (3v − w) + c3 (2w) = c1 u + (c1 + 3c2 )v + (c1 − c2 + 2c3 )w.
Since u, v and w are linearly independent by assumption, we conclude that
c1 = 0
and
c1 + 3c2 = 0
and
c1 − c2 + 2c3 = 0.
(1)
Putting c1 = 0 in c1 + 3c2 = 0 implies c2 = 0. And putting c1 = 0 and c2 = 0 in c1 − c2 + 2c3 = 0
implies c3 = 0. Thus the vectors in {u + v + w, 3v − w, 2w} are linearly independent.
Alternatively, one can write (1) as matrix equation
   

c1
0
1 0 0





c2 = 0
Ac = 1 3 0
1 −1 2
c3
0
Computing the reduced row echelon form yields


1 0 0
rref(A) = 0 1 0 .
0 0 1
Thus rref(A) and A have rank 3 and the equation Ac = 0 admits only the trivial solution c = 0.
Grader’s comment: Some students worked with the transposed matrix AT . For questions
involving only the rank of a matrix this does not matter. But generally, it does matter if one
uses the transposed matrix or the non-transposed one; in particular, if one sets up a ‘basis change
matrix’ and performs explicit calculations.
Math 51, Spring 2011
Solutions to First Exam — April 21, 2011
Page 2 of 8
2. (12 points) Let Z be the plane in R3 containing the points (1, 0, 1), (0, 1, −1), and (1, 2, 3).
(a) Find a parametric representation of Z.
(4 points) We need a point the plane passes through, and two vectors in the plane:
 
     
     
1
0
1
−1
1
1
0













x0 = 0 ,
v= 1 − 0 = 1 ,
w = 2 − 0 = 2
1
−1
1
−2
3
1
2
The parametric representation of the plane is thus:
 

 
 
−1
0
 1

0 + t  1  + s 2 s, t ∈ R


1
−2
2
(b) Give an equation for Z of the form ax + by + cz = d. (Here a,b,c,d are scalars, and x,y,z are the
usual variables for coordinates of points in R3 .)
(4 points)
     
6
0
−1





We find a normal n to the plane, where n = v × w from above: n = 1 × 2 = 2  .
−2
2
−2
The equation of the plane is then:
     
6
x
1
 2  · y  − 0 = 0 ⇐⇒ 6(x − 1) + 2(y − 0) − 2(z − 1) = 0
−2
z
1
⇐⇒ 3x + y − z = 2 .
(c) Find the coordinates of a point P in Z having the property that the vector from the origin to P
is perpendicular (normal) to Z.
 
a

(4 points) Denote the point we are looking for by b . Because the vector with the same entries
c
is perpendicular to the plane Z, it must be parallel to Z’s normal vector n. Thus, we must have:
 
 
6
a
 b  = t  2  for some t.
c
−2
But because the point is in the plane, it must satisfy the equation of the plane as follows:
3a + b − c = 2 ⇐⇒ 3(6t) + 2t − (−2t) = 2
⇐⇒ 22t = 2.
6
11
 2 
 11 
2
− 11

Thus t =
1
, so the point we are looking for is
11

.
Math 51, Spring 2011
Solutions to First Exam — April 21, 2011
3. (8 points) Compute, showing all steps, the reduced row

0 0 −1 4
1 2 3 4

2 4 6 2
3 6 10 8

0
1

2
3

0 −1 4 1 swap
2 3 4 3
 swap
4 6 2 6
6 10 8 8

1
0

2
3

1
0

0
0

1
0

0
0

1
0

0
0

1
0

0
0

1
0

0
0
echelon form of the matrix

1
3

6
8

2 3 4 3
0 −1 4 1

4 6 2 6 −2I
6 10 8 8 −3I

2 3
4
3
0 −1 4
1
 ·(−1)
0 0 −6 0 
0 1 −4 −1

2 3 4
3
−3II
0 1 −4 −1

0 0 −6 0 
0 1 −4 −1 −II

2 0 16 6
0 1 −4 −1

0 0 −6 0  ·(− 16 )
0 0 0
0

2 0 16 6
−16III
0 1 −4 −1
 +4III
0 0 1
0
0 0 0
0

2 0 0 6
0 1 0 −1

0 0 1 0
0 0 0 0
Page 3 of 8
Math 51, Spring 2011
Solutions to First Exam — April 21, 2011
Page 4 of 8
4. (12 points) Suppose that all we know about the 3 × 4 matrix A is that its entries are all nonzero, and
that its reduced row echelon form is


1 0 2 0
rref(A) = 0 1 −1 0
(Note: this is not the matrix A!)
0 0 0 1
(a) Find a basis for the null space of A; show your reasoning.
(4 points) We’re able to solve this part without knowing A, because N (A) = N (rref(A)) (why?).
We proceed in the standard manner: using rref(A), we can see that the third column does not
contain a pivot, so x3 is a freee variable. We now express the pivot variables in terms of the free
variable:

 
  


x1
−2
−2x3


x
+
2x
=
0
1
3


1
x2   x3 
 
 =

and so
x2 − x3
=0
x3   x3  = x3  1 




x4
=0
0
0
x4
 
−2 


 

1
and we get that
 1  is a basis for N (A).





