1) (10)
Discuss the rank of A varies with t
1 1 0
A = 3
t
1
2 t −1 1
2 − 3 0 1
1
For an 3x5 matrix A = 2
1 − 4 − 1 3
− 2 − 1 1 − 2 3
2) (20)
a)
Find a basis and the dimension of the row space and the column space of A
b)
Find an orthonormal basis and the dimension of the null space of A
c)
d)
Find an orthonormal basis and the dimension of the left null space of A
Conclusion: Rank, nullity of A?
Hint: Follow the example 5 of 4.8
3) (10)
Given the set W = (3t − s, t + 2s,2t + s)
a)
Determine whether W a subspace in R 3 . (You must show the set satisfies a
sub-space properties)
b)
4) (10)
5) (10)
Find the basis and the dimension. Describe this subspace geometrically.
Find a basis for orthonormal complement W ⊥ for a set W, a subspace in R 4
x1 − x2 − x3 + x4 = 0
W=
2 x1 + x2 + x3 + 2 x4 = 0
Find the coordinate vector of p( x) in P2 , relative to the basis S = p1 , p2 , p3 ,
where p( x) = 3 + 6 x − 10 x 2 , p1 ( x) = 2 − 4 x, p 2 ( x) = x + 3x 2 , p3 ( x) = 4 + 6 x 2
6) (10)
x x
x
2
a) (5) Find the standard coordinate basis = for the = and
y y S
y B1 − 1
3 − 4
B1 =
.
2 1
x
− 1 4
?
b) (5) What is the coordinate basis of in the basis B2 =
y
2 0
7) (10)
Are the following matrices linearly independent? Basis for M 22 ?
3 6
3 − 6
− 8
0 − 1 0
1 0
− 1 0 − 12 − 4 and − 1 2
8) (10)
Find the solution set for the equation AX = b where
0 − 1 0 2
x
0
a)
A=
X = ;b =
− 1 1 0 1
y
1
b)
Deduce from this, find that the null space of A
c)
What is the left null space A?
9) (5)
Given that the set {1, 1 + x + x 2 , p( x)} is a basis for AX = P2 , which of the following is a
possible value for p (x ) ?
(A) 0
(B) 1 + x
(C) −1
(D) 2 + x + x 2
10) (5) What is the transition matrix from S1 to S 2 given
1 4
1 3
S1 = , and S 2 = ,
− 2 − 4
3 8
3
Math 410, Linear Algebra
Dr. B Truong
Name: _______________________
LESSONS, CHAPTER 4
I)
Real Vector Space (VS) : V
In this section , we pay attention the the 10 axioms for VS.
Let V be a non-empty set of objects. Let u and v a pair of object in V and a scaler k
Pay very attention to the axioms
(1)
closure under addition (u + v) V
(6)
closure under scalar multiplication kuV
Examples of VS:
a)
Zero VS
b)
VS of R n
c)
VS{sequence of real numbers}
d)
VS of 2 2 matrices ( and m n matrices
e)
VS of {real valued functions}
Examples of collectios which are VS:
a)
V = R 2 ; u = (u1 , u 2 ) and ku = (u1 ,0)
b)
V = {set of real vectors} ; u + v = uv and ku = u k
1I)
Subspace (SS) W of a VS V: W is non-empty set W V and
We need to show a SS by satisfying 2 conditions
(u + v) W
( ku) W
a)
b)
Examples: SS
a)
b)
c)
d)
e)
f)
g)
The Zero SS
Lines through origine in R 2 and R 3
SS of matrices M nn
Set { continuous functions in (-∞,∞)}
{Continuous derivatives functions}
SS of {all polynomials}
SS of {all polynomials of degrees ≤ n}
Examples: Not a SS
a)
b)
c)
d)
{Positive real functions}
{Set vectors (a,b) such that a,b > 0 }
{Invertible matrices M nn }
{Set of all polynomials of degrees = n}
Subspace of a homogeneous linear system AX = 0
with m equations and n unknowns is a SS of R n
Here A is a m n matrix and X is a m 1 vector
III)
Spanning sets:
Let V be a VS
(You will be familiarize the following notations:
v j and w V and k j = constants, j = 1,2....r
w = k j v j j = 1,2....r
We say
j=1
Is the linear combination with coefficients
Further more, if
S = (w1 , w2 ,..., wr ) V
Then the set W of all possible linear combination vectors in V is a SP of V
We say the SS W of V spanned by S
Thus,
W = span{w1 , w2 ,..., wr } or W = span{ S}
Example 1:
1)
Standard unit vectors in R n
e1 = (1,0....), e2 = (0,1,....),.....en = (0,0,...n) span
2)
3)
Rn
The vector v = k ( a, b) , k = scalar, span the line passing through (0.0)
The vectors u = (a, b), v = (c, d ) span the plane containing u and v
I.e: The plane = span u, v
4)
The polynomials 1, x, x 2 ....x n span the VS Pn
Say :
5)
6)
Example 2:
Pn = span 1, x, x 2 ....x n
Vectors u, v and w in R 2 we can find scalars C1 ,C 2 such that
The linear combination satisfied:
w = C1u + C2 v as long as u is not in span v ( or reverse)
Vectors u, v and w in R 3 some time, we can’t find scalars C1 , C2 such that
The linear combination satisfied:
w = C1u + C2 v since w is not in spanu, v ( or reverse)
Express w = (6,11,6) as the linear combination of
u1 = (2,1,4) , u2 = (1,−1,3) , u 3 = (3,2,5)
We set to solve the system:
2 1 4 c1 6
2 1 4
c1
6
1 − 1 3 c2 = 11 AC = b . Where A = 1 − 1 3 ; C = c2 and b = 11
3 2 5 c 6
3 2 5
c
6
3
3
1
We have C = A −1b = adj( A) .
