Linear Algebra (Part - 2) - Mathematics, Engineering Video Lecture - Engineering Mathematics

welcome viewers this is second lecture
on linear algebra in my first lecture we
have given the definition of a real
vector space to review it I first again
give you the definition of a real vector
space a triplet me a set and two
operations plus and star forms a vector
space provided certain axioms are
satisfied the first section is that the
operation plus and star we call it
scalar opera scalar multiplication are
closed in V the addition is commutative
and associative further the existence of
additive identity is there in the set V
and existence of additive inverse is
also there in the set V as far as the
scalar multiplication star is concerned
satisfied distributive properties with
respect to vector addition and star is
also associated and there exists a
scalar one if all these properties are
satisfied then V Plus star is a real
vector space we have discussed the
amount of examples in my earlier lecture
I will continue with few more examples
to fix up the ideas the example I take
now is from the set of inner degree
polynomials PN X of the form PN X is
equal to a naught plus a 1 X plus a 2 X
square plus a n xn where the X lies in
given interval I and all the
coefficients a naught a 1 a 2 a n
etcetera are real then this set P forms
a vector space with polynomial addition
and scalar multiplication which is
defined in this
Manor if I have two polynomials PN X and
QN X where the polynomial PNX is a
naught plus a 1 X plus n X is 2 power N
and another polynomial PNX of the form b
naught plus b 1 x plus BN x is 2 power n
then the addition of two polynomial is
defined as a naught plus B naught plus a
1 plus B 1 X and so on and the last term
is n plus BN x raised to power in the
idea is the coefficients are added a
naught B naught a 1 B plus B 1 is the
coefficient of x n plus BN is a
coefficient of X n now this is called
polynomial addition by the scalar
multiplication by a scalar C is defined
as C PN X is equal to C times the
polynomial and if you simplify it is C a
naught plus C a 1 X plus C n X n now
with these two operations we say that
this set forms a vector space for
proving this first thing we have to do
is we have to say that P is closed with
respect to two operations the addition
and scalar multiplication now by this I
mean to say that if I have a polynomial
PNX in P and a polynomial QN X in P then
their sum is also a polynomial in P so
if you look at this definition one can
easily observe that when we add the two
the right hand side is also polynomial
of degree n with real coefficients a
naught plus B naught a 1 plus B 1 etc so
this operation is closed with respect to
the set P is closed with respect to
addition now as far as scalar
multiplication is concerned the same is
true look at this this
see PNX is also polynomial of degree in
the only differences all the
coefficients are multiplied by C but
they still remain real that is why the
CP and X is a polynomial of degree N or
we say that P is closed with respect to
the two operations the next property is
that this addition is commutative as
well as associated now this can be
easily observed because P annexed
rescuin X if this is so then the
coefficients of Q and X plus P and X
will be V naught a naught b1 plus a1 BN
plus n and so on and this simply means
these are real coefficients so they are
committed so PN X plus QN X will be the
same as Q and X plus PN X similarly one
can easily establish that this operation
is associative as far as the additive
identity in the set is concerned one can
easily see that polynomial with zero
coefficients is the additive identity
and this belongs to the set of
polynomials
similarly the additive inverse of PN X
is minus PN X and that means the
additive inverse also exists in the set
P as far the distributive properties
they are concerned they can also be
easily checked because all the
coefficients are real numbers and they
satisfy and distributive properties so
these properties are satisfied then
comes the existence of one such that one
x pn x is pn x the effect of this scalar
multiplication with pn x is that all the
coefficients are multiplied by 1 and
multiplying a real number by one will
not change the coefficient and that is y
+ 1 x pn x is pn x and this means that
the set of polynomials P is a vector
space in the next example
we consider set queue of all
second-degree polynomials of the form P