0
(b) If the columns of A, in order, are a1 , a2 , a3 , a4 ∈ R3 , circle all sets of vectors below that give a
basis for the column space of A. You do not need to justify your answer(s).
{a1 }
{a1 , a2 }
{a1 , a3 }
{a1 , a2 , a3 }
{a2 }
{a2 , a3 }
{a1 , a2 , a4 }
{a3 }
{a4 }
{a1 , a4 }
{a2 , a4 }
{a1 , a3 , a4 }
{a3 , a4 }
{a2 , a3 , a4 }
{a1 , a2 , a3 , a4 }
(4 points) We’ve seen that one basis for C(A) is formed by taking the columns of A that correspond to the pivot-columns of rref(A). Thus, {a1 , a2 , a4 } is a basis for C(A). From this, it
immediately follows that dim C(A) = 3 and therefore no set of size 1, 2, or 4 can ever be a basis
for C(A); we can thus eliminate from further consideration all such sets in the above list.
We’re left to consider the three remaining three-element sets of column vectors listed above.
Now, recall that according to a result from Chapter 12 of the text, a three-element subset of a
three-dimensional subspace V is a basis for V if and only if it is linearly independent; this means
it’s sufficient for us to check whether these remaining choices are independent sets.
Next, note that since rref(A) has the same column dependencies as A (or equivalently using part
(a)), we know that
 
 
−2
0
1
 = 0 .
A
1
0
0
(ctd next page)
Math 51, Spring 2011
Solutions to First Exam — April 21, 2011
Page 5 of 8
Thus, using the column-mixing properties of the matrix-vector product, we have that
−2a1 + a2 + a3 = 0,
which means that the set {a1 , a2 , a3 } is linearly dependent, so it cannot be a basis for C(A).
Finally, each of the remaining two options is an independent set (and thus a basis); for example,
since
a3 = 2a1 − a2 ,
we may re-think of {a1 , a3 , a4 } as {a1 , 2a1 − a2 , a4 }. Using the fact that the basis {a1 , a2 , a4 }
is linearly independent, we may make an argument analogous to that from problem 1(c) of this
exam to see that {a1 , 2a1 −a2 , a4 } is also independent. A similar argument works for {a2 , a3 , a4 }.
(c) Suppose we also know that the second and third columns of A are, respectively,
 
 
1
9
a2 = −1 and a3 = 5
1
3
Use this information to find the first column a1 of A; give your reasoning.
 
   
9
5
1
1
(4 points) From above, we see that a1 = 12 (a2 + a3 ). Thus, a1 = −1 + 5 = 2 .
2
3
2
1
Math 51, Spring 2011
Solutions to First Exam — April 21, 2011
Page 6 of 8
5. (10 points) Each of the statements below is either always true (“T”), or always false (“F”), or sometimes true and sometimes false, depending on the situation (“MAYBE”). For each part, decide which
and circle the appropriate choice; you do not need to justify your answers.
(a) A set of 4 vectors in R5 is linearly independent.
T
F
MAYBE
One the one hand, consider any four-element subset of the standard basis for R5 ; on the other,
consider any four-element set containing 0 = (0, 0, 0, 0, 0).
(b) A set of 4 vectors in R5 spans R5 .
T
F
MAYBE
If this was possible, then Proposition 12.1 would imply that any set of 5 vectors in R5 is linearly
dependent, which is clearly false (for, consider the standard basis of R5 ).
(c) A set of 5 vectors in R4 is linearly independent.
T
F
MAYBE
T
F
MAYBE
See Proposition 8.4; every set of 5 vectors in R4 is linearly dependent.
(d) A set of 5 vectors in R4 spans R4 .
One the one hand, consider any basis for R4 with an extra vector thrown in; on the other,
consider any five different scalar multiples of a single nonzero vector.
(e) A set of 5 vectors which spans R5 is linearly independent.
T
F
MAYBE
F
MAYBE
F
MAYBE
This follows immediately from Proposition 12.3, since R5 has dimension 5.
(f) A set of 5 linearly independent vectors in R5 spans R5 .
T
This follows immediately from Proposition 12.3, since R5 has dimension 5.
(g) The span of 4 vectors in R5 is a 4-dimensional subspace.
T
One the one hand, consider any four-element subset of the standard basis for R5 ; on the other,
consider any four-element set containing 0 = (0, 0, 0, 0, 0).
(h) The span of 5 vectors in R4 is a 5-dimensional subspace.
T
F
MAYBE
R4 doesn’t have any 5-dimensional subspaces, because such a subspace would have a basis of size
5; but there are no linearly independent subsets of R4 of size 5 by Proposition 8.4.
(i) A set of 4 vectors in R4 forms a basis for R4 .
T
F
MAYBE
One the one hand, consider the standard basis of R4 , on the other, consider any four-element
set containing 0 = (0, 0, 0, 0).
(j) A basis for R4 contains exactly 4 vectors.
T
This follows immediately from Proposition 12.2, since R4 has dimension 4.
F
MAYBE
Math 51, Spring 2011
Solutions to First Exam — April 21, 2011
Page 7 of 8
6. (12 points) For each part, provide with reasoning an example of a matrix (A, B, or C, respectively)
that satisfies the given property, or briefly explain why no such matrix can exist.
 
