Math 308 C
Final Exam
Your Name
Spring 2015, 6-8-2015
Your Signature
Student ID #
1.
2.
3.
4.
5.
6.
7.
8.
Form
Bonus
P
35
25
17
17
20
7
12
5
3
(10)
141
Points
of
• No books are allowed. But you are allowed one sheet (10 x 8) of handwritten notes
(back and front). You may use a calculator.
• Place a box around
your final answer
to each question.
• If you need more room, use the back of each page and indicate to the grader how to
find the logic order of your answer.
• Raise your hand if you have questions or need more paper.
• For TRUE/FALSE problems, you just need to cross the right box. For each correct
answer, you will get 1 point, for each incorrect answer, -1 point is added. For no
answer you will get zero points. In each subsection of the TRUE/FALSE part, you
can never get less than zero points.
• In order to get points for formal correctness, underline vectors, use {}-brackets for
sets, declare parameters, mark equivalent matrices properly and keep a reasonable
order and neatness.
Do not open the test until everyone has a copy and the start of the test is announced.
GOOD LUCK!
1.(35 points) For each correct answer in the TRUE/FALSE part, you will get
1 point, for each incorrect answer, there will be one point subtracted, i.e. you
get -1 point. For no answer, you get 0 points. You can not get less than 0
points out of one subproblem (which are the problems, (a)-(g))
(a)
A homogeneous system can either have a unique solution
or infinitely many solutions.
The Gauss-Jordan algorithm leads to equivalent matrices.
A homogeneous system with at least one free variable has
infinitely many solutions.
Let A be a (4, 6)-matrix. Then any solution to Ax = 0 is
a vector in R6 .
TRUE
FALSE
TRUE
TRUE
FALSE
FALSE
TRUE
FALSE
Let 0 6= u. Then the span of {u} has infinitely many
elements.
Let S1 $ S2 ⊆ Rn . Then span(S1 ) $ span(S2 ).
If Ax = b has a solution, then b ∈ row(A).
The zero vector is always an element of the span of vectors.
Let m > n and let S = {u1 , u2 , . . . , un } ⊆ Rm . Then S is
linearly dependent.
For any vector u and c ∈ R, the set {u, cu} is linearly
dependent.
If [A|0] has exactly one solution, the columns of A are
linearly independent.
TRUE
FALSE
(b)
TRUE
TRUE
TRUE
TRUE
FALSE
FALSE
FALSE
FALSE
TRUE
FALSE
TRUE
FALSE
TRUE
FALSE
TRUE
FALSE
TRUE
TRUEE
FALSE
FALSE
TRUE
FALSE
TRUE
FALSE
(c)
Let S = span{u1 , u2 } and let u1 , u2 be linearly independent. Then dim(S) = 2.
Let A be an (n, n)-matrix with rank(A) = n. Then
nullity(A) = 0.
{u1 , u2 , u3 , u4 } ⊆ R3 is a linearly dependent set.
Let S ⊆ Rn be a subspace of dimension m. Then any basis
of S contains exactly m elements.
Let S1 , S2 ⊆ R3 be subspaces with dim(S1 ) = 1, dim(S2 ) =
2. Then S1 ⊆ S2 .
If A is a matrix in echelon form with no zero rows, then
the rows form a basis of row(A).
Let A be an (n, m)-matrix. Then nullity(A) ≤ m.
(d) For any homomorphism T there is a corresponding matrix
Let T : Rm → Rn be a homomorphism and let A be its
corresponding matrix. Then A is an (n, m)-matrix.
Let m > n and let T : Rm → Rn be a homomorphism.
Then T can be surjective, but not injective.
Let T be a homomorphism with corresponding (m, n)matrix A. Then {Au | u ∈ Rn } = range(T ).
Let T be a surjective homomorphism. Then the range of
T is equal to the domain of T .
Let T : Rn → Rn be a homomorphism with corresponding
matrix A. If [A|0] has infinitely many solutions then T is
not surjective.
TRUE
FALSE
A with T (u) = Au.
TRUE
FALSE
TRUE
FALSE
TRUE
FALSE
TRUE
FALSE
TRUE
FALSE
(e)
(f)
Let A be an (n, m)-matrix and B be an (m, k)-matrix.
Then AB is an (n, k)-matrix.
It is possible for a non-zero (n, n)-matrix that Ak = 0n,n
for some k ∈ N.
Let A be an (n, n)-matrix. Then det(A2 ) = (det(A))2 .
A square matrix A is singular, if and only det(A) = 0.
Let A be an (m, n)-matrix. Then rank(A) ≤ m.
Matrices in this part have dimension (n, n).
An eigenvector cannot be the zero vector, but an eigenvalue may be 0.
Let A be a matrix, u ∈ Rn , λ ∈ R such that Au = λu.
Then u is an eigenvector of A with eigenvalue λ.
Let A be a matrix with eigenvalue λ. Then the matrix
A − λIn is not invertible.
Let Eλ be an eigenspace for a matrix A.
Then 1 ≤ dim(Eλ ).
Let A be a matrix with characteristic polynomial
(λ − 1)2 (λ + 1).Then A is invertible.
TRUE
FALSE
TRUE
FALSE
TRUE
TRUE
TRUE
FALSE
FALSE
FALSE
TRUE
FALSE
TRUE
FALSE
TRUE
FALSE
TRUE
FALSE
TRUE
FALSE
The zero vector is orthogonal to every other vector of
the same Euclidean space.
Two non-zero vectors with all components being negative cannot be orthogonal.
TRUE
FALSE
TRUE
FALSE
(g)
2.(1+4+1+6+2+(1+1)+(2+1+1+1+4)=25 points) Let h be a real number. Consider the following linear homomorphism:




u1
hu1 + 4u2 + 2hu3
.
2u2 + hu3
Th :  u2  → 
u3
(4 − h)u1 + 5u2 + 3u3
(a) Find the corresponding matrix Ah , such that Th (u) = Au for all u ∈ R3 .


h 4 2h
0 2 h 
Ah = 
4−h 5 3
(b) Determine h such that Th /Ah is invertible.


h 4 2h
0 2 h ) = 6h − 5h2
det(
4−h 5 3
and det(A) = 0 ⇐⇒ h(6 − 5h) = 0 ⇐⇒ h = 0 or h = 6/5. So Th /Ah is invertible if and
only if h 6= 0 or h 6= 6/5.
For the rest of this problem, i.e. for (c)-(g), we set h = 1, so that Th becomes
T1 .
(c) Write down the matrix A1 that corresponds to T1 .


1 4 2
 0 2 1 
3 5 3
(d) Find the inverse of T1 .


1 4 2 1 0 0
 0 2 1 0 1 0 
3 5 3 0 0 1

1
4
2
1 0 0
2
1
0 1 0 
→  0
0 −7 −3 −3 0 1



1 4 2
1 0 0
 0 2 1
0 1 0 
0 0 1 −6 7 2

1 4 0 13 −14 −4
6 −6 −2 
→  0 2 0
0 0 1 −6
7
2


1 0 0
1 −2
0
 0 1 0
3 −3 −1 
0 0 1 −6
7
2

So

T −1

1 −2
0
: u 7→  3 −3 −1  u
−6
7
2
(e) Find u such that T1 (u) = [1, 0, −2]t .


 

1 −2
0
1
1
 3 −3 −1   0  = 
5 
−6
7
2
−2
−10
(f) Let R : R3 → R4 be a homomorphism that has

2
0
2
 0 −1
2
B=
 1 −1 −1
2
2
1




as matrix.
(i) What is the domain of R? R3
(ii) What is the codomain of R? R4
(g) Consider the concatenation R ◦ T1 , which maps u ∈ R3 to R(T1 (u)) (first T1 , then R).
(i) What is the domain of R ◦ T1 . Because we first apply T1 , the domain of R must be the
same as the domain of T1 which is R3 .
(ii) What is the codomain of R ◦ T1 ? As we lastly apply R, the codomain of R ◦ T1 is the
same as the codomain of R which is R4 .
(iii) Can R ◦ T1 be surjective? Justify briefly. As the domain is R3 and the codomain is
R4 , i.e. 3 < 4, the homomorphism cannot be surjective.
(iv) Find the associated matrix C such that R ◦ T1 (u) = Cu for every u ∈ R3 . We have
T1 (u) = Au and R(v) = Bv, we get the following:
R ◦ T1 (u) = R ◦ (Au) = BA(u).
So we need to multiply the two matrices:





2
0
2
8 18 10
1 4 2
 0 −1


8
5
2  
  6
BA = 
 1 −1 −1  · 0 2 1 =  −2 −3 −2
3 5 3
2
2
1
5 17 9.




3.(5+2+5+2+2+1=17 points) Consider the following matrix:


−1
1 −2
0
2 1
2 −4 −2 0  .
A =  0 −1
2 −1
2
4 −2 0
(a) Find the null space of A.