to X is equal to a naught plus a 1 X
plus a 2 X square such that and the
coefficients are related by the relation
a 1 is equal to a naught plus 1 so it's
not all polynomials but only those
polynomials which satisfy this relation
then this set Q is not a vector space
with respect to operation of polynomials
of addition and scalar multiplication
defined as P 2 X plus Q 2 X is equal to
a naught plus a 1 X plus a 2 X square
this is P 2 X plus V naught plus B 1 X
plus B 2 X square is Q 2 X it is defined
as a naught plus B naught plus a 1 plus
B 1 X plus a 2 plus B 2 X square that is
the coefficients are added together the
scalar multiplication C times P 2 X is
defined as C a naught plus a 1 X plus a
2 X square which is ca naught plus C 1 C
times a 1 X plus C a 2 X square
now one may notice that this Q will not
form a vector space as these two
operations are not closed but they say I
mean to say that if I add the two
polynomials P 2 X and Q 2 X belonging to
the set Q then the resulting polynomial
is a naught plus B naught plus a 1 plus
B 1 X plus a 2 plus B 2 X square
according to the definition which is
given to us a 1 is equal to a naught
plus 1 because P P 2 X belongs to the
set Q similarly B 1 is equal to B naught
plus 1 Q 2 belongs to Q and from this
one can see that a 1 plus B 1 is equal
to a naught plus B naught plus 2 but if
you apply this here then this property
requires this as
to be in the set q requires that a 1
plus B 1 is equal to a naught plus B
naught plus 1 instead of 2 so this
polynomial will not belong to the set Q
or we say this operation is not closed
in Q similarly C P to X is equal to C
naught plus a 1 X plus a 2 X square if
we take you to X to be 0 then C a naught
plus C a 1 X plus C a 2 X square and
this means C a 1 should be equal to ca
naught plus 1 but this is not true
because a 1 is equal to a naught plus 1
being cute P 2 X belonging to Q so this
property is also not satisfied so this
or is also not closed in Q and that
makes we kill not to be a sub not to be
a vector space now so far we were
discussing real vector spaces now we'll
be coming to complex vector spaces the
basic difference between the two is in
the real vector spaces the scalars are
taken from the set of real numbers but
in case of complex vector spaces scalars
are taken from set of complex numbers so
we say that complex vector spaces are
defined in the same way as real vector
spaces with the difference that scalars
are permitted to be complex numbers that
means all the axioms need to be
satisfied for this set of scalars I'll
illustrate with this example the set of
n tuples V is equal to x1 x2 xn where x1
x2 xn belongs to C is a vector space in
C n with respect to vector addition
which is defined as X 1 X 2 xn belonging
to V by 1 Y 2 y n belonging to V the sum
of the two is again an N tuple where
they are components
wise addition is taking place so this
sum is equal to X 1 plus y 1 comma X 2
plus y 2 and the last is xn plus y in
while the scalar multiplication is
defined as C star X 1 X 2 xn is equal to
component wise multiplication C X 1
comma C X 2 C X in the last term now you
may notice that Here I am considering
all these numbers belonging to C this is
required to satisfy the closure property
because if these are not if sees us and
complex number then see X 1 will also be
a complex number so if these X 1 X 2 xn
are real then the closure property will
not be satisfied so this set forms a
vector space in C n if you consider RN
real vector space then there will be a
problem so to satisfy the closure
property we consider set C this
coefficient this X 1 X 2 xn belongs to C
instead of R and this means that the set
V is nothing but C n so CN forms a
vector space with respect to these
operations the additive identity can be
checked easily because the zero vector
theta B is also in C n 0 0 0 is the
theta vector it is in C n so this is a
rest of the properties can easily be
checked so this forms a vector space now
in the next example I am considering M
by n matrices with complex entries
instead of real entries this example I
have taken earlier but those but at that
time denim these are real entries and we
have seen that they form a vector space
and now we will see that with complex
entries they also form a vector space
with respect to matrix addition and
scalar multiplication the next example
is the set P of all energy
the polynomial with complex coefficients
