1. Announcements
Quiz 1 after lecture.
Today (Jan 20, Wed) is the last day to drop this class with no
academic penalty (No record on transcript). No refunds.
2 Corrections made to yesterday's slide (change 20 to 16 and
R3-R2 to R3-R1)
2. Last Class...
Suppose we have a set of vectors v1 . . . vp in Rn . If the vector
equation x1 v1 + . . . + xp vp = 0 has ONLY THE TRIVIAL SOLUTION
we say that the set of vectors is linearly independent.
3. Last Class...
Suppose we have a set of vectors v1 . . . vp in Rn . If the vector
equation x1 v1 + . . . + xp vp = 0 has ONLY THE TRIVIAL SOLUTION
we say that the set of vectors is linearly independent.
In other words, if we can nd atleast ONE non-zero weight c1 , c2 ,
. . ., cp such that c1 v1 + . . . + cp vp = 0, then the set v1 . . . vp is a
LINEARLY DEPENDENT set
4. Linear Independence Check by Inspection
Some set of vectors have some obvious properties that will help you
decide whether they are linearly independent without doing any row
operations.
5. Linear Independence Check by Inspection
Some set of vectors have some obvious properties that will help you
decide whether they are linearly independent without doing any row
operations.
If you have only one vector v and if it is not the zero vector
then v is linearly independent. (x1 v = 0 has only trivial
solution)
6. Linear Independence Check by Inspection
Some set of vectors have some obvious properties that will help you
decide whether they are linearly independent without doing any row
operations.
If you have only one vector v and if it is not the zero vector
then v is linearly independent. (x1 v = 0 has only trivial
solution)
The zero vector is linearly dependent. (The equation x1 0 = 0
has many solutions.
7. Linear Independence Check by Inspection
Some set of vectors have some obvious properties that will help you
decide whether they are linearly independent without doing any row
operations.
If you have only one vector v and if it is not the zero vector
then v is linearly independent. (x1 v = 0 has only trivial
solution)
The zero vector is linearly dependent. (The equation x1 0 = 0
has many solutions.
If two vectors are given and one vector is a multiple of the
other then the two vectors are LINEARLY DEPENDENT
8. Linear Independence Check by Inspection
Some set of vectors have some obvious properties that will help you
decide whether they are linearly independent without doing any row
operations.
If you have only one vector v and if it is not the zero vector
then v is linearly independent. (x1 v = 0 has only trivial
solution)
The zero vector is linearly dependent. (The equation x1 0 = 0
has many solutions.
If two vectors are given and one vector is a multiple of the
other then the two vectors are LINEARLY DEPENDENT
If a set has more vectors than there are entries in each vector,
then the set is LINEARLY DEPENDENT.
9. Linear Independence Check by Inspection
Some set of vectors have some obvious properties that will help you
decide whether they are linearly independent without doing any row
operations.
If you have only one vector v and if it is not the zero vector
then v is linearly independent. (x1 v = 0 has only trivial
solution)
The zero vector is linearly dependent. (The equation x1 0 = 0
has many solutions.
If two vectors are given and one vector is a multiple of the
other then the two vectors are LINEARLY DEPENDENT
If a set has more vectors than there are entries in each vector,
then the set is LINEARLY DEPENDENT.
If a set of vectors contains the zero vector, the set is linearly
dependent.
10. Examples
Problem 16 section 1.7 Determine by inspection whether the
vectors are linearly independent. Justify your answer.
4 6
−2 , −3
6 9
11. Examples
Problem 16 section 1.7 Determine by inspection whether the
vectors are linearly independent. Justify your answer.
4 6
−2 , −3
6 9
Solution The second vector is 1.5 times the rst vector. So these
vectors are linearly dependent.
12. Examples
Problem 18 section 1.7 Determine by inspection whether the
vectors are linearly independent. Justify your answer.
4 −1 2 8
, , ,
4 3 5 1
13. Examples
Problem 18 section 1.7 Determine by inspection whether the
vectors are linearly independent. Justify your answer.
4 −1 2 8
, , ,
4 3 5 1
Solution There are 4 vectors and each vector has 2 entries each. So
this set is linearly dependent.
14. Examples
Problem 20 section 1.7 Determine by inspection whether the
vectors are linearly independent. Justify your answer.
1 −2 0
4 , 5 , 0
−7 3 0
15. Examples
Problem 20 section 1.7 Determine by inspection whether the
vectors are linearly independent. Justify your answer.
1 −2 0
4 , 5 , 0
−7 3 0
Solution The zero vector is a part of this set. So the set is linearly
dependent.
18. What happened here?
The matrix A acted on a vector x from R4 and produced a
new vector b in R2 . Or, A transforms x into b.
