Linear Transformation (Part - 1) - Video Lecture - Engineering

good morning viewers today we are going
to discuss linear transformation this
lecture includes the definition we'll
give some examples of linear
transformation then we will discuss some
important results regarding this and
then it will introduce the concept of
length rank and nullity and finally some
results related to rank and nullity to
start with I will first give a
definition of linear transformation for
this let us consider two real vector
spaces V and W a lu transformation T
from V into W is a function
I say T's from V to W which assigns a
unit vector T alpha in W for each alpha
in V such that it satisfies the
properties the first property is that T
is additive by this I mean to say that
when teens operated on alpha plus beta
for each alpha and V times V then it
will be transformed to T alpha plus T
beta by this we say that T is additive
the second property is T is homogeneous
but if I mean to say that when T is
applied on C alpha alpha is a vector and
C is a scalar in R then it becomes C
times T alpha so TC alpha is equal to CT
alpha and if these two properties are
satisfied then such a transformation T
is called a linear transformation I
introduced the principle of
superposition by this I mean to say that
if I have two vectors alpha beta and V
then T of alpha plus B beta is equal to
a of T alpha plus B of T beta where a
and B are scalar say are the way I have
defined this transformation
so that T from V to V into W is a
mapping such that T of C alpha plus beta
is equal to C times T alpha plus T beta
for alpha beta in V in fact this will
become an alternative definition for
linear transformation so instead of
satisfying to properties linear property
and homogeneous property the two
properties are combined into one single
property and defined in this manner now
a linear transformation from V to V is
also called a linear operator on V to
illustrate what I have done let us
consider two vector spaces V and W by
this I mean to say we consist of a set
of vectors two operators one is addition
of vectors another a scalar
multiplication defining V as well as in
W and they satisfy certain properties so
that way V and W are two vector spaces
then a linear transformation from V into
W is denoted by T V provided a vector in
X goes to TX in W and a vector in Y goes
to T Y in W and they satisfy the
property the homogenous property and the
linear property which I have defined
earlier now in this case I say V is the
domain of the Lua transformation YW is
co domain for the transformation defect
this set which consists of images of X
and V in V all x and y AV that set is
called range of T V and definitely this
is a subset of W later on we will prove
that it may be a vector space now let us
take an example we have a transformation
T from
cube to r2 which is defined as T of X Y
Z a vector goes to X Z in r2 so this is
a vector in r3 and this a vector in r2
so if we define respect this
transformation in this particular manner
then we'll show that T is a linear
transformation now if it is a linear
transformation one has to prove that T
is additive by deciding to say that if I
take alpha and beta two vectors in r3
then T of alpha plus beta is T of alpha
plus T of beta for each alpha beta V so
let us consider alpha as x1 y1 z1 and
beta as x2 y2 z2 so these are two
vectors in r3 now T of this is equal to
T of this vector because sum of two
vectors is X 1 plus X 2 this is y 1 plus
y 2 and z1 plus z2 so sum of these two
vectors is equal to this so T of this is
equal to this vector
now this transformation says that the
first component will become the first
component of T alpha and the third
component will become the second
component so that way XT of X 1 plus X 2
y 1 plus y 2 Z 1 plus Z 2 will become X
1 plus X 2 and z1 plus z2
now this will simplify to X 1 red 1 plus
X 2 Z 2 and by definition of the
transformation T X well that one is T of
x1 y1 z1 and x2 Z 2 is P of x2 y2 z2 and
that is how we have proved that T of 2 T
of alpha plus beta is equal to T of
alpha plus T of beta and that proves the
additive property now we prove that is
homogeneous for this purpose we have to
show that T of CL Phi is equal to C
times T alpha C being the scale
now to prove this letter consider alpha
is x1 y1 z1 and the scalar C then T of C