they form also vector space with respect
to a polynomial addition and
multiplication by complex scalars these
are some easy straightforward examples
and one can easily work them out now the
next topic is subspaces so first the
definition of subspaces we say that let
V be a real vector space with vector
addition plus and scalar multiplication
star then a subset W of V is a subspace
of the vector space if W itself is a
real vector space with respect to the
two operations plus and star so W is a W
itself is a vector space with respect to
the two operations now this is an
example if may be a vector space then V
is a subspace of V we can say that V is
a subset of V and since V is a vector
space so this a trivial example that V
is a subspace of V the another trivial
example is the singleton zero is a
subspace of V because this is additive
identity if it belongs to V then it will
be a subspace these are two examples
further it is observed that addition is
commutative as well as associative in W
because all the members in W they belong
to V so if it is associative and
commutative in V then it has to be in W
so that is straightforward the next
thing is scalar multiplication again the
same logic is applied it is distributive
in V so it has to be distributive in W
also because W is a subset of V the next
thing is existence of one in R is also
implied
so if you have to prove that W is a
subspace of a given vector space V then
we need to check the following four
conditions the first condition is that
if alpha beta belongs to W then the
vector alpha plus beta must also belong
to W the idea is alpha beta belongs to W
but the sum alpha plus beta Maybeline V
but may not be in W so one has to check
that for each pair alpha and beta in W
it's some also belong to W that is this
addition is closed in W the closure
property is satisfied second is additive
identity existing W such that alpha plus
theta is equal to alpha for all alpha in
W so if the the additive identity must
also be in W without that it will not be
a vector space the next thing is that
for each alpha belonging to W there
exists minus alpha in W such that alpha
plus minus alpha is equal to the null
vector theta so if alpha belongs to W
its inverse must also belong to W so
this is also a requirement and that is
to be fulfilled by W to be a vector
space and the last property is that
scalar multiplication is closed in W for
example if C belongs to R and alpha
belongs to W then C star alpha must
belong to W so the closure property with
respect to scalar multiplication must be
closed in W so all these four properties
are satisfied rest of the properties are
privily satisfied because they are
satisfied for V so this W will form a
vector space now one can notice that
these four conditions can be translated
into a single condition like C alpha
plus beta belongs to W for alpha beta
belongs to W and C belonging to R so if
this single condition is satisfied then
all these four conditions are implied so
let us prove this
so the first thing is a closure property
so if I take C is equal to one because
this property satisfied for all C
belonging to R in particular if C is
equal to one then alpha plus beta
belongs to W for alpha beta belongs to W
so if this property is satisfied for all
C then it will be satisfied for C is
equal to 1 and this amounts to the
closure property of vector addition in W
so alpha plus beta belongs to W for all
alpha beta belonging to W so if this
property is satisfied so this property
is implied let us see the second
property additive identity here if alpha
is equal to beta and if C is equal to
minus 1 if you consider this then in
this property if u consists of 5 is
equal to beta so beta minus beta is
nothing but null vector so if C alpha
beta belongs to W for all values of
alpha beta and C in R then beta minus
beta is equal to 0 will also belongs to
W so there this shows that there exists
additive identity in W so existence of
additive identity in W is established in
terms of this condition then the next is
additive inverse this we have given this
condition so if I take C is equal to
minus 1 and beta is equal to 0 the
existence of beta is equal to 0 I have
already established so if we take beta
is equal to 0 and C is equal to minus 1
substitute it here it is nothing but
minus alpha belonging to W so if alpha
is there minus alpha is also in W and
that establishes the additive inverse in
W so thief of properties have been
