MTH 309 Y
EXAM 1 REVIEW
• Vectors
1) R� = the set of all column vectors with � entries.
2) Operations on vectors in R�:
– addition
– multiplication by scalars
3) Geometric interpretation of vectors and vector operations.
4) Linear combinations of vectors.
5) Span of a set of vectors.
6) Linear independence of a set of vectors.
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MTH 309 Y
EXAM 1 REVIEW
• Matrices
1) Operations on matrices:
–
–
–
–
–
addition A + B
multiplication by scalars �A
multiplication of matrices AB
matrix transpose AT
matrix inverse A−1
2) Properties of matrix algebra
À
Ã
Õ
AB �= BA
(AB)T = BT AT
(AT )T = A
(A + B)T = AT + BT
Œ
œ
–
—
(AT )−1 = (A−1)T
(AB)−1 = B−1A−1
(A−1)−1 = A
(A + B)−1 �= A−1 + B−1
3) Column space Col(A) and null space Nul(A) of a matrix.
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MTH 309 Y
EXAM 1 REVIEW
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• Matrix transformations and linear transformations
1) An � × � matrix A defines a matrix transformation
TA : R� −→ R��
TA (v) = Av
2) Composition of linear transformations = matrix multiplication:
TA ◦ TB = TAB
3) Linear transformation is a function T : R� → R� such that
(i) T (u + v) = T (u) + T (v)
(ii) T (�v) = �T (v)
4) Every matrix transformation is a linear transformation.
5) Every linear transformation T : R� → R� is a matrix transformation:
T = TA
�
�
where A = T (e1) � � � T (e�) is the standard matrix of T .
MTH 309 Y
EXAM 1 REVIEW
4
• Linear equations
1) Three equivalent forms of equations
⎧
⎨ �11�1 + · · · + �1��� = �1
...
...
...
system of linear equations
⎩
��1�1 + · · · + ����� = ��
⎡
⎤
⎡
⎤
⎡
⎤
�11
�1�
�1
�1 ⎣ ... ⎦ + · · · + �� ⎣ ... ⎦ = ⎣ ... ⎦
��
��1
���
⎡
�11
⎣ ...
��1
⎡ ⎤
⎤ ⎡ ⎤
� � � �1�
�1
�1
... ⎦ · ⎣ ... ⎦ = ⎣ ... ⎦
��
��
� � � ���
2) How to solve matrix equations:
row reduction (= Gauss-Jordan elimination)
3) Related notions:
–
–
–
–
elementary row operations
reduced echelon form of a matrix
leading ones
pivot positions and pivot columns
vector equation
matrix equation
MTH 309 Y
EXAM 1 REVIEW
5
• Sample True/False questions and answers.
For each of the statements given below decide if it is true or false. If
you decide that it is true justify your answer. If you think it is false
give a counterexample.
a) If a system of linear equations has more variables that equations
then it must have infinitely many solutions.
b) If u� v� w are vectors in R� such that the set {u� v� w} is linearly
independent then the set {u� v} is also linearly independent.
c) If A, B are �� invertible matrices then A+B is also an invertible
matrix.
d) If T : R � → R� is a linear transformation then T (0) = 0 where
0 is the zero vector.
e) If v1, v2, v3 are vectors in R2 such that the set {v1� v2� v3} is
linearly dependent then v1 ∈ Span(v2� v3).
f) If A is a 4 × 5 matrix then the equation Ax = 0 has infinitely
many solutions.
g) If v, w are solutions of the matrix equation Ax = b then the vector
v + w is in the null space Nul(A).
MTH 309 Y
EXAM 1 REVIEW
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Here are solutions to the sample true/false questions from the previous page. You should try to answer all questions by yourself
before reading these solutions.
a) FALSE: the system which has more variables than equations may have zero
solutions.
Example:
�
�1 + �2 + �3 = 1
�1 + �2 + �3 = 2
has no solutions.
b) TRUE: if the set {u� v} were linearly dependent then the equation
�1 v1 + �2 v2 = 0
would have a non-trivial solution �1 = �1 , �2 = �2 . This would mean that the
equation
�1 v1 + �2 v2 + �3 v3 = 0
also has a non-trivial solution: �1 = �1 , �2 = �2 , �3 = 0, so the set {u� v� w} would
be linearly dependent.
�
�
�
�
1 0
−1 0
c) FALSE: for example A =
� B=
are invertible matrices but
0 1
0 −1
�
�
0 0
is not invertible.
A+B =
0 0
d) TRUE: since T is a linear transformation we have
T (0) = T (0 + 0) = T (0) + T (0)
Subtracting T (0) from both sides we obtain 0 = T (0).
MTH 309 Y
e) FALSE: take e.g.
EXAM 1 REVIEW
v1 =
�
1
0
�
�
v2 =
�
0
1
�
�
7
v3 =
�
0
2
�
Then the set {v1 � v2 � v3 } is linearly dependent but v1 is not in Span(v2 � v3 ).
f) TRUE: the equation Ax = 0 is consistent since it has at least one trivial solution
x = 0. Moreover, A has more columns than rows which means that one of the
columns does not contain a pivot position, and so it corresponds to a free variable.
Therefore we must have infinitely many solutions.
f) FALSE: take e.g.
�
�
0 0
A=
�
1 0
b=
�
0
1
�
�
v=
�
1
0
�
�
w=
�
Both vectors v and w are solutions of the equation Ax = b, but
� �
0
A(v + w) =
2
1
1
�
�
Thus A(v + w) is not the zero vector, and so v + w is not in Nul(A).