times x1 y1 z1 is equal to T of C x1 cy
1 CZ 1 what I have done is I have taken
this scalar C inside this so this is
equal to C of X 1 and C times Z 1 so T
of this 3 dimensional vector is this 2
dimension vector and from this we can
take C outside and what we have is C
times x1 z1 and that means C times T of
this is equal to C times this vector
that means this vector in r3 maps to a
vector in r2 so C times T x1 y1 z1 is
equal to C times x1 z1 and that proves
the homogeneous property and hence it is
a linear transformation in this we have
very defined a linear transformation
from R 3 to R 3 and we will again show
that it is a linear transformation so
the transformation is T of X Y Z is R
times rx ry and our set now to prove
this again we have to first show the
linear property so to see times x1 y1 z1
plus x2 y2 z2 now I am actually
combining the two properties the
homogenous as well as the editor
property so I say T time T of C x1 y1 z1
plus x2 y2 z2 is equal to T times C X 1
plus X 2 C y1 plus y2 C is at 1 plus Z 2
so basically I have combined these two
vectors in this form now using the
definition for the linear transformation
this expression can be simplified to R
times ax 1 plus 6x2 are times C y1 plus
y2 r times Caesar Delta Z 2
they say effective
transformation is that each component is
our times the visual value so this
left-hand side is now C times our X 1 ry
1 RZ 1 plus R X 2 R by 2 R Z 2 so sum of
two vectors and that means the
transformation T applied and C times
first vector plus the second vector is
equal to C times the transformation
applied on the first vector plus
transformation applied on the second
vector
this proves that T is a linear
transformation now I will give you
metrical interpretation to the examples
which we have taken so far the first
example is the linear coefficient T
defined from R cube into R 2 is T times
X Y Z is equal to X Z if we call it a
projection transformation let us
consider this vector this is the vector
three dimensional vector X Y Z it's this
is x axis y axis z axis this is the X
component this is the Y component and
this is the this is the red component
now when T this vector when we consider
T of this vector then what we have is
accept so what is X with this is X and
this is y so this is a vector which is
the projection of this vector so this
transformation means that this vector is
projected this this vector has a
projection this so when we apply linear
transformation on this vector what we
get is this projected vector so in the
second example we consider the vector
alpha in R 3 as X Y Z this is the vector
X Y Z it has three components x y and z
now when we apply linear transformation
on
then it becomes our EXO our exes this is
our X this is our Y and this is all said
so each of these is our x these values
so our x ry RZ are these components and
then the vector will be this vector so
this vector will be transformed to this
vector and what we have seen that this
vector is rotated to this vector when
you apply this linear transformation and
that is why we call this transformation
as rotation now in the process if R is
less than 1 it is a positive number
lying between 0 & 1 then we say that
this is contraction and when R is
greater than 1 we say it is in dilation
then in the next example T of x1 y1 z1
is equal to 2 X 1 by 1 plus 1 and third
component 3 third ones is a
transformation from R 3 to R 3 this
transformation is not a linear
transformation so what we have to do is
that this linear property is not
satisfied for this transformation so let
us consider T applied on C times first
vector the second vector a and the
domain set domain the test phase so it
is T times C X 1 plus X 2 if I combine
the two second company with C Y 1 plus y
2 and C Z 1 plus Z 2 now
when K is operated on this vector then
according to this definition the first
component is 2 times this so it is 2006
1 plus X 2 the second component second
component will map to Y 1 plus 1 so C 1
M plus 1 by 2 becomes C y 1 plus y 2
plus 1 and the third component see that
one positive it becomes 3 times this
component so season
then Quebec 2 is 3 times C Z 1 plus Z 2
so T of this vector is this so let us
simplify this so T 2.6 1 plus 2 times X
2 is C times 2 X 1 plus 2 X 2 but when
it comes to this you see my ll plus y 2
plus 1 it is sea level this Y 1 plus 1