established and the fourth one is that
major property for scalar multiplication
so to satisfy to check this property
with respect to this given general
condition then beta is equal to 0 if we
substitute then C alpha belongs to W
that means for C in belonging to R and
alpha belonging to W C alpha is also in
W and that means closure property
for scalar multiplication is also
satisfied if this condition is satisfied
so we can put a remark here that if W is
a subspace of V then C alpha plus beta
is in W for each pair of alpha beta in W
and if this is there then we can say
that W is a subspace so this is given in
the form of the theorem states that a
non empty set W of V is a subspace of V
if for each pair of I'll find beta in W
and each scalar C in R this linear
combination C alpha plus beta is again
in W so if this condition is satisfied
then W is a subspace or if W is a
subspace then C alpha plus beta is again
in W so we have proved this theorem this
property is also used as a definition
for subspaces so this is a very powerful
property and many times we use it as a
definition for subspaces now I will
explain the definition with the help of
an example so example is taken from R
cube so if we consider V is equal to R Q
and W is the set of all triplets where X
1 and X 2 belongs to R but the third
component is 0 then W is a subspace with
respect to usual operations we know that
R cube is a subspace with respect to
usual addition and scalar multiplication
we have press tablished that W is also a
subspace in this so to prove this we
consider C alpha plus beta so I am using
that very definition that's if alpha and
beta belonging to W then C alpha plus
beta belongs to W then it becomes a
subspace so let us consider C alpha plus
beta alpha happens to be X 1 X 2 0
belonging to the set W by 1 by 2
is nothing but beta in W then if you add
the two it is C X 1 comma C X 2 comma 0
plus y 1 by 2 0 and finally we will have
C X 1 plus y 1 C X 2 plus y 2 and 0 now
C X 1 plus y 1 belongs to R because x1
and y1 and C all they are real so this
also belongs to our C X 2 plus y 2 also
belongs to our and the third component
is 0 so this vector belongs to W so C
alpha plus beta belongs to W 4 given
alpha beta in W and that means this W is
a subspace that's it now the next
example I consider again another subset
of our cube here W 1 is equal to X 1 X 2
1 such that X 1 X 2 belongs to R so
instead of 0 now I'm having one here but
one can see that W 1 is not a subspace
of our cube so with respect to usual
operations in our cube so let us
consider W 1 as being given here now
consider see X 1 X 2 comma 1 plus y 1
plus y 2 comma 1 so we like to check
whether C alpha plus beta also belongs
to W 1 or not so because of this
left-hand side and the right-hand side
according to the definition of the two
operations it becomes C X 1 plus y 1 C X
1 plus y 1 second component is C X 2
plus y 2 and the third component is 1
plus 1 is equal to C plus 1 so this is
not a subspace because we want here to
be 1 now this will be 1 only when C is
equal to 0 for general value of C this
will not be 1 and that is why this
right-hand side will not belong to W 1
or this operation is not this scalar
this a linear combination does not
belong to the
one and we say that w1 is not a subspace
this is another example taken from the
set of matrices like V is a set of all n
by n matrices and W is a set of all n by
n symmetric matrices then W is a
subspace with respect to usual addition
and multiplication we have already seen
that V is a vector space when V is the
set of square matrices and we have the
usual addition and multiplication
operations usual addition and scalar
multiplication operations in V so it's a
vector space but we have to see that
this set W of symmetric matrices whether
it is a vector subspace or not so let us
start with we've is a set of n by n
matrices it's a vector space now we
consider set W set of n by square
matrices which are symmetric we clearly
know that W is a subset of V this is the
very first condition W to be a subspace
of V that W should be a first subset of
me so this we have established now let
us consider two members in W alpha as a
IJ and square matrix beta is bij another
square matrix such that a IJ is equal to
AJ why because they belong these two
matrices belong to W and they are
symmetric matrices so H J must be equal
to a GI similarly bij is equal to VG I
now consider the linear combinations C