−1
1 −2
0
2 1
−1
1 −2
0
2 1 0



0 −1
2 −4 −2 0 0 →
0 −1
2 −4 −2 0
[A|0] =
2 −1
2
4 −2 0 0
0
0
0
0
0 1
The backward substitution gives then:
x6 = 0, x5 = t1 , x4 = t2 , x3 = t3 , x2 = 2t3 − 4t2 − 2t1 and x1 = −4t2 . Hence the
of A is

 
  
0
−4
0
 −2   −4   2 

 
  
 0   0   1 



  
null(A) = span{
 ,  1  ,  0 }.
0

 
  
 1   0   0 
0
0
0

0
0 .
0
null space
(b) Verify that the vector v = [−4, −8, 0, 1, 2, 0]t is an element of null(A).Any element
u of null(A) satisfies Au. So


−4

  −8 

−1
1 −2
0
2 1 
 0 

 = [0, 0, 0]t .
 0 −1
2 −4 −2 0  

1

2 −1
2
4 −2 0 
 2 
0
(c) Find a basis of null(A), that contains the vector v = [−4, −8, 0, 1, 2, 0]t . We can use
Algorithm II for finding a basis.




−4
0 −4 0
1
0
1 0
 −8 −2 −4 2 
 −4 −1 −2 0 




 0

 0
0
0 1 
0
0 1 



 1
 →  0

0
1
0
0
0
0




 2
 2
1
0 0 
1
0 0 
0
0
0 0
0
0
0 0








1
0
1 0
0 −1
2 0
0
0
0 1
0
0
0 0
0
1 −2 0
0
0
0 0












→ 




1
0 1 0
0 −1 2 0 

0
0 0 1 

0
0 0 0 

0
0 0 0 
0
0 0 0
We see that the first, second and fourth columns are pivot columns, hence we choose
accordingly the basis vectors from the original vectors. Thus, a basis is

 
  
−4
0
0
 −8   −2   2 

 
  
 0   0   1 



  
{
 ,  0  ,  0 }
1

 
  
 2   1   0 
0
0
0
(d) What is the nullity of A? What is the rank of A?
As the basis we computed has three vectors, the nullity of A is 3. By the rank-nullity
theorem, the rank is 6 − 3 = 3.
(e) Consider the homomorphism T : R6 → R3 , u 7→ Au.
(i) Interpret null(A) in terms of T . The nullity of A is just the kernel of T .
(ii) Is T surjective? Justify your answer by using (d).Yes, T is surjective. The reason is
that the rank of A is 3, which means that the dimension of the column space is 3. But the
column space of A is equal to the range of T . The codomain of T is R3 , so that the range
which has dimension 3, must be equal to R3 , hence surjective.
4.(6+3+5+2+1=17 points) Consider the following matrix:


1
0
0
0
 0 −2 −1
0 


 0
0
1
1 
0 −2 −2 −1
(a) Determine the characteristic polynomial χA of A. Show all your work! (Key, so that
you can continue: The characteristic polynomial is χA = λ4 + λ3 − λ2 − λ.) We use cofactor
expansion, which gives:
χA =
=
=
=
=
=
(λ − 1)[(−2 − λ)(1λ)(−1λ) + 2 + 2(−2λ)]
(λ − 1)[(−2 − λ)(1λ)(−1λ) − 2 − 2λ]
(λ − 1)[(−2 − λ)(1λ)(−1λ) + 2(−1 − λ)]
(λ − 1)(−1 − λ)[(−2 + 2λ − λ + λ2 + λ)]
(λ − 1)(−1 − λ)(1 + λ)
−(λ − 1)(1 + λ)2 λ
(b) Determine the eigenvalues for A. The eigenvalues of A are the roots of χA , which are
0,-1,1.
(c) Compute the eigenspace for the eigenvalue λ = −1. What is the dimension of this
eigenspace? We need to find the nullspace of [A − (−1)I|0].




2
0
0 0 0
2
0
0 0 0


 0 −1 −1 0 0 
 →  0 −1 −1 0 0 

 0
 0
0
2 1 0 
0
2 1 0 
0 −2 −2 0 0
0 −2 −2 0 0
Backward substitution gives then x4 = t, x3 = −1/2t, x2 = 1/2t, x1 = 0 (x1 is not a free
variable but simply zero!). Hence




0
0
 1/2 
 1 



E−1 = span{
 −1/2 } = span{ −1 }.
1
2
Because this span is linearly independent, it is a basis at the same time. From that we see
that dim(E−1 ) = 1.
(d) What is the only possible dimension of the eigenspace with eigenvalue λ = 0? Answer
this question with the help of χA and justify your answer.
The multiplicity of λ = 0 as a root of χA is 1. Hence dim(E0 ) ≤ 1. But any eigenspace
must have at least dimension one (because the zero vector never is an eigenvector!) , the
dimension of E0 must be equal to 1.
(e) Based on the knowledge about the eigenvalues of this matrix, what can be said about
the determinant of A? Justify your answer.As A has eigenvalue 0, the negation of the big
theorem implies, that det = 0.