A
7
− 11 3
We find that the det(A) = 2 and adj( A) = 4 − 2 − 2
5
− 1 − 3
7 6
− 11 3
9 9 / 2
1
1
1
C = A −1b = adj( A) b C = 4 − 2 − 2 11 = − 10 = − 5
A
2
2
− 1 − 3 6
5
1 1/ 2
That means :
6
2 1
3
9 1
11 = 1 − 5 − 1 + 2
6 2 4 3 2 5
1 12
as a linear combination of
Express C =
8 − 2
1 4
2 0
; B =
A =
3 0
1 2
Example3:
We have to find x, y such that
1 4
2 0 1 12
+ y
=
x
3 0
1 2 8 − 2
Or solve ( by equating all elements of both sides)
x + 2 y = 1 (1)
4 x + 0 y = 12 (2)
3x + y = 8 (3)
0 x + 2 y = −2 (4)
x = 3; y = −1
The equation (2) and (4) give
These values also satisfy the eq (1) and (3). Thus we have
1 12 1 4 2 0
= 3
− 1
C = 3 A − B or
8 − 2 3 0 1 2
IV)
Linear Independent:
r
Let v j 1 V a VS ; j = 1,2....r
Form a vector equations for a linear combination as follows
c v
j
j
=0
j
Conclusion:
v j 1r is said to be Linearly Independent if all
a)
cj = 0 j
stands for for all
b)
Otherwise, if ck 0 for some k the set v j 1 is Linear Dependent
r
Note:
r
a)
The set v j 1 in which vk = 0 for some k , is Linear Dependeny (L/D)
b)
In R n , any 2 vectors u and v such that u kv are Linearly Independent (L/I)
To show the v j 1 in R n or M ij or set of polynomials Pn always solve the equation
r
c v
j
j
=0
j
c M
j
j
= 0 or
j
c
j
fj =0
j
Usefull facts:
The unit vector in Pn is (1,0,0....0), (0,1,0.....0),....(0,0,0,0,0..1)
So the function f ( x) = 2 + 2 x + x 3 can be written as
x0
1
x
f ( x) = (2 2 0 3) 2 = 2 + 2 x + 0 x 2 + 3x 3
x
x3
So the vector (2 2 0 3) represents the function f ( x) = 2 + 2 x + x 4
Definition:
If the set of function f j 1 ( x) is continuous up to (n-1) th derivative
Then the determinamt
n
f1
f1 '
•
•
•
W ( f1 , f 2 ,...., f n ) =
f1
( n −1)
f2
f2 '
•
•
•
f2
( n −1)
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
fn
•
•
•
•
fn
( n −1)
is called the Wronskian of f1 , f 2 ,...., f n
If W ( f1 , f 2 ,...., f n ) 0 on (−,) then f j 1 ( x) are L/I
n
Conclusion: a)
Example 4:
b)
If W ( f1 , f 2 ,...., f n ) = 0 on (−,) then f j 1 ( x) are L/D
a)
Whether sin x and cos x are L/I?
n
W ( x, cos( x)) =
sin x cos x
=1
cos x − sin x
L/I on on (−,)
b)
Whether functions
f1 ( x) = −1 + 2 x − x 2 f 2 ( x) = x + x 2 f 3 ( x) = 2 + 2 x 2 are L/I
− 1 + 2x − x 2
W ( f1 , f 2 , f 3 ) =
2 − 2x
−2
x + x2
1 + 2x
2
2 + 2x 2
4x
4
(− 1 + 2x − x )(4(1 + 2 x) − 2(4x)) − ( x + x )(4(2 − 2x) + 8x)
=
2
2
+ (2 + 2 x 2 )(2(2 − 2 x) + 2(1 + 2 x) )
=8
So we conclude that the set f j 1 ( x) are linearly independent
3
Another way to do this problem is now, we consider these functions in P2 as vectors
And find the Wronskian accordingly
− 1
0
2
2
2
f1 ( x) = −1 + 2 x − x = 2 f 2 ( x) = x + x = 1 , f 3 ( x) = 2 + 2 x = 0
− 1
1
2
−1 0 2
W (v1 , v2 , v3 ) = 2 1 0 = (−2 + 0 + 4) − (−2 + 0 + 0) = 4 0 L / I
−1 1 2
2
Note that the Wronskian are not the same.
Math 410, Linear Algebra
Dr. B Truong
Name: _______________________
LESSONS, CHAPTER 4 CONTINUE
I)
Coordinates and Basis and Change of Basis:
To make it easy, say the vector
a a
u = = is a vector in standard base
b b S
1 0
0 1
a a
1 0 a
Ie = =
b b S 0 1 b
Now, it is the same vector, but in different base, say B1 or in any base B j , then we can
express u as
a a
1 0 a
= =
= B j u j =
b b B j 0 1 b
C1
C 2 B
(1)
j
Now, we have to determine C1 , C2 .
1 0 a
= B j u j =
0 1 b
C1
C
= Bk u k = Bk 3 and so on.....
C 2 B
C 4 B
j
j
And we are abe to determine C3 ,C 4 too.
we want to transform u S
uS
to u B1 we say:
to u B1 ( S ) ( B1 ) , then
B1−1
To make it easy
u S u S is to transfor m ( Su S ) = ( B1u B1 )
then the transition matrix is TS 1 = B1−1
NOTES Basically we have the transition matrix from base
−1
B j to Bk is T jk = Bk B j
(3)
(2)
2 2
1 − 1
3 1
and B2 =
Example 1: Given u = = . Write u in B1 =
1 1 S
1 3
4 2
We have to find C1 , C2 , C3 and C4 such that
2
1 − 1 C1
1 3 C 3
1 = 1 3 C = − 1 5 C
S
2 B1
4 B2
(a)
(b)
(4)
For (a) in (4)
We have
−1
u B1 = B1 Su S
−1
C1
2
1 − 1 C1
1 − 1 2
−1 2
=
=
B
=
1
C
1
1
S 1 3 C 2 B1
S 1 3 1 S
2 B1
C1
1 3 1 2
1 7 7/4
C = 4 − 1 1 1 = 4 − 1 = − 1 / 4
S
2 B1
So ve have
C1 = 7 / 4 and C2 = −1/ 4
For part (b) in (4), we have two options, we either
i)
transform from S B2
ii)
transform from B1 B2
If we choose option i), the transition matrix TS 2 is B2−1 S = B2−1
If we choose option ii), the transition matrix T12 = B2−1 B1
Now: i)
If we choose option i),
The the transition matrix T2 S is B2−1 S = B2−1
TS 2 = B
−1
2
3 1
=
4 2
−1
=
1 2 − 1
2 − 4 3
therefore
C 3
2
3 1 C 3
2
1 2 − 1 2
1 3
1 = 4 2 C C = TS 2 1 = 2 − 4 3 1 = 2 − 5
S
4 B2
S
S
B2
4 B2
ii)
If we choose option ii),
The the transition matrix T12 is B2−1 B1
T12 = B2−1 B1 =
1 2 − 11 − 1 1 1 − 5
=
2 − 4 3 1 3 2 − 1 13
Thus
C 3
1 1 − 5 7 / 4 1 1 − 5 7 1 12 1 3
C = T12u B1 = 2 − 1 13 − 1 / 4 = 8 − 1 13 − 1 = 8 − 20 = 2 − 5 =
B2
4 B2
II)
Dimension:
To make it easy, dimension of a set vectors in R n of a finite dimension spacet is the
numbers of vectors which are linearly independent
Example 2:
a)
2 4 1
1 , 1 2 has Dim =3 ( they are L/I) cgeck it by find the det ≠0
1 2 3
b)
2 4 1
1 , 2 5 has Dim =2, since v2 = 2v1 L / D
1 2 3
c)
2 4 − 2
1 , 2 − 1 has Dim =1, since v2 = 2v1 L / D
1 2 − 1
c)
d)
e)
This Spannig set spnans a line.