IJ plus B IJ so because the typical
element of this matrix CA IJ plus B IJ
now this is equal to AI J becomes a GI
so it is seed a GI plus B GI so we can
say that as C alpha plus beta is also
symmetric matrix so the operation is
closed and W is a subspace this is
another example of taken from the set of
map
but this time w is not a subspace so let
us consider V to be two there are two
matrices and W is again set of all two
by two matrices but the condition is
that this is non singular matrices so we
have a bigger set considering all square
matrices of order two while W is only
both square matrices of order two which
are non singular then W is not a
subspace of V where V is a vector space
with respect to usual vector matrix
addition and scalar multiplication so
one can say that the null matrix is an
additive identity of V this belongs to V
that is a two by two matrix it's a
vector space so null matrix has to be
there in V but if we have null matrix
also in W then it is non singular matrix
it is not a non singular matrix so if
null matrix does not belong to W because
null matrix has determinant zero null
matrix is a singular matrix so null
matrix does not belong to W but W has to
be a vector space in its own right then
additive identity has to be there in W
and because of this reason W will not be
a subspace of V so W is not a subspace
of V so in the next example we are again
considering a triplet X 1 comma X 2
comma X C they all belong to R with the
condition that X 1 is equal to X 2 then
s is a subspace in our queue for usual
operations I have different types of
sets some forms subspace some do not
form subspaces in this example if this
condition is satisfied then they form a
subspace so let us consider see the
solution again consider
the combination see alpha plus beta
which is equal to C X 1 comma X 2 comma
X 3
this triplet belonging to s plus y 1 y 2
y 3 belonging to s then simplifying
right inside will have C X 1 plus y 1
comma C X 2 plus y 2 comma C X 3 plus y
3 now since X 1 X 2 X 3 and y 1 y 2 y 3
belongs to s so they must satisfy the
given constraint that is x1 is equal to
x2 and y1 is equal to y2 and using this
condition we can see that see X 1 plus y
1 will also be the same as see X 2 plus
y 2 that means this constraint will be
satisfied by this right-hand side of C
alpha plus beta and therefore C alpha
plus beta belongs to W and hence it is a
subspace in the next example W is equal
to X 1 comma X 2 comma X 3 but this
constraint is now different X 1 is equal
to X 2 is equal to 3 X 3 we have to
check whether it forms a subspace of r3
or not the solution is on the same lines
we consider an alpha and W which is a
comma 2 a comma 3 a because this
condition has to be maintained and beta
is equal to B comma 2 B comma 3 B B the
two vectors in W then C alpha plus beta
is C times the alpha vector plus the
beta vector simplifying it will have C a
plus B comma 2 times a plus B comma 3
times a plus B and that means the
condition X 1 is equal to 2 X 2 is equal
to 3 X 3 is also satisfied for C alpha
plus beta and that means C alpha plus
beta also belongs to W and this proves
that W is a subspace X 2 square plus X 3
square and we'll see that it is not a
space now the solution in the solution
we consider an example where alpha is 13
comma 2 comma 3 and Vita is 25 comma 4
comma 3 are two vectors in W one can
easily see that they satisfy this
constraint 13 is equal to X 2 square
plus X 3 square 4 plus 9 and in this
case 25 is equal to 4 square plus 3
square that is 16 plus 9 is 25 so these
two satisfy this condition but if you
consider alpha plus beta then the sum of
this these two vectors is 38 is equal to
6
38 comma 6 comma 9 but this vector does
not belong to W one can easily check
that 38 is not equal to 6 square plus 9
square that means if alpha and beta are
these two vectors then there is some
alpha plus beta will not belong to W now
this is a usual way to show that a set
is not a subspace so if you have to show
that a set is a subspace then you have
to consist the alpha plus beta and try
to show that see alpha plus beta also
belongs to W but if you know if you have
to show that it is not a subspace then
we'll take some typical examples which
show that see alpha plus beta is not
belonging to W so this one simple
example is enough to show that W is not
a subspace
now this is slightly different example
here I am considering the system of
equations represented in the matrix form