plus y 2 see Y 1 plus 1 and Y 2 and this
is C can be taken out is 3/3 1 plus cz 2
so this this vector is nothing but C
times P of x1 y1 z1 plus this vector but
this vector is not T of this vector
because plus 1 is missing from here so
what we can say is the linear property
is not satisfied or we can say that this
is not equal to C T of X 1 Y 1 plus z1
plus T of X 2 y 2 Z 2 so this
transformation is not a linear
transformation apart from these example
there are some simple examples of do the
transformation from V to V R the first
is the identity transformation we
denoted by I such that every vector in
alpha maps to itself since a
transformation from V to V so every else
vector I'll find we will not do itself
such a transformation is identity
transformation and one can easily see
that it is actually satisfy this
property like I L suppose beta is a
times I alpha plus I beta and that means
I applied and this is alpha cos beta and
I apply them this is a alpha plus beta
so both the sides are equal and one can
say that identity transformation is a
linear transformation the another simple
example is the vo transformation we
denoted by theta and we define it as
that theta of alpha is equal to 0 that
means every alpha in this vector space V
will map to theta additive identity so
this is a 0
transformation also feed every reptile
alpha will map to zero so this also is
this also satisfy will be a property
that can be very easy to prove and that
means that Selenia transformation this
is another example here we define the
unit assumption T from r2 to r3 and it
is defined as TX is equal to X for X in
R 2 and T X in R 3 where the matter
where a is the matrix of 3 by 2 order so
let us define them at this
transformation T as X and value vector
and r2 will map to this vector in r3 now
we show that T is a linear
transformation now again the method of
proof is the same we start with C alpha
plus beta and the electricty alpha cos
beta in V we operate T on this and what
we have is T times C X 1 by 1 plus X 2
by 2 set 2 dimensional vector so alpha
is X 1 y 1 and B times X 2 by 2 then
this is operated on T when T is operated
on this vector then we should have T
times C X 1 plus X 2 C y 1 plus y 2 C Z
1 plus Z 2 and that means 1 0 1 1 0 1
this is the definition of this linear
transformation and when you multiply it
is C X 1 plus X 2 this multiplied by
this is C X 1 plus X 2 and plus C y 1
plus y 2 and then this is equal to C Y 1
plus y 2 and then T C alpha plus beta
one can see that this is equal to C
times X 1 second company X 1 plus y 1
entire component is y 1 and plus from
here we can write down X 2 this X 2 plus
y 2 the second component and what we
have is y 2 so T C alpha plus beta is
equal to C times 1
1 1 0 1 X 1 bar 1 + 1 0 1 1 0 1 X 2 y 2
so if we use that this is actually
equivalent to this and this means that
this is C times T alpha + T beta and
that proves that it's a linear
transformation now I will discuss some
important results in the form of
theorems channel 1 says that the T is a
linear transformation from the vector
space V into W then the first result is
that P of theta 1 is equal to theta 2
where theta 1 is identity vector V and
theta 2 is the identity vector W the
second result says that T of alpha minus
beta is equal to t alpha minus T beta
now the first result says that the
identity vector of V under this
transformation will nap to identity
vector of W now and this says that this
we have additive property now this is
this up with the subtraction now let us
consider the proof for the first
property so we can write del t theta 1
what is theta 1 theta 1 is an identity
vector in V so when identity vector is
added into identity vector what we have
is identity so theta 1 can be written as
theta 1 plus theta 1 so plus theta 1
plus theta 1 since these new year
becomes 2 theta 1 plus T theta 1 and
that means T theta one is metal epoch
theta 2 because T theta L is equal to
theta 1 plus D theta theta 2 theta 1
this is possible only when P theta 1 is
equal to theta 2 and this shows that the
oddity of we will map to identity of W
to prove the second property we consider
T of C alpha plus beta is equal to CT
meter plus Delta because T is a linear
transformation but what we can do is we
can simply take C is equal to minus one
and that means T of alpha minus beta is
equal to P alpha minus T beta so proof
