Math 220 * Fall 2022 * Practice Exam #4
Dec 12
Instructor: David Walmsley
NAME:
The exam will cover sections 7.1,7.2,7.3, and selected parts from 10.1 and 10.2. The directions for the exam will be similar to previous exam directions. The problems on this practice
exam might involve more computation and be a little longer than the actual exam. This should
give you extra practice. You will be expected to solve modest systems of equations by hand
(i.e. row reduction).
1. Determine if each set is a subspace. If it is not a subspace, give an example to show that
a specific subspace condition fails. If it is a subspace, prove this by either proving the
conditions of being a subspace hold or by showing the set is a span of some vectors.
x
(a) S =
:y≥x
y
Not closed under scalar multiplication.
(b) S = {ax2 + b : a, b ∈ R}
Is a subspace: S = span(1, x2 ).
a b
(c) S =
:a=b+c
0 c
1 0
1 1
Is a subspace: S = span(
,
)
0 0
0 1
a b
2
(d) S =
:a=b +c
0 c
Not closed under addition or scalar mult: the matrix ⃗u =
1 1
1 1
2 2
the sum
+
=
does not.
0 0
0 0
0 0
1 1
0 0
belongs to S but
2. Determine if each set is linearly independent. If the set is dependent, express on the
vectors as a linear combination of the others.
(a) {3x2 − x + 2, 4x + 1, 2x2 + 4x + 7}


3 0 2
Set up the equation in the definition. Need to row reduce  −1 4 4 .
2 1 7
(b) {x2 − 4x − 1, −x2 + 5x + 2, 2x2 − 5x + 1}
Similar to last problem.
1 2
−1
2
1 0
(c)
,
,
3 4
2 −1
0 0


1 −1 1
 2
2 0 
.
Need to row reduce 
 3
2 0 
4 −1 0
√
(d) {x, x2 , x}
√
We want to find the solutions set to c0 x + c1 x2 + c2 x = 0. Pick at least 3 values of
x to create a system of equations:
x = 1: c0 + c1 + c2 = 0
x = 4: 4c0 + 16c1 + 2c2 = 0
x = 9: 9c0 + 81c1 + 3c2 = 0
Should get c0 = c1 = c2 = 0 as the only solution, hence independent.
3. In each case, determine if ⃗v belongs to the span given.
1
(a) ⃗v = x2 + x + 1, span(x2 + 2x + 1, 3x2 − x + 1, −x2 + 5x + 1)


1
3 −1 1
5 1 . Should get inconsistent system.
Row reduce  2 −1
1
1
1 1
(b) ⃗v = 12x2 − 11x + 2, span(x2 + 2x + 1, 3x2 − x + 1, −x2 + 5x + 1).


1 0
2 −3
5 , which says that ⃗v = −3⃗v1 + 5⃗v2 .
Should row reduce to  0 1 −1
0 0
0
0
1 4
1 2
−1
2
1 0
(c) ⃗v =
, span
,
,
5 3
3 4
2 −1
0 0




1 −1 1 1
1 0 0 1

 2
 0 1 0 1 
1 0
4
2
2
0
→
, so ⃗v = 1 2 + −1

+
 3
 0 0 1 1 
2 0 5 
3 4
2 −1
0 0
4 −1 0 3
0 0 0 0
4. Determine if it’s possible (without actually checking the details) for each set S to be
basis for V . Explain your thinking in one or two sentences. [Hint: Theorem 7.16 might
be helpful.]
       
0
1
1 
 1
(a) S =  1  ,  1  ,  2  ,  0  , V = R3 .


0
1
1
1
Not possible. Any basis for R3 must have 3 vectors. 4 or more vectors must be
linearly dependent.
   
0 
 1



1 , 1  , V = R3 .
(b) S =


0
1
No, need 3 vectors in any basis for R3 . 2 vectors cannot span all of R3 .
      
0
1 
 1
(c) S =  1  ,  1  ,  0  , V = R3 .