2
1
 1 
 0 



5.(5+3+5+4+3=20 points) Let S = span{s1 = 
 2  , s2 =  −1 }. (a) Find a
1
−1
⊥
basis B for S .
We must find the null space of
t
s1 0
.
st2 0
So let’s solve:
2 1
2
1 0
2 1 2 1 0
→
1 0 −1 −1 0
0 1 4 3 0
Backward substitution gives x3 = t1 , x4 = t2 , x2 = −4t1 − 3t2 , x1 = t1 + t2 . Hence

 

1
1
 −4   −3 
 

B = {
 1  ,  0 }.
0
1
(b) Show that the vector v = [−1, −1, 4, −5]t is orthogonal to S. We calculate the dot
product of v with every vector of the defining span:
 

2
−1
 1   −1 
 

 2  4  = 0
1
5
and



1
−1
 0   −1 



 −1   4  = 0,
−1
5
hence orthogonal.
(c) Write v as a linear combination of the vectors in B from (a).
We need to find the coefficients of the linear system given by



 

1
1
−1
 −4 

 

 + c2  −3  =  −1  ,
c1 
 1 
 0   4 
0
1
5
in other words, we need to solve

1
1 −1
 −4 −3 −1 


 1
0
4 
0
1
5

This leads to the solution c1 = 4, c2

1
 −4
4
 1
0
= −5. In summary,


 
1
−1

 −3   −1
 − 5
 

 0 = 4
1
5




(d) Show that for any vector u and any real number c ∈ R the equality
||cu|| = |c| · ||u||
√
√
√
holds.We have ||cu|| = cu · cu = c2 u · u = |c| u · u = |c| · ||u||.
(e) Find the condition on two vectors w1 , w2 ∈ Rn such that their distance is equal to 0.
The only vector that has length 0 is the zero vector. Hence: 0 = ||w1 − w2 || if and only if
w1 − w2 = 0, which is when w1 = w2 .
6. (3+4=7 points) Let
2 −2
A=
.
−1
1
(a) Verify that the characteristic polynomial χA of A is χA = λ2 − 3λ.
det(A) = (2 − λ)(1 − λ) − 2 = λ2 − 3λ.
(b)
λ by A and compute
A2 − 3 · A =
Compute
χA (A), i.e.
substitute
the term.
2 −2
2 −2
2 −2
6 −6
6 −6
0 0
−3
=
−
=
−1
1
−1
1
−1
1
−3
3
−3
3
0 0
7.(4+1+1+4+1+1=12 points) Consider the Euclidean space R3 .
(a) List all possible dimensions of the subspaces of R3 . Any subspace has dimension smaller
or equal to the overlaying subspace, hence the possible dimensions are 0,1,2,3
(b) How many different subspaces of R3 of dimension 3 are there?In R3 only R3 has
dimension 3, hence there is 1 such space.
(c) How many different subspaces of R3 of dimension 1 are there?All lines have dimension
3, hence there are infinitely many.
(d) Give an example of two different one dimensional subspaces of R3 .
 
1
span{ 0 }
0
and


0
span{ 1 }.
0
(e) As which kind of object can we geometrically interpret one dimensional subspaces in
R3 . It is the lines that have dimension 1 in R3 .
(f) Which subspace(s) has/have only finitely many elements?There is only one subspace
that has only finitley many elements, and this is the zero space.
8.(2+3=5 points)
Let u1 , u2 , u3 , u4 , u5 be vectors in R6 and consider the set of vectors
U := {u1 , u2 , 2u1 − u2 , u3 , u4 − 3u5 , u5 }.
Now let A be the matrix that has the vectors of U as columns.
(a) What are the dimensions of the matrix A? It is a 6x6-matrix, hence the dimensions
are (6, 6).
(b) What is the determinant of A. Justify your answer. As the second and the fourth
columns are linear combinations of the other columns, we have linear dependent columns,
which means by the Big Theorem, that det(A) = 0.