So
the basis for example a) is Span{v1 , v2 , v3 }
the basis for example a) is Span{v1 , v3 }
the basis for example a) is Span{v1}
Base on examples above
Let W be subspace of a finite dimensional vector space V, then
i)
W is finite dimensional
ii)
DimW ≤ Dim V
iii)
Dim W = Dim V W = V
Example 2:
a)
2 2 1
3 1 − 1
B1 = 1 , − 1 2 , B2 = 1 , 1 0
1 1 1
− 5 − 3 2
Find the transition matrix from B1 to B2 . As discussed above
5
2
3
2
1 − 1 2 2 1
3
1
= B2−1 B1 = 1
1
0 1 −1 2 = − 2 − 3 −
2
− 5 − 3 2 1 1 1
5
1
6
−1
T1→2
−1
b)
2 2 1 − 5 9
− 5
−1
If W = 8 , then W B1 = B1 W = 1 − 1 2 8 = − 9
1 1 1 − 5 − 5
− 5
c)
Thn we can compute WB1 as
d)
5
7
2
3
−
2 9 2
1
23
WB1 = T1→2W1 = − 2 − 3 − − 9 =
2 2
5
1
6 − 5 6
By direct computing
WB2
IV
1 − 1
3
=B W = 1
1
0
− 5 − 3 2
−1
2
−1
1
− 5
8 = −1
− 5
1
1
2
1
2
2
1
7
−
2 − 5 2
1
23
− 8 =
2 2
1 − 5 6
Row space, Column space and Null space
To be short:
For finte dimentional space. Let Amn be a matrix containing m vectors in R n
1)
The Row Space ( Row ( A) ) is the space spanned by the remaining horizontal vectors
v j after elementary ro elementary operation. We say Row ( A) = Span{v j }
2)
The Column Space ( Col ( A) ) is the space spanned by the remaining Vertical
vector vk after elementary ro elementary operation. We say Col ( A) = Span{vk }
3)
The Null Space ( Null ( A) ) is the solution of AX = 0
For further easy thought
1)
The Row Space ( Row ( A) ) is space of independent horizontal vectors
2)
The Column Space ( Col ( A) ) is the space independent vertical vectors
3)
The Null Space ( Null ( A) ) is the solution of AX = 0 , X R n
Null space of A is spanned by independent vectors in X
Theorem:
The vector b is said to be in the col(A) if there is asolution x such that Ax = b
Or we can say that Ax = b is consistent if b Col ( A)
Example 3:
Assume A be a 4 by 6 matrix. After row elimination, we have
1
0
A is row equivalent to 0
0
0
2 −2 −3
0 1
3
0 0
1
0 0
0
0 0
0
0 1
2 0
2 − 2
0 0
0 0
1 0 0
2 0 0
− 2 1 0
Row ( A) = span , or any 3 original L/I row vectors
− 3 3 1
0 2 2
1 0 − 2
1 − 2 − 3
0 1 3
Col ( A) = span 0 , 0 , 1 any 3 original L/I column vectors
0 0 0
0 0 0
For the Null (A), pay to the attention that this is X R 6 . If we solve the equation
AX = b
The number survival vectors are 3. The solution space is involved 3 parameters
Let
Then
x 2 = s, x5 = t , x 6 = w
x4 = −2 x5 + 2 x6 = −2t + 2w
x3 = −3x4 − 2 x5 = −3(−2t + 2w) − 2t = 4t − 6w
x1 = −2 x2 + 2 x3 + 3x4 − x6 = −2s + 2(4t − 6w) + 3(−2t + 2w) − w = −2s + 2t − 7w
− 2s + 2t − 7 w 0 2
− 7
s
− 2 0
0
4t − 6w 0 4
− 7
= s + t + w or
Null ( A) =
− 2t + 2w 0 − 2
2
0
t
0 1
0 0
1
w
0 2 − 7
− 2 0 0
0 4 − 6
Null ( A) = Span , ,
0 − 2 2
0 1 0
0 0 1
CHECK: All vectors in Null(A) are orthogonal to all rows of equivalence of A or
orthogonal to all the vectors of the original matrix A
V)
The Rank The Nullity
Definition:
1)
The Rank of a matrix A Rank ( A) is the number of vectors in Row(A) or in Col(A)
2)
The Nullity of a matrix A is the number of vectors in Null(A)
For the Example 3 above, the Rank ( A) = 3. This is also the “Dimension of A”
(Some textbook denotes Rank ( A) = RA , The Null Space A = ( A) and Nullity = )
Note that, for matrix Amn
DimA = Rank ( A)
1)
2)
3)
4)
The Row(A) is a subspace in R n
The Col (A) is a subspace in R m
The Null (A) is a subspace in R n
The Rank(A)+ Nullity = n
If W is a subspace in R n then
1)
W ⊥ is a subspace in R n
2)
The only common vector of W and W ⊥ is 0
3)
The orthogonal complement of W ⊥ is W
We also see that
1)
The Null (A) is the orthogonal complement of Row (A)in R n
2)
The Null ( A T ) is the orthogonal complement of Col (A)in R m
Math 410, Linear Algebra
Dr. B Truong
Name: _______________________
LESSONS, CHAPTER 4 CONTINUE
I)
Coordinates and Basis and Change of Basis:
To make it easy, say the vector
a a
u = = is a vector in standard base
b b S
1 0
0 1
a a
1 0 a
Ie = =
b b S 0 1 b
Now, it is the same vector, but in different base, say B1 or in any base B j , then we can
express u as
a a
1 0 a
= =
= B j u j =
b b B j 0 1 b
C1
C 2 B
(1)
j
Now, we have to determine C1 , C2 .
1 0 a
= B j u j =
0 1 b
C1
C
= Bk u k = Bk 3 and so on.....