0
1
1
It’s possible!
(d) S = {x2 + x, 2x2 − 3x, 4x2 + x + 1, x3 }, V = P 3
It’s possible, since a basis for P 3 must have 4 vectors.
(e) S = {x3 + x2 + x + 1, x2 − x + 1, x − 1}, V = P 3
Not possible, not enough vectors.
5. Remove vectors from S = {x2 + x, x + 1, x2 + 2x + 1, x2 + 1, x2 + x + 1} to yield a basis
for V = P 2 .




1 0 1 1 1
1 0 1 0 1/2
 1 1 2 0 1  →  0 1 1 0 1/2 , so the first, second, and fourth vectors are
0 1 1 1 1
0 0 0 1 1/2
linearly independent. Any 3 independent vectors will form a basis for P 2 because it has
dimension 3, so remove the third and fifth vectors from the set.
2
Z
1
6. Consider the “weighted” inner product ⟨f, g⟩ =
(x2 + 1)f (x)g(x) dx.
−1
(a) Determine whether or not each set S1 = {1, x, x2 } and S2 = {1, x, 8x2 −
orthogonal set.
S1 is not since ⟨1, x2 ⟩ =
̸ 0, but S2 is.
11
}
5
is an
(b) Determine the lengths ∥x∥ and ∥x2 ∥.
p
p
∥x∥ = ⟨x, x⟩, ∥x2 ∥ = ⟨x2 , x2 ⟩
(c) Use Gram-Schmidt on the set S1 to yield an orthogonal basis for {v1 , v2 , v3 } for
S = P 2.
I got v1 = 1, v2 = x, and I should have gotten v3 = x2 − 25 .
(d) Determine an orthonormal basis for S.
Just divide each vector in the previous part by its length.
(e) Determine projS x4 and projS (x2 + 2x).
24/35
2/9 − 4/25 + 6/35 2 2
+ 0x +
(x − ), and since x2 + 2x is a
I got projS x4 =
8/3
136/525
5
poly of degree 2 or less, we should get projS (x2 + 2x) = x2 + 2x.
(f) Use the facts that ⟨ex , v1 ⟩ = 3.23, ⟨ex , v2 ⟩ = 1.19, ⟨ex , v3 ⟩ = 0.54 to determine
projS ex .
3.23
1.19
0.54 2 11
projS ex =
+
x+
(x − )
8/3
16/15
24/35
40
(g) Determine nonzero values of a and b for which x3 + ax2 + b is orthogonal to x2 .
Set up the equation ⟨x3 + ax2 + b, x2 ⟩ = 0, and determine what
that2 inner
product
2
2
2
is in terms of a and b. I get the inner product to be a 7 + 5 + b 5 +3 . Setting
b 52 + 32
this equal to zero, b can be anything, but a must then equal − 2 2 . So b = 1,
+5
7
2
2
+
a = − 52 23 works.
+5
7
7. SOMETIMES/ALWAYS/NEVER. Determine if the statement is sometimes/always/never
true. Assume S is a subspace (of nonzero dimension) and ⃗u,⃗v are nonzero vectors.
(a) If ⃗u and ⃗v are orthogonal, then proj⃗v ⃗u = ⃗0.
Always, since ⟨⃗u, ⃗v ⟩ = 0.
(b) If projS ⃗u = ⃗u, then ⃗u is in S.
Always, since this means ⃗u is a linear combination of vectors in S.
(c) For a vector ⃗u and subspace S, projS (projS ⃗u) = projS ⃗u.
Always true, since projS ⃗u is a vector in S.
(d) If ⃗u and ⃗v are linearly independent, then they are orthogonal.
1
0
1
1
2
Sometimes.
and
are orthogonal and independent in R .
and
0
1
0
1
are independent but not orthogonal.
3
(e) ⃗v − proj⃗v ⃗u is orthogonal to ⃗u.
Sometimes. ⃗v − proj⃗v ⃗u is some scalar multiple of ⃗v , so this statement is only true if
⃗u and ⃗v are orthogonal.
(f) If ⟨⃗u, ⃗v ⟩ = 3, then ⟨2⃗u, −4⃗v ⟩ = −24.
Always true by the properties of inner products.
4