of the second property is also simple
now some more results to the new
transformation from being to W then for
any vector alpha 1 alpha 2 alpha N in V
and the scalar C 1 C 2 C L C M then T of
C 1 alpha 1 plus C 2 alpha 2 plus C n
alpha n is equal to C 1 T alpha 1 plus C
2 steel 4 2 plus C NT L fine what we are
going to do is we have n vectors here we
have all scalars what we have done is we
have taken the linear combination of the
vectors alpha 1 alpha 2 alpha N and the
scalar C 1 C 2 C n now to have this
vector now since we have mu as a vector
space so if alpha 1 alpha 2 alpha N they
belong to me so this linear combination
will also belong to V so this is also
vector V and T is a linear
transformation then this theta this
vector is equal to C 1 T alpha 1 plus C
2 T alpha 2 plus C n TL Phi n we have if
we have only two vectors or we can say m
is equal to 2 then T of C 1 alpha 1 plus
C 2 alpha 2 is equal to C 1 t alpha 1
plus C 2 T alpha 2 the superposition
property but actually these are n
vectors so to prove this what we can do
is we can consider C 1 alpha 1 plus C 2
alpha 2 plus C ll find we can take this
as one vector and we are C 1 alpha 1 1
vector plus another vector near property
can be applied so we have C 1 alpha 1
plus the second vector now the second
vector the same thing can be applied to
the second vectors if we take Li and
what will have a final result that T of
C 1 alpha 1 plus C 2 alpha 2 plus C and
alpha n is equal to C 1 t alpha 1 plus C
2 T alpha 2 plus C n TL 4 so basically a
generalization of what we had earlier so
for it is the but the world was for
vectors this is for n vectors now we
lose our vector space which will be
having misses so let us consider s is a
basis for the vector V then given the
linear transformation T and alpha
belonging to V then 2 alpha will be
completed at women by its image s T
alpha 1 T alpha 2 T alpha n the idea is
that what will happen to these vectors
alpha 1 alpha 2 alpha N so we have this
is a basis is I am dimensional vector V
is an dimensional vector space then T
alpha T TL for being the image of a
vector in V then the unit vector will be
determined in terms of T alpha 1 alpha 2
t alpha n so let us prove this say f is
a basis for B and alpha belongs to V
then alpha which is a newer combination
of these vectors as being the basis can
be written as alpha is equal to C 1
alpha 1 plus C 2 alpha 2 plus CN alpha n
then we see the linear transformation so
L T of alpha is equal to C 1 T alpha 1
plus C 2 T alpha 2 plus C NT alpha n now
this is the result which we I have just
established in my last 0 then T alpha is
written mine by T alpha 1 T alpha 2 TL
fine how see this vector alpha
determined by the scalar C 1 C 2 C N and
alpha masked this alpha 1 maps to t
alpha 1 alpha 2 lab so T alpha 2 alpha
and X 2 T alpha N so C 1 C 2 C 2 C n are
uniquely determined so T alpha will also
be uniquely determined and hence we can
say that TL v determined by T alpha 1 T
alpha 2 n TL fine now
we introduce more concepts to start with
the image T of linear transformation T
from V to W is defined as the set
consisting of all that moves such that T
V is equal to W for some Reba longing to
V sometimes this is also called as range
of T the tunnel T of transformation T is
defined as the set consisting of all
vectors V in V such that T V is equal to
identity in W that way we define two
sets in HT of HT and kernel T these two
sets are related with the linear
transformation T clearly in HT is a
subset of W and kernel T is a subset of
V in fact we will prove that in if T is
a subspace of W by moving the vector
space
kernel T is a subspace of V so let us
show it pictorially we have two vector
spaces V and W X belongs to V and under
this transformation T this vector X will
go to Y now this is the image set which
is consisting of all boys which has some
point X in the set V so V is an image of
X all these wise include the range of
image set which is contained in W now
for the kernel T this is the domain in
which the vectors will not do the
identity vector theta in W so all these
vectors in this domain will map to theta
so parallel T is the set of all X a V
such that TX is equal to theta so this
is a subset of V and this is a subset of
W
now we say that a linear transformation
T is said to be 1 1 if alpha not equal
to beta belonging to V implies the T
