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Math 51 — Spring 2009 — Exam 1 Solutions
Problem 1. (15 pts.) Mark the following statements as either TRUE or FALSE. If a statement is false, give a simple counterexample. If a statement is true, give a justification. Read the questions very carefully!
Three points per part: 1 point for the answer and 2 points for the justification (except in parts (b), (c), (d), where these point values were swapped).
a) If the reduced row echelon form of a matrix A has a pivot in every column, then the null space N (A) of A is empty.
TRUE FALSE
We know that the null space of any matrix always contains the zero vector, so in particular, it is never empty. (In fact, under the given assumption on A, then the null space of A contains only the zero vector; it does not contain any nonzero vectors.) Notes: A very common mistake was to confuse the set containing only the zero vector with the empty set.
b) The cross product of two vectors in R^3 has length zero if and only if the two vectors are linearly dependent.
TRUE FALSE
Since ‖v × w‖ = ‖v‖‖w‖ sin θ, we see that if v × w has length 0, then either v or w is the zero vector, or sin θ = 0. If v = 0 or w = 0 , then {v, w} is linearly dependent since any set containing the zero vector is linearly dependent. If sin θ = 0, then this means that v and w lie on the same line, so again {v, w} is linearly dependent. On the other hand, if v, w are dependent, then one of them is a scalar multiple of the other, say v = cw. But then v × w = (cw) × w = c(w × w) = c( 0 ) = 0 , and so v × w has length 0. Notes: Many tried to prove this true simply by giving a single example — but one example is never enough to show that a statement about arbitrary vectors is true.
c) The dot product of two vectors is zero if and only if the two vectors are orthogonal.
TRUE FALSE
This is the definition of orthogonal (see page 26 of the Levandosky text).
d) For any k × n matrix A, dim N (A) + dim C(A) = n. TRUE FALSE
This is the statement of the Rank-Nullity Theorem.
e) For any v 1 , v 2 in Rn, if dim Span {v 1 , v 2 } = 0, then v 1 = v 2. TRUE FALSE
The only subspace of Rn^ of dimension 0 is { 0 }, so the given assumption tells us that Span {v 1 , v 2 } = { 0 }. But v 1 , v 2 ∈ Span {v 1 , v 2 }, so we must have v 1 = v 2 = 0 , and therefore v 1 = v 2.
Problem 2. (12 pts.) Consider the following system of equations:
x + 3y + z = b 1 2 x − z = b 2 3 x + 3y + 2z = b 3
where x, y and z are unknowns, and b 1 , b 2 and b 3 are real numbers.
a) For what values of b 1 , b 2 and b 3 is
x = 0 y = 1 z = 2
a solution to the above system?
(4 points) Setting x = 0, y = 1 and z = 2 in the given system yields b 1 = 0 + 3 · 1 + 2 = 5 b 2 = 2 · 0 − 2 = − 2 b 3 = 3 · 0 + 3 · 1 + 2 · 2 = 7
b) For what values of b 1 , b 2 and b 3 is
x = 0 y = 1 z = 2
the unique solution to the above system?
(4 points) From part (a) we know that only (b 1 , b 2 , b 3 ) = (5, − 2 , 7) gives (x, y, z) = (0, 1 , 2) as a solution. To check uniqueness of this solution, we find the null space of the coefficient matrix (call it A). Row reducing gives:
1 0 −^12
0 0 −^13
Thus N (A) = { 0 }, and so for the values (b 1 , b 2 , b 3 ) = (5, − 2 , 7) , the system has a unique solution.
c) For what values of b 1 , b 2 and b 3 does the above system of equations have no solutions?
(4 points) Since there are no zero rows in the reduced row echelon form above, there cannot be any “0 = 1” contradiction in the reduced system. Thus, for all values of (b 1 , b 2 , b 3 ), the above system of equations has a solution; for no values of (b 1 , b 2 , b 3 ) is there no solution.
Problem 4. (10 pts.) Complete each of the following sentences: a) A collection of vectors v 1 ,... , vk is defined to be linearly dependent if
(5 points)
... there exist real numbers c 1 ,... , ck, not all of which are 0, such that
c 1 v 1 + · · · + ckvk = 0
Alternate solution:
... one of the vectors, vi, may be written as a linear combination of the other vectors:
vi = c 1 v 1 + · · · + ci− 1 vi− 1 + ci+1vi+ 1 + · · · + ckvk
b) A basis for a subspace V is defined to be
(5 points)
... a set B of vectors in V such that both - Span B = V , and - B is linearly independent.
Problem 5. (12 pts.) Suppose A =
(^) and rref(A) =
(You don’t have to verify this!) a) Give a basis for N (A).
(4 points) Using rref(A), the vectors x such that Ax = 0 are of the form
x =
x 1 x 2 x 3 x 4 x 5
− 3 x 3 − 2 x 5 − 4 x 3 + 2x 5 x 3 − 3 x 5 x 5
= x 3
The two vectors on the right-hand side above are plainly linearly independent and span
N (A); consequently the set B =
forms a basis for N (A).
b) Give a basis for C(A).
