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This is the First Midterm Solved Exam of Linear Algebra and Multivariable Calculus which includes Intersection, Independent Set, Explanation, Dot Product etc. Key important points are: Dividing, Linearly Independent, Set Of Vectors, Defined, Linear Transformation, Function, Careful, Reduced Row, Echelon Form, Matrix
Typology: Exams
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(a) Vectors v and w in R
7 are defined to be orthogonal if
(4 points)
v · w = 0.
(b) A basis for a subspace V of R
n is defined to be
(4 points)... a linearly independent set of vectors {v 1 ,... , vk} such that
span(v 1 ,... , vk) = V.
(c) A function T : R
n → R
m is called a linear transformation if
(4 points)
T(x + y) = T(x) + T(y) for all x, y ∈ R
n , and
T(cx) = cT(x) for all c ∈ R
n and x ∈ R
n .
(a) Compute, showing all steps, the reduced row echelon form of the matrix
(7 points) First, R2 - 3R1, R3 - R1, R4 - 2R1 gives
Multiplying R4 by -1, exchanging R3 and R4, and dividing R2/6, gives,
R3 - R2 gives:
R2 + R3 and 6*R3 gives:
R1 + 2R2, multiplying R2 by -1 and dividing R3 by 13/6 gives:
R1 - 5R3, R2 +3R3 gives the RREF:
(b) Fill in the blanks (no reasoning needed): Rank of A: 3 Nullity of A: 2
One point for one answer correct; three points for both answers correct.
As usual, we’ll write N (A) and C(A), respectively, for the null space and column space of A.
(a) Find, with reasoning, a basis for N (A).
(4 points) Note that A is already in row reduced echelon form, i.e. rref(A) = A. Thus, the free
variables are x 3 and x 5. Note that x ∈ N (A) if and only if:
x 1 = − 4 x 3 + 3x 5
x 2 = −x 3 − 2 x 5
x 4 = 0.
Thus,
x 3
: x 3 , x 5 ∈ R.
And a basis for N (A) is given by:
(b) Find all solutions to the equation Ax =
(4 points) Ax =
(^) if and only if:
x 1 = − 4 x 3 + 3x 5 + 3
x 2 = −x 3 − 2 x 5 + 5
x 4 = − 7.
Hence, the set of solutions is:
: x 3 , x 5 ∈ R.
For quick reference, here again is the matrix: A =
(c) Find a basis for N (A) that contains the vector
, or state why no such basis exists.
(4 points) Let v =
Note that, Av =
Thus,
Thus, there is a basis for N (A) containing v. Since dim N (A) = 2, v will form a basis for N (A)
along with any other vector in N (A) that is not a scalar multiple of v. Thus, one basis of N (A)
is given by,
(d) Find a basis for C(A) that contains
, or state why no such basis exists.
(4 points) Note that dim C(A) = 3. Also C(A) ⊂ R
3
. Thus C(A) = R
3
. Hence the vector
is in C(A) and one possible basis is:
Note: In part c and d, 2 points for saying that the vector is in N (A)/C(A) respectively. And
2 more points for finding the basis correctly. The most common mistake in this part is that
students show a different basis and prove that the span contains the concerned vectors. The
problem asks to find a basis containing the vector.
2 → R
4 is a linear transformation, and that
and T
(a) Find T
and T
(4 points) We have the following two equalities:
and
Since T is a linear transformation, this gives
(b) Let b ∈ R
4
. Find one or more conditions on b that determine precisely whether b is equal to
T(x) for some x ∈ R
2 ; that is, whether b belongs to im(T). (Your answer should be given in the
form of one or more equations involving the components b 1 , b 2 , b 3 , b 4 of b.)
(4 points) By part (a), we know that the matrix corresponding to T is
. The system
of equations T(x) = b gives rise to the following augmented matrix, and we find its reduced row
echelon form:
0 4 b 1
− 3 2 b 2
1 0 b 3
1 − 1 b 4
1 0 b 3
0 1 −b 4 + b 3
0 0 b 1 − 4 b 3 + 4b 4
0 0 b 2 + b 3 + 2b 4
Thus if x =
x
y
, then T(x) = b if and only if x = b 3 and y = −b 4 +b 3 and
b 1 − 4 b 3 + 4b 4 = 0,
b 2 + b 3 + 2b 4 = 0.
Therefore b is in the image of T if and only if the last two equations above are true.
Note: it actually works to solve this part by using the augmented matrix
4 8 b 1
− 1 4 b 2
1 0 b 3
0 − 2 b 4
, but the reason this works is that the vectors
and
form a basis
for R
2
. A complete solution should state this fact as a justification for the method used.
times 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.
In all these statements, the vector e 1 =
(in R
2 ), and similarly e 2 =
(a) Given a 2 × 5 matrix A, the equation Ay = e 1 has no solutions y in R
5
. T F MAYBE
(Equivalently, the issue is whether or not e 1 lies in C(A).) It might be that A has rank 2, i.e.
that C(A) = R
2 , in which case the equation has a solution; however, if A is the zero matrix,
then C(A) contains only the zero vector, and the above equation does not have a solution.
(b) Given a 5 × 2 matrix B, the equation Bz = Be 1 has infinitely many
solutions z in R
2 .
The equation has at least one solution, z = e 1 , but whether it has infinitely many is equivalent
to whether N (B) is non-trivial, i.e., whether B has positive nullity. This depends, since B could
have rank 2 (and nullity 0); but if B is the zero matrix, then B has rank 0 (and nullity 2).
(c) Given vectors v 1 , v 2 , v 3 in R
2 with the property that each of the sets
{v 1 , v 2 }, {v 2 , v 3 }, and {v 1 , v 3 }
is linearly independent, the set {v 1 , v 2 , v 3 } is also linearly independent.
No set of three vectors in R
2 is ever linearly independent, since dim(R
2 ) = 2 (see Prop. 12.1).
(d) Given vectors w 1 , w 2 , w 3 in R
5 with the property that each of the sets
{w 1 , w 2 }, {w 2 , w 3 }, and {w 1 , w 3 }
is linearly independent, the set {w 1 , w 2 , w 3 } is also linearly independent.
For example, let w 1 = (1, 0 , 0 , 0 , 0) and w 2 = (0, 1 , 0 , 0 , 0). On the one hand, consider w 3 =
w 1 +w 2 , in which case the three two-element sets are each independent but the three-element set
is not; on the other hand, consider w 3 = (0, 0 , 1 , 0 , 0), in which case all the sets are independent.
(e) Given nonzero v ∈ R
2 , the set
V = {x ∈ R
2 | ‖x + v‖
2 = ‖x‖
2
2 } is a subspace of R
2 .
The key point is that V is exactly the set of those vectors orthogonal to v: this is because
‖x + v‖
2 = (x + v) · (x + v) = x · x + x · v + v · x + v · v = ‖x‖
2
2
which equals ‖x‖
2
2 if and only if x · v = 0. (Aside: compare with Prop. 4.7.) Now,
(i) 0 ∈ V because 0 · v = 0; and
(ii) if x 1 , x 2 ∈ V , then (x 1 + x 2 ) · v = x 1 · v + x 2 · v = 0 + 0 = 0, so x 1 + x 2 ∈ V ; and
(iii) if x ∈ V and c is a scalar, then (cx) · v = c(x · v) = c(0) = 0, so cx ∈ V.
Thus, V is a subspace of R
2 .