C 2 B
C 4 B
j
j
And we are abe to determine C3 ,C 4 too.
we want to transform u S
uS
to u B1 we say:
to u B1 ( S ) ( B1 ) , then
B1−1
To make it easy
u S u S is to transfor m ( Su S ) = ( B1u B1 )
then the transition matrix is TS 1 = B1−1
NOTES Basically we have the transition matrix from base
−1
B j to Bk is T jk = Bk B j
(3)
(2)
2 2
1 − 1
3 1
and B2 =
Example 1: Given u = = . Write u in B1 =
1 1 S
1 3
4 2
We have to find C1 , C2 , C3 and C4 such that
2
1 − 1 C1
1 3 C 3
1 = 1 3 C = − 1 5 C
S
2 B1
4 B2
(a)
(b)
(4)
For (a) in (4)
We have
−1
u B1 = B1 Su S
−1
C1
2
1 − 1 C1
1 − 1 2
−1 2
=
=
B
=
1
C
1
1
S 1 3 C 2 B1
S 1 3 1 S
2 B1
C1
1 3 1 2
1 7 7/4
C = 4 − 1 1 1 = 4 − 1 = − 1 / 4
S
2 B1
So ve have
C1 = 7 / 4 and C2 = −1/ 4
For part (b) in (4), we have two options, we either
i)
transform from S B2
ii)
transform from B1 B2
If we choose option i), the transition matrix TS 2 is B2−1 S = B2−1
If we choose option ii), the transition matrix T12 = B2−1 B1
Now: i)
If we choose option i),
The the transition matrix T2 S is B2−1 S = B2−1
TS 2 = B
−1
2
3 1
=
4 2
−1
=
1 2 − 1
2 − 4 3
therefore
C 3
2
3 1 C 3
2
1 2 − 1 2
1 3
1 = 4 2 C C = TS 2 1 = 2 − 4 3 1 = 2 − 5
S
4 B2
S
S
B2
4 B2
ii)
If we choose option ii),
The the transition matrix T12 is B2−1 B1
T12 = B2−1 B1 =
1 2 − 11 − 1 1 1 − 5
=
2 − 4 3 1 3 2 − 1 13
Thus
C 3
1 1 − 5 7 / 4 1 1 − 5 7 1 12 1 3
C = T12u B1 = 2 − 1 13 − 1 / 4 = 8 − 1 13 − 1 = 8 − 20 = 2 − 5 =
B2
4 B2
II)
Dimension:
To make it easy, dimension of a set vectors in R n of a finite dimension spacet is the
numbers of vectors which are linearly independent
Example 2:
a)
2 4 1
1 , 1 2 has Dim =3 ( they are L/I) cgeck it by find the det ≠0
1 2 3
b)
2 4 1
1 , 2 5 has Dim =2, since v2 = 2v1 L / D
1 2 3
c)
2 4 − 2
1 , 2 − 1 has Dim =1, since v2 = 2v1 L / D
1 2 − 1
c)
d)
e)
This Spannig set spnans a line.
So
the basis for example a) is Span{v1 , v2 , v3 }
the basis for example a) is Span{v1 , v3 }
the basis for example a) is Span{v1}
Base on examples above
Let W be subspace of a finite dimensional vector space V, then
i)
W is finite dimensional
ii)
DimW ≤ Dim V
iii)
Dim W = Dim V W = V
Example 2:
a)
2 2 1
3 1 − 1
B1 = 1 , − 1 2 , B2 = 1 , 1 0
1 1 1
− 5 − 3 2
Find the transition matrix from B1 to B2 . As discussed above
5
2
3
2
1 − 1 2 2 1
3
1
= B2−1 B1 = 1
1
0 1 −1 2 = − 2 − 3 −
2
− 5 − 3 2 1 1 1
5
1
6
−1
T1→2
−1
b)
2 2 1 − 5 9
− 5
−1
If W = 8 , then W B1 = B1 W = 1 − 1 2 8 = − 9
1 1 1 − 5 − 5
− 5
c)
Thn we can compute WB1 as
d)
5
7
2
3
−
2 9 2
1
23
WB1 = T1→2W1 = − 2 − 3 − − 9 =
2 2
5
1
6 − 5 6
By direct computing
WB2
IV
1 − 1
3
=B W = 1
1
0
− 5 − 3 2
−1
2
−1
1
− 5
8 = −1
− 5
1
1
2
1
2
2
1
7
−
2 − 5 2
1
23
− 8 =
2 2
1 − 5 6
Row space, Column space and Null space
To be short:
For finte dimentional space. Let Amn be a matrix containing m vectors in R n
1)
The Row Space ( Row ( A) ) is the space spanned by the remaining horizontal vectors
v j after elementary ro elementary operation. We say Row ( A) = Span{v j }
2)
The Column Space ( Col ( A) ) is the space spanned by the remaining Vertical
vector vk after elementary ro elementary operation. We say Col ( A) = Span{vk }
3)
The Null Space ( Null ( A) ) is the solution of AX = 0
For further easy thought
1)
The Row Space ( Row ( A) ) is space of independent horizontal vectors
2)
The Column Space ( Col ( A) ) is the space independent vertical vectors
3)
The Null Space ( Null ( A) ) is the solution of AX = 0 , X R n
Null space of A is spanned by independent vectors in X
Theorem:
The vector b is said to be in the col(A) if there is asolution x such that Ax = b
Or we can say that Ax = b is consistent if b Col ( A)
Example 3:
Assume A be a 4 by 6 matrix. After row elimination, we have
1
0
A is row equivalent to 0
0
0
2 −2 −3
0 1
3
0 0
1
0 0
0
0 0
0
0 1
2 0
2 − 2
0 0
0 0
1 0 0
2 0 0
− 2 1 0
Row ( A) = span , or any 3 original L/I row vectors
− 3 3 1
0 2 2
1 0 − 2
1 − 2 − 3
0 1 3
Col ( A) = span 0 , 0 , 1 any 3 original L/I column vectors
0 0 0
0 0 0
For the Null (A), pay to the attention that this is X R 6 . If we solve the equation
AX = b
The number survival vectors are 3. The solution space is involved 3 parameters
Let
Then
x 2 = s, x5 = t , x 6 = w
x4 = −2 x5 + 2 x6 = −2t + 2w
x3 = −3x4 − 2 x5 = −3(−2t + 2w) − 2t = 4t − 6w
x1 = −2 x2 + 2 x3 + 3x4 − x6 = −2s + 2(4t − 6w) + 3(−2t + 2w) − w = −2s + 2t − 7w
− 2s + 2t − 7 w 0 2
− 7
s
− 2 0
0
4t − 6w 0 4
− 7
= s + t + w or
Null ( A) =
− 2t + 2w 0 − 2
2
0
t
0 1
0 0
1
w
0 2 − 7
− 2 0 0
0 4 − 6
Null ( A) = Span , ,
0 − 2 2
0 1 0
0 0 1
CHECK: All vectors in Null(A) are orthogonal to all rows of equivalence of A or
orthogonal to all the vectors of the original matrix A
V)
The Rank The Nullity
Definition:
1)
The Rank of a matrix A Rank ( A) is the number of vectors in Row(A) or in Col(A)
2)
The Nullity of a matrix A is the number of vectors in Null(A)
For the Example 3 above, the Rank ( A) = 3. This is also the “Dimension of A”
(Some textbook denotes Rank ( A) = RA , The Null Space A = ( A) and Nullity = )
Note that, for matrix Amn
DimA = Rank ( A)
1)
2)
3)
4)
The Row(A) is a subspace in R n
The Col (A) is a subspace in R m
The Null (A) is a subspace in R n
The Rank(A)+ Nullity = n
If W is a subspace in R n then
1)
W ⊥ is a subspace in R n
2)
The only common vector of W and W ⊥ is 0
3)
The orthogonal complement of W ⊥ is W
We also see that
1)
The Null (A) is the orthogonal complement of Row (A)in R n
2)
The Null ( A T ) is the orthogonal complement of Col (A)in R m
Math 410, Linear Algebra
Dr. B Truong
Name: _______________________
LESSONS, CHAPTER 4
I)
Real Vector Space (VS) : V
In this section , we pay attention the the 10 axioms for VS.