X is equal to theta so X is a column
vector and theta is null vector so if I
consider a as a square a as a
rectangular matrix of order M by N and X
is a column vector with n rows then week
one can define the subset W as all the
solution vectors which satisfies X is
equal to theta then we have to prove
that W is a subspace so we have a the
set of all rectangular matrices of order
M by n with respect to usual operations
of matrix addition and scalar
multiplication they form a subspace the
form of vector space but as far as this
W is concerned W is a subset of K but
this is not this will not be a subspace
to prove this we consider X as X 1 X 2
xn a column vector such that ax is equal
to theta y is equal to Y 1 y 2 y n such
that a y is equal to theta now because
the linear combination Z is equal to CX
plus y and we will see that a said
whether it belongs to this given set or
not so we consider KZ is equal to a
times column vector C X plus y which is
equal to C times ax plus ay Y since CX
is X is Theta so this is C theta is 0
and a y is Theta so C theta plus theta
is also theta so a that is equal to
theta so this red is also a solution
vector so it forms a subspace so W is a
subspace in particular the subspace W is
called the null space of a
now we define linear combination a
vector beta in V is said to be a linear
combination of vectors alpha 1 alpha 2
alpha N in V provided there exists
scalars c1 c2 CN in R such that beta is
expressed as C 1 alpha 1 plus C 2 times
alpha 2 plus C 3 times alpha 3 and C n
alpha n in short we write it as
summation I is equal to 1 to n CI alpha
is a compressed form of representing
this linear combination let us
illustrate this with example in this
example we have to show that the vector
1 2 3 is a linear combination of vectors
1 0 0 0 1 0 and 0 0 1 so what we are
supposed to do is you have to express
the given vector 1 2 3 in as a linear
combination of the vectors 1 0 0 0 1 0 0
0 1 that means there exist scalars a B
and C such that 1 2 3 is equal to a
times the first vector plus B times the
second vector plus C times the third
vector so if you add them up then we'll
have a 0 0 Plus this is 0 B 0 plus 0 0 C
and this gives me ABC so if there are
two vectors so if we create the two
vectors left from left hand side and
right hand side there will be equal
component wise and that means a will be
1 B will be 2 and C will be 3 so we can
find we have found K is equal to 1 B is
equal to 2 C is equal to 3 if we
substitute these values here this will
nothing but 1 2 3 in short we have
represented 1 2 3 as
1 x 1 0 0 plus 2 times 0 1 0 plus 3
times 0 0 1 so we can say that 1 2 3 has
been represented as linear combination
of the given three vectors 1 0 0 0 1 0
and 0 0 1 in the next example again we
are having vectors 5 minus 1 1 and you
have to express it as a linear
combination of vectors 0 1 minus 1 1 1 0
and 1 0 2 the procedure is the same we
consider left hand side the given vector
right hand side the linear combination
of the 3 given vectors 0 1 minus 1 1 1 0
and 1 0 2 this right hand side is
simplified it is C 1 plus it's the first
component is C 2 plus C 3 the second
component is C 1 plus C 2 no component
is coming from this then we'll have
minus 1 into C 1 so it is minus C 1 plus
2 times C 3 no component is coming from
the second vector if you equate a
left-hand side and right-hand side
component wise then C 2 plus C 3 is
equal to 5 and C 1 plus this next is C 1
plus C 2 is equal to minus 1 in the last
component is minus C 1 plus twice C 3 is
1 so we have to solve these three
equations to get the values of C 1 C 2
and C 3
in the earlier example the things were
simple and we can straight away get the
constants but in this case we are
getting three equations to solve them
simultaneously now to solve these three
equations one can observe that if you
add these two equations we'll have first
equations remain the same and the
addition of these two equations give
rise to C 2 plus 2 C 3 is equal to 0 and
that means we'll have two equations in
two
unknown c-2 and c-3 so to simplify it if
I multiply this equation by two the and
subtract so this and this will get
cancelled and what we have is C 2 is
equal to 10 from this equation this will
get cancelled C 2 is equal to 10 and
once you have C 2 is equal to 10 you can
substitute it here to get C 3 is equal
to minus 5 and once we get C 2 nc3 one
can get a C 1 from this equation and