19. What happened here?
The matrix A acted on a vector x from R4 and produced a
new vector b in R2 . Or, A transforms x into b.
The matrix A acted on a vector u from R4 and produced the
zero vector 0 in R2 . Or, A transforms u into 0.
20. What happened here?
The matrix A acted on a vector x from R4 and produced a
new vector b in R2 . Or, A transforms x into b.
The matrix A acted on a vector u from R4 and produced the
zero vector 0 in R2 . Or, A transforms u into 0.
Solving Ax = b is equivalent to nding all vectors x in R4 that are
transformed into the vector b in R2 when acted upon by A
21. Pictorially
Multiplication
by A
b
x
0
0
u Multiplication R2
4
R by A
22. Pictorially
Multiplication
by A
b
x
0
0
u Multiplication R2
4
R by A
23. Pictorially
Multiplication
by A
b
x
0
0
u Multiplication R2
4
R by A
24. Remember functions from calculus?
The correspondence between x and Ax is a function from one set
of vectors to the other.
25. Remember functions from calculus?
The correspondence between x and Ax is a function from one set
of vectors to the other.
Again, function just transforms one real number into another.
26. Transformation, Domain etc
A transformation (or function or mapping) T from Rn to Rm is a
rule that assigns to each vector x in Rn a vector T (x) in Rm .
27. Transformation, Domain etc
A transformation (or function or mapping) T from Rn to Rm is a
rule that assigns to each vector x in Rn a vector T (x) in Rm .
The set Rn is called Domain of T .
28. Transformation, Domain etc
A transformation (or function or mapping) T from Rn to Rm is a
rule that assigns to each vector x in Rn a vector T (x) in Rm .
The set Rn is called Domain of T .
The set Rm is called Co-Domain of T .
29. Transformation, Domain etc
A transformation (or function or mapping) T from Rn to Rm is a
rule that assigns to each vector x in Rn a vector T (x) in Rm .
The set Rn is called Domain of T .
The set Rm is called Co-Domain of T .
The notation T : Rn → Rm means the domain is Rn and the
co-domain is Rm .
30. Transformation, Domain etc
A transformation (or function or mapping) T from Rn to Rm is a
rule that assigns to each vector x in Rn a vector T (x) in Rm .
The set Rn is called Domain of T .
The set Rm is called Co-Domain of T .
The notation T : Rn → Rm means the domain is Rn and the
co-domain is Rm .
For x in Rn , the vector T (x) is called the image of x.
31. Transformation, Domain etc
A transformation (or function or mapping) T from Rn to Rm is a
rule that assigns to each vector x in Rn a vector T (x) in Rm .
The set Rn is called Domain of T .
The set Rm is called Co-Domain of T .
The notation T : Rn → Rm means the domain is Rn and the
co-domain is Rm .
For x in Rn , the vector T (x) is called the image of x.
Set of all images T (x) is called the Range of T .
33. Example 2, Section 1.8
0. 5 0 0
1
a
Let A = 0 0.5 0 , u= 0 , v = b
0 0 0.5 −4 c
Dene T : R3 → R3 by T (x) = Ax. Find T (u) and T (v).
34. Example 2, Section 1.8
0. 5 0 0
1
a
Let A = 0 0.5 0 , u= 0 , v = b
0 0 0.5 −4 c
Dene T : R3 → R3 by T (x) = Ax. Find T (u) and T (v).
Solution The problem is just asking you to nd the products Au
and Av
(0.5)(1) + 0.0 + 0.(−4) 0. 5
Au = 0.1 + (0.5).0 + 0.(−4) = 0
0.1 + 0.0 + (0.5).(−4) −2
(0.5)(a) + 0.b + 0.c 0. 5a
Av = 0.a + (0.5).b + 0.c = 0.5b
0.a + 0.b + (0.5).c 0. 5c
35. Example 4, Section 1.8
1 −3 2 6
Let A = 0 1 4 , b= −7
3 −5 −9 −9
Let T be dened by by T (x) = Ax. Find a vector x whose image
under T is b and determine whether x is unique.
36. Example 4, Section 1.8
1 −3 2 6
Let A = 0 1 4 , b= −7
3 −5 −9 −9
Let T be dened by by T (x) = Ax. Find a vector x whose image
under T is b and determine whether x is unique.
Solution The problem is asking you to solve Ax = b. In other words,
write the augmented matrix and solve.
1 −3 2 6
R3-3R1
0 1 −4 −7
3 −5 −9 −9
39. Example 4, Section 1.8
1 0 0 −5
0 1 0 −3
0 0 1 1
−5
So, x = −3.
1
40. Example 4, Section 1.8
1 0 0 −5
0 1 0 −3
0 0 1 1
−5
So, x = −3. All columns have pivots which means no free
1
variables. So the solution is unique.