Let V be a non-empty set of objects. Let u and v a pair of object in V and a scaler k
Pay very attention to the axioms
(1)
closure under addition (u + v) V
(6)
closure under scalar multiplication kuV
Examples of VS:
a)
Zero VS
b)
VS of R n
c)
VS{sequence of real numbers}
d)
VS of 2 2 matrices ( and m n matrices
e)
VS of {real valued functions}
Examples of collectios which are VS:
a)
V = R 2 ; u = (u1 , u 2 ) and ku = (u1 ,0)
b)
V = {set of real vectors} ; u + v = uv and ku = u k
1I)
Subspace (SS) W of a VS V: W is non-empty set W V and
We need to show a SS by satisfying 2 conditions
(u + v) W
( ku) W
a)
b)
Examples: SS
a)
b)
c)
d)
e)
f)
g)
The Zero SS
Lines through origine in R 2 and R 3
SS of matrices M nn
Set { continuous functions in (-∞,∞)}
{Continuous derivatives functions}
SS of {all polynomials}
SS of {all polynomials of degrees ≤ n}
Examples: Not a SS
a)
b)
c)
d)
{Positive real functions}
{Set vectors (a,b) such that a,b > 0 }
{Invertible matrices M nn }
{Set of all polynomials of degrees = n}
Subspace of a homogeneous linear system AX = 0
with m equations and n unknowns is a SS of R n
Here A is a m n matrix and X is a m 1 vector
III)
Spanning sets:
Let V be a VS
(You will be familiarize the following notations:
v j and w V and k j = constants, j = 1,2....r
w = k j v j j = 1,2....r
We say
j=1
Is the linear combination with coefficients
Further more, if
S = (w1 , w2 ,..., wr ) V
Then the set W of all possible linear combination vectors in V is a SP of V
We say the SS W of V spanned by S
Thus,
W = span{w1 , w2 ,..., wr } or W = span{ S}
Example 1:
1)
Standard unit vectors in R n
e1 = (1,0....), e2 = (0,1,....),.....en = (0,0,...n) span
2)
3)
Rn
The vector v = k ( a, b) , k = scalar, span the line passing through (0.0)
The vectors u = (a, b), v = (c, d ) span the plane containing u and v
I.e: The plane = span u, v
4)
The polynomials 1, x, x 2 ....x n span the VS Pn
Say :
5)
6)
Example 2:
Pn = span 1, x, x 2 ....x n
Vectors u, v and w in R 2 we can find scalars C1 ,C 2 such that
The linear combination satisfied:
w = C1u + C2 v as long as u is not in span v ( or reverse)
Vectors u, v and w in R 3 some time, we can’t find scalars C1 , C2 such that
The linear combination satisfied:
w = C1u + C2 v since w is not in spanu, v ( or reverse)
Express w = (6,11,6) as the linear combination of
u1 = (2,1,4) , u2 = (1,−1,3) , u 3 = (3,2,5)
We set to solve the system:
2 1 4 c1 6
2 1 4
c1
6
1 − 1 3 c2 = 11 AC = b . Where A = 1 − 1 3 ; C = c2 and b = 11
3 2 5 c 6
3 2 5
c
6
3
3
1
We have C = A −1b = adj( A) .
A
7
− 11 3
We find that the det(A) = 2 and adj( A) = 4 − 2 − 2
5
− 1 − 3
7 6
− 11 3
9 9 / 2
1
1
1
C = A −1b = adj( A) b C = 4 − 2 − 2 11 = − 10 = − 5
A
2
2
− 1 − 3 6
5
1 1/ 2
That means :
6
2 1
3
9 1
11 = 1 − 5 − 1 + 2
6 2 4 3 2 5
1 12
as a linear combination of
Express C =
8 − 2
1 4
2 0
; B =
A =
3 0
1 2
Example3:
We have to find x, y such that
1 4
2 0 1 12
+ y
=
x
3 0
1 2 8 − 2
Or solve ( by equating all elements of both sides)
x + 2 y = 1 (1)
4 x + 0 y = 12 (2)
3x + y = 8 (3)
0 x + 2 y = −2 (4)
x = 3; y = −1
The equation (2) and (4) give
These values also satisfy the eq (1) and (3). Thus we have
1 12 1 4 2 0
= 3
− 1
C = 3 A − B or
8 − 2 3 0 1 2
IV)
Linear Independent:
r
Let v j 1 V a VS ; j = 1,2....r
Form a vector equations for a linear combination as follows
c v
j
j
=0
j
Conclusion:
v j 1r is said to be Linearly Independent if all
a)
cj = 0 j
stands for for all
b)
Otherwise, if ck 0 for some k the set v j 1 is Linear Dependent
r
Note:
r
a)
The set v j 1 in which vk = 0 for some k , is Linear Dependeny (L/D)
b)
In R n , any 2 vectors u and v such that u kv are Linearly Independent (L/I)
To show the v j 1 in R n or M ij or set of polynomials Pn always solve the equation
r
c v
j
j
=0
j
c M
j
j
= 0 or
j
c
j
fj =0
j
Usefull facts:
The unit vector in Pn is (1,0,0....0), (0,1,0.....0),....(0,0,0,0,0..1)
So the function f ( x) = 2 + 2 x + x 3 can be written as
x0
1
x
f ( x) = (2 2 0 3) 2 = 2 + 2 x + 0 x 2 + 3x 3
x
x3
So the vector (2 2 0 3) represents the function f ( x) = 2 + 2 x + x 4
Definition:
If the set of function f j 1 ( x) is continuous up to (n-1) th derivative
Then the determinamt
n
f1
f1 '
•
•
•
W ( f1 , f 2 ,...., f n ) =
f1
( n −1)
f2
f2 '
•
•
•
f2
( n −1)
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
fn
•
•
•
•
fn
( n −1)
is called the Wronskian of f1 , f 2 ,...., f n
If W ( f1 , f 2 ,...., f n ) 0 on (−,) then f j 1 ( x) are L/I
n
Conclusion: a)
Example 4:
b)
If W ( f1 , f 2 ,...., f n ) = 0 on (−,) then f j 1 ( x) are L/D
a)
Whether sin x and cos x are L/I?