that gives me a 5-1 1 as minus 11 times
0 1 minus 1 that is the value of C 1
plus C 2 which is 10 1 1 0 plus C 3
minus 5 1 0 2 so this vector 5 minus 1
by 1 is expressed as linear combination
of these 3 given vectors
now the next definition is span of
vectors that is say we have been given a
set W of vectors alpha 1 alpha 2 alpha n
be the set of vectors in a vector space
V then the set s of all linear
combinations of alpha 1 alpha 2 alpha n
is called the span of vectors alpha 1
alpha 2 alpha N so all the linear
combination of this vector will form the
set s and that set will be called the
span of vectors and we denote the span
of vectors as s W that means the span
generated from W now since W is a subset
of V and s W is linear combination of
vectors of W that means s W may be a
subset of V and since all the linear
combinations are in s W so s W is a
subspace of V now we are in a position
to state a theorem we say that if W is a
non-empty subset of that
space then SW is the smallest subset of
V containing W is the smallest it's not
only the subset but it is a subspace
containing W now to prove this since
every vector in since every vector W in
W is also in s double because as W is a
set of linear combinations of vectors of
W so this W in W is also will be in SW
and that simply means that W will be a
subset of s W now how do say that there
be in others this know as W let us say
is not a smallest subspace let us say
there has to be another subspace T says
that W is a subset of T now we say that
for such a subspace steal any linear
combination of W is also in T because
it's a subspace so W is a subspace W is
a subset of s W which is a subset of T
because all that collinear combinations
are in s tableau so they will be a
subset of T and that means that s W is
the smallest subspace containing W now
the next evolution is that set W is a
spanning set of V if every vector in V
can be written as a linear combination
of vectors of alpha 1 alpha 2 alpha in
of W now it may happen that s W is equal
to V the span of W is V then alpha 1
alpha 2 alpha n actually span V or in
other words using this definition one
can say that W is equal to alpha 1 alpha
2 alpha N is a spanning set for the
vector space V now to determine whether
a set W of vector spans the given space
V then any arbitrary vector beta in V
must be expressed as a linear
combination of vectors of W
now this realistic with the example the
example is to show that the vectors 1 0
0 0 1 0 and 1 1 1 spells are cube so let
us transfer any an arbitrary vector a be
seen r3 then it can be expressed as a
linear combination of these three
vectors C 1 1 0 0 plus C 2 0 1 0 plus C
3 1 1 1 so let us simplify this comes
out to be C 1 plus C 3 the first
component the second component is C 2
plus C 3 and third component is simply C
3 equating right-hand side and the
left-hand side we get C is equal to C 3
the last component and then from here B
is equal to C 2 plus C 3 and here it is
a is equal to C 1 plus C 3 so this C 3
can easily be seen to be C and once you
get the value of C 3 s C C 2 can be
calculated as B minus C and from this
equation if you put the value of C 3 as
C then C 1 comes out to be a minus C so
for given values a B C 1 can uniquely
determine the scalars C 1 as a minus C C
2 as B minus C C 3 s C thus we can say
that these three vectors 1 0 0 0 1 0 and
1 1 1 spans the vector space R 3 however
in this example we have to show that the
vector the vectors 1 1 0 and 0 1 1 does
not span the vector space R 3 so let us
see the solution we again we consider an
arbitrary vector a
we see and consider it to be a linear
combination of given vectors so we write
down ABC is equal to c1 times 1 1 0 plus
c2 times 0 1 1 simplified it is ABC is
equal to c1 comma C 1 plus C 2 and
finally C 2 and this means a is equal to
C 1 B is equal to C 1 plus C 2 and C is
equal to C 2 and simplifying this we'll
get B is equal to a plus C now for the
vector 2 1 3 belonging to R 3 cannot be
expressed as a linear combination of
these three vectors because if it has to
satisfy if it has if it has to be
represented as this combination then it
must satisfy this constraint but if you
substitute the value of B as 1 a as 2
and CS 3 then one can notice that 1 is
not equal to 2 plus 3 that simply means
that this vector 2 1 3 cannot be
represented as a linear vector linear
combination of these two vectors so the
vectors 1 1 0 and 0 1 1 fail to spend r3
that's the conclusion however the span