1
x
(a) The linear system A 1 =  3  has infinitely many solutions.
x2
−1
(4 points) First, note that for the system to make sense,
 the
 matrix A should be a 3×2 matrix.
1
It would have infinitely many solutions if the vector  3  is in C(A) and nullity(A) > 0. The
−1
nullity condition implies that rank(A) < 2; i.e., that the two columns of A are linearly dependent.
Such a matrix exists and possible choices of A include




1 0
1
2
 3 0 and  3
6  among many other options.
−1 0
−1 −2
 
1
x
(b) The linear system B 1 =  3  has exactly one solution.
x2
−1
(4 points) Again, thematrix
 B should be a 3×2 matrix. The system would have a unique
1
solution if the vector  3  is in C(B), and also nullity(B) = 0. The nullity condition implies
−1
that rank(B) = 2; i.e., that the two columns of B are linearly independent. Such a matrix exists
and possible choices of B include




1 0
1 1
 3 0 and 3 0  among many other options.
0 −1
−1 0
 
1
x
(c) The linear system C 1 =  3  has no solutions.
x2
−1


1
(4 points) If C is a 3 × 2 matrix, the system would have no solution if  3  is not in the column
−1
space of C. Such a matrix C exists and can be constructed easily by including a row of zeros
to give an ”inconsistent” equation. For example, if the last row of C is zero, the system would
imply (0)x1 + (0)x2 = −1, which is impossible. Possible choices of C include




0 0
1 0
0 0 and 0 1 among many other options.
0 0
0 0
Math 51, Spring 2011
Solutions to First Exam — April 21, 2011
Page 8 of 8
7. (12 points)
(a) Suppose S : R2 → R3 is a linear transformation such that
 
 
2
−1
1
2
S
= 3 and S
= 2 
0
1
1
2
Find the matrix of S.
(4 points) Recall that the matrix corresponding to a linear transformation is given by


|
|
S(e1 ) S(e2 )
|
|
so to figure out the matrix, we need only compute S(e1 ) and S(e2 ). Conveniently, S(e1 ) is given
to us. And we can figure out S(e2 ) as follows: by linearity,
0
2
1
S(e2 ) = S
=S
−2
1
1
0
 
   
−1
2
−5
2
1
=  2  − 2 3 = −4
=S
− 2S
1
0
2
1
0
Thus, the matrix A corresponding to S is


2 −5
A = 3 −4
1 0
(b) Let T : R2 → R2 be the linear transformation that projects vectors onto the line y = x. Find the
matrix of T.
(4 points) We recall that for projections in R2 , the matrix of ProjL is given by the expression
2
1
v1 v1 v2
kvk2 v1 v2 v22
where v = (v1 , v2 ) is any spanning vector for the line L that we are projecting onto. In this
particular example, L is y = x, and this is spanned by (among other choices) the vector (1, 1).
1 1 1
.
So, letting v = (1, 1) = (v1 , v2 ), we compute that the matrix B of T is B =
2 1 1
(Alternatively, we could use the formula ProjL (v) = (v · u)u to find the columns of B; here
u = √12 (1, 1) is a unit vector that spans L. This formula for ProjL in Rn holds for any n!)
(c) For S and T as above, find the matrix of S ◦ T or explain why it cannot be defined.
(4 points) The linear transformation S ◦ T : R2 → R3 is defined, because the codomain of T and
the domain of S coincide (both are R2 ). Or, at the matrix level, S ◦ T has matrix given by the
product AB (using the names above); this product exists because A is 3 × 2 and B is 2 × 2.
 3





− 2 − 32
2 −5 −3 −3
1 1


We do the product (note it’s 3 × 2): AB = 21 3 −4
= 12 −1 −1 = − 21 − 12 
1 1
1
1
1 0
1
1
2
2