n
W ( x, cos( x)) =
sin x cos x
=1
cos x − sin x
L/I on on (−,)
b)
Whether functions
f1 ( x) = −1 + 2 x − x 2 f 2 ( x) = x + x 2 f 3 ( x) = 2 + 2 x 2 are L/I
− 1 + 2x − x 2
W ( f1 , f 2 , f 3 ) =
2 − 2x
−2
x + x2
1 + 2x
2
2 + 2x 2
4x
4
(− 1 + 2x − x )(4(1 + 2 x) − 2(4x)) − ( x + x )(4(2 − 2x) + 8x)
=
2
2
+ (2 + 2 x 2 )(2(2 − 2 x) + 2(1 + 2 x) )
=8
So we conclude that the set f j 1 ( x) are linearly independent
3
Another way to do this problem is now, we consider these functions in P2 as vectors
And find the Wronskian accordingly
− 1
0
2
2
2
f1 ( x) = −1 + 2 x − x = 2 f 2 ( x) = x + x = 1 , f 3 ( x) = 2 + 2 x = 0
− 1
1
2
−1 0 2
W (v1 , v2 , v3 ) = 2 1 0 = (−2 + 0 + 4) − (−2 + 0 + 0) = 4 0 L / I
−1 1 2
2
Note that the Wronskian are not the same.
Math 410, Linear Algebra
Dr. B Truong
Name: _______________________
Distance
The distance problem is general can be found without using formula. Because, the formula is
derived the similar way. Similar problems are treated the same way. In 2 dimension, we can use
a cross product or the dot product ( projection)
Example 1
Find the distance from the point (1,3,6) to the line
P(1,3,6)
x − 2 y +1 z −1
=
=
=t
2
2
5
D
H
u
α
Q(2,−1,1)
(L)
v = (2,2,5)
n
In 3-D the point H of intersection of the line passing P and orthogonal to the line L is
hard to find ( if we can say that it is hard).
Now assume the point H is found.
Let u = PQ = (−1,4,1) and α is the angle between u an v
We want to find the distance from P to H: D = PH
The vector parallel to line L is v = (2,2,5) and we choose a point on it say Q(2,−1,1)
Use a cross product u v = u v sin . So that sin =
uv
u v
We also have
sin =
Thus
D
u
D uv
=
u
u v
sin =
D=
uv
v
Now
i
j k
u v = − 1 4 1 = (20 − 2,2 + 5,−2 − 8) = (18,7,−10)
2 2 5
D=
uv
v
=
(18,7,−10)
=
(2,2,5)
473
33
473
33
=
*** In 2-D H can be found and D is the projection of the vector u onto n
Example 2
Find the distance from P (5,1) to the line y − 2 x = −2
The line in parametric form is
x −1
y = 2x − 2 =
=t
1
2
The vector v // to the line is v = (1,2) . The vector n is orthogonal to v is
n = (−2,1)
Choose a random point Q on the line i,e P (5,1)
P(5,1)
D
(L) v = (1,2)
u
Q(0,−2)
H
n = (−2,1)
u = PQ = (5,3)
We have
D = Proj n u =
In the example 1: D =
un
n
uv
v
=
=
(5,3) • (−2,1)
(−2,1)
(5,3) (1,2)
(1,2)
=
=
−7
5
=
7 5
5
(0,0,10 − 3)
5
=
7
5
=
7 5
5
Math 410, Linear Algebra
Dr. B Truong
Name: _______________________
Distance
The distance problem is general can be found without using formula. Because, the formula is
derived the similar way. Similar problems are treated the same way. In 2 dimension, we can use
a cross product or the dot product ( projection)
Example 3:
Find the distance from P(4,1,−2) to the plane 3x − 2 y + 5 z = 6
P
u
n
QQ
Q
The distance from P to the plane can be understood as the length of the projection vector
of vector u ( connecting P to ANY point Q on the plane) onto the normal vector n of the plane
n = (2,−3,5)
We can choose any point on the plane, (satisfying the plane equation) i,e Q = (1,1,1)
and v = (2,−3,5)
So that u = PQ = (3,0,−3)
Here
Proj n u =
un
un
Dis tan ce = Proj n u =
=
n
n
6 − 15
4 + 9 + 25
=
9
38
Math 410:
Chapter 3 and 4
Use the Dot product and the cross product to find the distance from a point to a line (l)
Math 410, Linear Algebra
Dr. B Truong
Name: _______________________
LESSONS, CHAPTER 4 CONTINUE
I)
Coordinates and Basis and Change of Basis:
To make it easy, say the vector
a a
u = = is a vector in standard base
b b S
1 0
0 1
a a
1 0 a
Ie = =
b b S 0 1 b
Now, it is the same vector, but in different base, say B1 or in any base B j , then we can
express u as
a a
1 0 a
= =
= B j u j =
b b B j 0 1 b
C1
C 2 B
(1)
j
Now, we have to determine C1 , C2 .
1 0 a
= B j u j =
0 1 b
C1
C
= Bk u k = Bk 3 and so on.....
C 2 B
C 4 B
j
j
And we are abe to determine C3 ,C 4 too.
we want to transform u S
uS
to u B1 we say:
to u B1 ( S ) ( B1 ) , then
B1−1
To make it easy
u S u S is to transfor m ( Su S ) = ( B1u B1 )
then the transition matrix is TS 1 = B1−1
NOTES Basically we have the transition matrix from base
−1
B j to Bk is T jk = Bk B j
(3)
(2)
2 2
1 − 1
3 1
and B2 =
Example 1: Given u = = . Write u in B1 =
1 1 S
1 3
4 2
We have to find C1 , C2 , C3 and C4 such that
2
1 − 1 C1
1 3 C 3
1 = 1 3 C = − 1 5 C
S
2 B1
4 B2
(a)
(b)
(4)
For (a) in (4)
We have
−1
u B1 = B1 Su S
−1
C1
2
1 − 1 C1
1 − 1 2
−1 2
=
=
B
=
1
C
1
1
S 1 3 C 2 B1
S 1 3 1 S
2 B1
C1
1 3 1 2
1 7 7/4
C = 4 − 1 1 1 = 4 − 1 = − 1 / 4
S
2 B1
So ve have
C1 = 7 / 4 and C2 = −1/ 4
For part (b) in (4), we have two options, we either
i)
transform from S B2
ii)
transform from B1 B2
If we choose option i), the transition matrix TS 2 is B2−1 S = B2−1
If we choose option ii), the transition matrix T12 = B2−1 B1
Now: i)
If we choose option i),
The the transition matrix T2 S is B2−1 S = B2−1
TS 2 = B
−1
2
3 1
=
4 2
−1
=
1 2 − 1
2 − 4 3
therefore
C 3
2
3 1 C 3
2
1 2 − 1 2
1 3
1 = 4 2 C C = TS 2 1 = 2 − 4 3 1 = 2 − 5
S
4 B2
S
S
B2
4 B2
ii)
If we choose option ii),
The the transition matrix T12 is B2−1 B1
T12 = B2−1 B1 =
1 2 − 11 − 1 1 1 − 5
=
2 − 4 3 1 3 2 − 1 13
Thus
C 3
1 1 − 5 7 / 4 1 1 − 5 7 1 12 1 3
C = T12u B1 = 2 − 1 13 − 1 / 4 = 8 − 1 13 − 1 = 8 − 20 = 2 − 5 =
B2
4 B2
II)
Dimension:
To make it easy, dimension of a set vectors in R n of a finite dimension spacet is the
numbers of vectors which are linearly independent
Example 2:
a)
2 4 1
1 , 1 2 has Dim =3 ( they are L/I) cgeck it by find the det ≠0
1 2 3
b)
2 4 1
1 , 2 5 has Dim =2, since v2 = 2v1 L / D
1 2 3
c)
2 4 − 2
1 , 2 − 1 has Dim =1, since v2 = 2v1 L / D
1 2 − 1
c)
d)
e)
This Spannig set spnans a line.