of these two vectors that is 1 1 0 and 0
1 1 is the vector space V such that a
comma a plus C comma C where a and C
belongs to R so this is a vector space
so this is the span of this but the
vector which we have conserved will not
fall in this vector space V now we have
another theorem and according to this if
we have two subspaces s 1 and s 2 of the
vector space V then the intersection of
the two
subspaces is non-empty and it is a
subspace now to prove this it is
observed that the intersection of s1 and
s2 is non-empty why because s1 is a
subspace s2 is a subspace so theta has
to belong to both s1 and s2 so theta
must belong to s 1 intersection s 2 and
that is why s 1 intersection s 2 is
non-empty now we consider two cases the
first case is if s 1 intersection s 2 is
a singleton theta if that is the case
then it is a subspace of V because this
we have already established but in the
second case if there is there are
vectors other than theta in it so let us
say there are two vectors alpha and beta
belonging to this intersection s 1
intersection s 2 then let us consider C
alpha plus beta that means we are trying
to apply the basic definition of
subspaces so for alpha beta belonging to
a set if C alpha plus beta also belong
to the set then that set form say
subspace that is what we are going to
apply so let us consider say alpha plus
beta also belongs to s 1 intersection s
2 now since s 1 as s 2 are subspaces and
that means s 1 intersection s 2 is also
a subspace C alpha plus beta belongs to
s 1 C alpha plus beta belongs to s 2 so
they belong to s 1 intersection s 2 also
and that means s 1 intersection s 2 is a
subspace that proves the theorem however
if you consider the union of the two
sets it may not be a subspace now to
prove this we have to establish by
example so this is an example which we
consider in example s 1 is equal to sets
a comma 0 a belongs to R so it's a
subset of R 2 s 2 is 0 comma B belongs
to R now we have to show that s 1 Union
- is not a subspace now the first thing
is in the solution we show that s1 and
s2 are subspaces one can easily see that
a comma zero a belongs to R is a vector
space as two is zero comma B is also a
vector space so s1 and s2 are subspaces
of r2 because this is a subset of r2
this is also a subset of r2 so s1 and s2
are subspaces let us consider what is s
1 Union s 2 s 1 1 and s 2 is all sets
all vectors which are either of the form
a comma 0 or of the form 0 comma B and a
B belongs to R so that all this type of
sets belongs to s 1 Union s 2 now for
this we consider alpha a 0 1 belonging
to s 1 and I am taking B is equal to 1
beta is equal to 1 comma 0 belonging to
s 2 1 is a now alpha beta belongs to s 1
Union s 2 let us take this C alpha plus
beta is C 0 comma 1 plus 1 comma 0 is
equal to if you look at this it is C 0
plus 1 it is 1 C plus 0 is 0 1 comma C
where C is not 0 for c is equal to 0
things will be ok but for C nonzero this
property may not be satisfied so if C is
not 0 then 1 comma C does not belong to
you s 1 Union s 2 so s 1 Union s 2 is
not a subspace
so by s 1 intersection s 2 will be
subspaces for s 1 s 2 subspaces but s 1
Union s 2 need not be a subspace
now this result will generalize in the
form of theorem three and what we say is
that if s 1 s 2 SN are subspaces of V
then the first result says that s 1
intersection s 2 intersection s n is
also a subspace although we have proved
for n is equal to 2 similarly s 1 Union
s 2 Union s n need not be a subspace we
have generalized it the result we have
proved for s 1 a union s 2 but it can be
further generalized to finite n
subspaces the result has been
established for n is equal to 2 the
viewers can prove these results and the
hint is they can use the mathematical
induction to prove these results now
towards the end of this lecture let me
summarize what we have done today I have
started with review of the first lecture
in which I have discussed I have given
the definition of vector spaces and then
I have taken some examples I have taken
example of polynomials some of the
polynomials they form vector spaces some
of the polynomials in they when they are
put under certain constraints then they
do not form vector spaces discuss
compensators pieces then I have defined
subspaces after that we have discussed
linear combination the span of vectors
and the next turn I'll be discussing
direct sum and dependent and dependent
vectors and basis that's all thank you