(4 points) A basis for C(A) is given by taking the columns in A corresponding to those
columns in rref(A) with pivots. Thus C =
is a basis for^ C(A).
c) Given that A ·
, find all solutions to the system A · x =
(4 points) The set S of all solutions to the above system may be obtained by translating the
subspace N (A) by any particular solution vector in S. But the vector
already lies in S,
as stated in the problem. Thus, the solutions to the above system may be described as
S =
x =
s, t ∈ R
Problem 7. (15 pts.) Suppose v =
(^). Let V be the set of vectors x in R^3 such that x × v = 0 ; that is:
V =
x^ ∈^ R
(^3) x ×
(For the cross product you may use the formula:
x 1 x 2 x 3
×
y 1 y 2 y 3
x 2 y 3 − x 3 y 2 x 3 y 1 − x 1 y 3 x 1 y 2 − x 2 y 1
If you find that geometric or algebraic properties of the cross product are more useful than the above formula, you may use these properties as well.) a) Show that V is a subspace of R^3.
(5 points) There are several possible solutions. (If you’re not sure how to compose a math- ematical proof, you can mimic what is below in other circumstances.) First solution: V is a subspace because it satisfies the three following properties:
- V contains 0 because 0 × v = 0.
- V is closed under addition because if x, y both belong to V , then (x + y) × v = x × v + y × v = 0 + 0 = 0 ; hence x + y belongs to V.
- V is closed under scalar multiplication, because if x is in V , then for any scalar c, (cx) × v = c(x × v) = c 0 = 0 , so cx is in V.
Note that the above solution could be written without using the formula for the cross product: it uses the distributivity properties of the cross product only. Second solution: One can claim that x × v = 0 if and only if x is collinear to v (as v is not the zero vector), but strictly speaking this claim should be justified (no credit was deducted in any case). Once one knows this, one can say that this implies that V is the line spanned by v, i.e. V = Span(v), which is automatically a subspace of R^3 — you don’t have to prove this, as this was seen in the course. (Now, to justify the claim, one can say that ‖x × v‖ = ‖x‖‖v‖ sin θ, where θ is the angle between x and v, so the cross product is zero exactly when θ is a multiple of π radians. This means exactly that x, v are collinear.) One can alternatively solve the system (using the formula for the components of x × v):
− 4 x 2 + 2 x 3 = 0 4 x 1 − x 3 = 0 − 2 x 1 + x 2 = 0
and find solutions to be of the form^ x^3
which is the line spanned by v; i.e., again showing that V = Span(v).
Third solution: One could say that the above system, once introduced, means that x is in V if and only if it is in the null space of a matrix A, which can be explicitly defined (and will be done in part (b)). Then, as one knows that N (A) is a subspace for any matrix A, this proves that V itself is a subspace of R^3.
b) Give a matrix A such that V = N (A).
(5 points) From the formula, x ∈ V if and only if 0 = x × v =
− 2 x 3 + 4x 2 − 4 x 1 + x 3 2 x 1 − x 2
. But the latter
vector equals
x 1 x 2 x 3
(^). So V = N (A) for the matrix A =
Notes: Many answers contained a bunch of matrices, corresponding to operations on rows. Make sure to answer the question. Your answer should clearly contain: “one can take A =.. .”. Justify your answers (some credit has been given to non-justified answers.) Another matrix was possible, but clearly it should contain three columns, and one has to be careful that its rank is not too big. Many responses wanted to find A so that v is in N (A), without saying why v should be in V!
c) Find a basis for V.
(5 points) From part (b), it is sufficient to find a basis of N (A). We first find that
rref(A) =
(^). It follows (using the usual method) that N (A) = Span
so a basis of V is:
. (If in part (a) you gave solution 2, you could say directly
that {v} is a basis of the line V , of course.)
Notes: Same as above. We often had to extract the non-justified answers out of the verbiage. Many times, it was noticed that v is in V , but rarely justified why v spans V! Another common mistake was to look for a basis of C(A), which has nothing to with V.
Interesting question, by the way: Can you identify C(A)? Meaning: what is the link between a vector in C(A) and V? (Hint: orthogonality...)
c) Find a basis B for N (A).
(4 points) To find rref(A) we can save some time by starting with the same row operations we already did in part (b):
This gives pivot variables v 1 , v 2 , v 3 and free variable v 4 satisfying equations
v 1 − v 4 = 0 v 2 − 2 v 4 = 0 v 3 + 4v 4 = 0
with parametric form
v 1 v 2 v 3 v 4
v 4 2 v 4 − 4 v 4 v 4
=^ v^4
Therefore we can take B =
^.
d) Express each element of the basis B as a linear combination of the elements of B′.
(4 points) Note that by part (a) we know this is possible, since the basis vector of N (A) lies in N (A′) and, by definition, Span(B′) = N (A′). We solve the vector equation
c 1
+^ c^2
for unknowns^ c^1 and^ c^2.
The row reduction is particularly easy if we use the last two rows to zero out the first two:
giving solutions c 1 = −4 and c 2 = 1. So
An even quicker way to get the answer, given that we know the system is solvable by part (a), is to immediately read off the coefficients from the last two rows of equations, which are c 1 + 0c 2 = −4 and 0c 1 + c 2 = 1.