So
the basis for example a) is Span{v1 , v2 , v3 }
the basis for example a) is Span{v1 , v3 }
the basis for example a) is Span{v1}
Base on examples above
Let W be subspace of a finite dimensional vector space V, then
i)
W is finite dimensional
ii)
DimW ≤ Dim V
iii)
Dim W = Dim V W = V
Example 2:
a)
2 2 1
3 1 − 1
B1 = 1 , − 1 2 , B2 = 1 , 1 0
1 1 1
− 5 − 3 2
Find the transition matrix from B1 to B2 . As discussed above
5
2
3
2
1 − 1 2 2 1
3
1
= B2−1 B1 = 1
1
0 1 −1 2 = − 2 − 3 −
2
− 5 − 3 2 1 1 1
5
1
6
−1
T1→2
−1
b)
2 2 1 − 5 9
− 5
−1
If W = 8 , then W B1 = B1 W = 1 − 1 2 8 = − 9
1 1 1 − 5 − 5
− 5
c)
Thn we can compute WB1 as
d)
5
7
2
3
−
2 9 2
1
23
WB1 = T1→2W1 = − 2 − 3 − − 9 =
2 2
5
1
6 − 5 6
By direct computing
WB2
IV
1 − 1
3
=B W = 1
1
0
− 5 − 3 2
−1
2
−1
1
− 5
8 = −1
− 5
1
1
2
1
2
2
1
7
−
2 − 5 2
1
23
− 8 =
2 2
1 − 5 6
Row space, Column space and Null space
To be short:
For finte dimentional space. Let Amn be a matrix containing m vectors in R n
1)
The Row Space ( Row ( A) ) is the space spanned by the remaining horizontal vectors
v j after elementary ro elementary operation. We say Row ( A) = Span{v j }
2)
The Column Space ( Col ( A) ) is the space spanned by the remaining Vertical
vector vk after elementary ro elementary operation. We say Col ( A) = Span{vk }
3)
The Null Space ( Null ( A) ) is the solution of AX = 0
For further easy thought
1)
The Row Space ( Row ( A) ) is space of independent horizontal vectors
2)
The Column Space ( Col ( A) ) is the space independent vertical vectors
3)
The Null Space ( Null ( A) ) is the solution of AX = 0 , X R n
Null space of A is spanned by independent vectors in X
Theorem:
The vector b is said to be in the col(A) if there is asolution x such that Ax = b
Or we can say that Ax = b is consistent if b Col ( A)
Example 3:
Assume A be a 4 by 6 matrix. After row elimination, we have
1
0
A is row equivalent to 0
0
0
2 −2 −3
0 1
3
0 0
1
0 0
0
0 0
0
0 1
2 0
2 − 2
0 0
0 0
1 0 0
2 0 0
− 2 1 0
Row ( A) = span , or any 3 original L/I row vectors
− 3 3 1
0 2 2
1 0 − 2
1 − 2 − 3
0 1 3
Col ( A) = span 0 , 0 , 1 any 3 original L/I column vectors
0 0 0
0 0 0
For the Null (A), pay to the attention that this is X R 6 . If we solve the equation
AX = b
The number survival vectors are 3. The solution space is involved 3 parameters
Let
Then
x 2 = s, x5 = t , x 6 = w
x4 = −2 x5 + 2 x6 = −2t + 2w
x3 = −3x4 − 2 x5 = −3(−2t + 2w) − 2t = 4t − 6w
x1 = −2 x2 + 2 x3 + 3x4 − x6 = −2s + 2(4t − 6w) + 3(−2t + 2w) − w = −2s + 2t − 7w
− 2s + 2t − 7 w 0 2
− 7
s
− 2 0
0
4t − 6w 0 4
− 7
= s + t + w or
Null ( A) =
− 2t + 2w 0 − 2
2
0
t
0 1
0 0
1
w
0 2 − 7
− 2 0 0
0 4 − 6
Null ( A) = Span , ,
0 − 2 2
0 1 0
0 0 1
CHECK: All vectors in Null(A) are orthogonal to all rows of equivalence of A or
orthogonal to all the vectors of the original matrix A
V)
The Rank The Nullity
Definition:
1)
The Rank of a matrix A Rank ( A) is the number of vectors in Row(A) or in Col(A)
2)
The Nullity of a matrix A is the number of vectors in Null(A)
For the Example 3 above, the Rank ( A) = 3. This is also the “Dimension of A”
(Some textbook denotes Rank ( A) = RA , The Null Space A = ( A) and Nullity = )
Note that, for matrix Amn
DimA = Rank ( A)
1)
2)
3)
4)
The Row(A) is a subspace in R n
The Col (A) is a subspace in R m
The Null (A) is a subspace in R n
The Rank(A)+ Nullity = n
If W is a subspace in R n then
1)
W ⊥ is a subspace in R n
2)
The only common vector of W and W ⊥ is 0
3)
The orthogonal complement of W ⊥ is W
We also see that
1)
The Null (A) is the orthogonal complement of Row (A)in R n
2)
The Null ( A T ) is the orthogonal complement of Col (A)in R m
Math 410, Linear Algebra
Dr. B Truong
Name: _______________________
LESSONS, CHAPTER 4
I)
Real Vector Space (VS) : V
In this section , we pay attention the the 10 axioms for VS.
Let V be a non-empty set of objects. Let u and v a pair of object in V and a scaler k
Pay very attention to the axioms
(1)
closure under addition (u + v) V
(6)
closure under scalar multiplication kuV
Examples of VS:
a)
Zero VS
b)
VS of R n
c)
VS{sequence of real numbers}
d)
VS of 2 2 matrices ( and m n matrices
e)
VS of {real valued functions}
Examples of collectios which are VS:
a)
V = R 2 ; u = (u1 , u 2 ) and ku = (u1 ,0)
b)
V = {set of real vectors} ; u + v = uv and ku = u k
1I)
Subspace (SS) W of a VS V: W is non-empty set W V and
We need to show a SS by satisfying 2 conditions
(u + v) W
( ku) W
a)
b)
Examples: SS
a)
b)
c)
d)
e)
f)
g)
The Zero SS
Lines through origine in R 2 and R 3
SS of matrices M nn
Set { continuous functions in (-∞,∞)}
{Continuous derivatives functions}
SS of {all polynomials}
SS of {all polynomials of degrees ≤ n}
Examples: Not a SS
a)
b)
c)
d)
{Positive real functions}
{Set vectors (a,b) such that a,b > 0 }
{Invertible matrices M nn }
{Set of all polynomials of degrees = n}
Subspace of a homogeneous linear system AX = 0
with m equations and n unknowns is a SS of R n
Here A is a m n matrix and X is a m 1 vector
III)
Spanning sets:
Let V be a VS
(You will be familiarize the following notations:
v j and w V and k j = constants, j = 1,2....r
w = k j v j j = 1,2....r
We say
j=1
Is the linear combination with coefficients
Further more, if
S = (w1 , w2 ,..., wr ) V
Then the set W of all possible linear combination vectors in V is a SP of V
We say the SS W of V spanned by S
Thus,
W = span{w1 , w2 ,..., wr } or W = span{ S}
Example 1:
1)
Standard unit vectors in R n
e1 = (1,0....), e2 = (0,1,....),.....en = (0,0,...n) span
2)
3)
Rn
The vector v = k ( a, b) , k = scalar, span the line passing through (0.0)
The vectors u = (a, b), v = (c, d ) span the plane containing u and v
I.e: The plane = span u, v
4)
The polynomials 1, x, x 2 ....x n span the VS Pn
Say :
5)
6)
Example 2:
Pn = span 1, x, x 2 ....x n
Vectors u, v and w in R 2 we can find scalars C1 ,C 2 such that
The linear combination satisfied:
w = C1u + C2 v as long as u is not in span v ( or reverse)
Vectors u, v and w in R 3 some time, we can’t find scalars C1 , C2 such that
The linear combination satisfied:
w = C1u + C2 v since w is not in spanu, v ( or reverse)
Express w = (6,11,6) as the linear combination of
u1 = (2,1,4) , u2 = (1,−1,3) , u 3 = (3,2,5)
We set to solve the system:
2 1 4 c1 6
2 1 4
c1
6
1 − 1 3 c2 = 11 AC = b . Where A = 1 − 1 3 ; C = c2 and b = 11
3 2 5 c 6
3 2 5
c
6
3
3
1
We have C = A −1b = adj( A) .
A
7
− 11 3
We find that the det(A) = 2 and adj( A) = 4 − 2 − 2
5
− 1 − 3
7 6
− 11 3
9 9 / 2
1
1
1
C = A −1b = adj( A) b C = 4 − 2 − 2 11 = − 10 = − 5
A
2
2
− 1 − 3 6
5
1 1/ 2
That means :
6
2 1
3
9 1
11 = 1 − 5 − 1 + 2
6 2 4 3 2 5
1 12
as a linear combination of
Express C =
8 − 2
1 4
2 0
; B =
A =
3 0
1 2
Example3:
We have to find x, y such that
1 4
2 0 1 12
+ y
=
x
3 0
1 2 8 − 2
Or solve ( by equating all elements of both sides)
x + 2 y = 1 (1)
4 x + 0 y = 12 (2)
3x + y = 8 (3)
0 x + 2 y = −2 (4)
x = 3; y = −1
The equation (2) and (4) give
These values also satisfy the eq (1) and (3). Thus we have
1 12 1 4 2 0
= 3
− 1
C = 3 A − B or
8 − 2 3 0 1 2
IV)
Linear Independent:
r
Let v j 1 V a VS ; j = 1,2....r
Form a vector equations for a linear combination as follows
c v
j
j
=0
j
Conclusion:
v j 1r is said to be Linearly Independent if all
a)
cj = 0 j
stands for for all
b)
Otherwise, if ck 0 for some k the set v j 1 is Linear Dependent
r
Note:
r
a)
The set v j 1 in which vk = 0 for some k , is Linear Dependeny (L/D)
b)
In R n , any 2 vectors u and v such that u kv are Linearly Independent (L/I)
To show the v j 1 in R n or M ij or set of polynomials Pn always solve the equation
r
c v
j
j
=0
j
c M
j
j
= 0 or
j
c
j
fj =0
j
Usefull facts:
The unit vector in Pn is (1,0,0....0), (0,1,0.....0),....(0,0,0,0,0..1)
So the function f ( x) = 2 + 2 x + x 3 can be written as
x0
1
x
f ( x) = (2 2 0 3) 2 = 2 + 2 x + 0 x 2 + 3x 3
x
x3
So the vector (2 2 0 3) represents the function f ( x) = 2 + 2 x + x 4
Definition:
If the set of function f j 1 ( x) is continuous up to (n-1) th derivative
Then the determinamt
n
f1
f1 '
•
•
•
W ( f1 , f 2 ,...., f n ) =
f1
( n −1)
f2
f2 '
•
•
•
f2
( n −1)
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
fn
•
•
•
•
fn
( n −1)
is called the Wronskian of f1 , f 2 ,...., f n
If W ( f1 , f 2 ,...., f n ) 0 on (−,) then f j 1 ( x) are L/I
n
Conclusion: a)
Example 4:
b)
If W ( f1 , f 2 ,...., f n ) = 0 on (−,) then f j 1 ( x) are L/D
a)
Whether sin x and cos x are L/I?
n
W ( x, cos( x)) =
sin x cos x
=1
cos x − sin x
L/I on on (−,)
b)
Whether functions
f1 ( x) = −1 + 2 x − x 2 f 2 ( x) = x + x 2 f 3 ( x) = 2 + 2 x 2 are L/I
− 1 + 2x − x 2
W ( f1 , f 2 , f 3 ) =
2 − 2x
−2
x + x2
1 + 2x
2
2 + 2x 2
4x
4
(− 1 + 2x − x )(4(1 + 2 x) − 2(4x)) − ( x + x )(4(2 − 2x) + 8x)
=
2
2
+ (2 + 2 x 2 )(2(2 − 2 x) + 2(1 + 2 x) )
=8
So we conclude that the set f j 1 ( x) are linearly independent
3
Another way to do this problem is now, we consider these functions in P2 as vectors
And find the Wronskian accordingly
− 1
0
2
2
2
f1 ( x) = −1 + 2 x − x = 2 f 2 ( x) = x + x = 1 , f 3 ( x) = 2 + 2 x = 0
− 1
1
2
−1 0 2
W (v1 , v2 , v3 ) = 2 1 0 = (−2 + 0 + 4) − (−2 + 0 + 0) = 4 0 L / I
−1 1 2
2
Note that the Wronskian are not the same.
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