alpha and T beta they are not the same
that means if we are two different
elements in V then their images cannot
be the same they have to be different
that is if T is one-one then with T
alpha level 2 T beta then alpha and beta
have to be the same allele
transformation T is said to be on - when
range of T is equal to W and another
definition if we take the linear
transformation from V into W is an
isomorphism if it is one-one onto the
vector space V and W are said to be
isomorphic if there is an isomorphism of
V into W again we show it here we have
two different elements two different
vectors in V they map to different
vectors in double W so X maps to t x and
y maps to T where this is true for any
combination of x and y in V then we say
that the linear transformation is 1 1 in
the other example here we have two x and
y in V but both map to say element in W
that is x ly are different in V but they
are same in W and that means this
transformation T is not 1 1 now in this
case we have a vector space V and a
vector space W T is a linear
transformation X maps to this element E
X and W by maps to this element E Y and
W and this is the set in which we will
have images in which the elements in V
will map to this set that means if this
set L said W
become the same then it becomes an onto
transformation that means there is no
element in this w which is not an image
of this every element here is in in
which is in image of some point you V
that is onto transformation on the now
we can prove these theorems that
consider linear transformation T then
image T is a subspace of W and kernel T
is a subspace of W a subspace of V so in
HT is a subspace of W and kernel T is a
subspace of V now first prove the first
result that image T is a subspace of W
so to prove this we consider alpha beta
belonging to HT then if CL type of bit
also the last image T then image T is a
substance is a very definition of
subspace so we can say alpha and beta
belonging to image T so so they belong
to an HT that means there must be some
reason in the fact V so that alpha is
equal to t v 1 and 7 v 2 in V so that
beta is equal to T V 2 so let us
consider alpha and beta in this way so C
alpha cos beta is equal to C of T V 1
plus T of V 2 and since T is a linear
transformation so C alpha cos beta is
equal to T of C V 1 plus V 2 now V is a
subspace therefore C V 1 plus V 2 we
also belongs to V let us call this
vector s v3 then C alpha plus beta is
equal to T of V 3 that means C alpha of
DT is C alpha that B is an image of V 3
and that means C alpha cos beta also
belongs to in HT and this proves that in
HT is a subspace and the second part of
the theorem is that kernel T is a
subspace of V
so we consider two vectors in Colonel T
V 1 and V 2 and will prove that C C 1 V
1 plus V 2 is also in colonel T that's
the definition of subspace so we start
with v 1 v 2 belonging to penalty so
they belong to penalty this implies that
T V 1 is equal to identity theta and T V
2 is equal to theta or sister is a
linear transformation then TX 0 V 1 plus
V 2 is equal to 0 T V 1 plus T V 2 and
the place T of serial V 1 plus V 2 is
also theta and this proves that C 1 V 1
plus V 2 also belongs to thermal T and
that is ll T is a subspace now is the
remark that penalty is never empty
because T theta T F theta is equal to
theta so there is always a vector theta
equal L T which is not 2 theta itself so
penalty will never empty theta will
always belong to convert if there's any
more members in kernel T but this will
fatah will always be in kernel T and it
is never empty then circular masses if
penalty is Theta only the dimension of
penalty is equal to 0 this is another
remark then we have an important result
regarding dimension of penalty and
dimension of image T of HT and it says
that if T is a linear transformation
from V into W then dimension of V is
equal to dimension of penalty plus
dimension of image T now to prove this
result we proved it in two different
cases the first case is that T is a zero
transformation so if T is a via
transformation
then alpha belongs to B implies that T
alpha is equal to theta and parallel T
in that case will be V and image T will
be theta itself and the emission of
image T in that case will be zero or
dimension external T is equal to
dimension of V is equal to M so the
emission of physical dimension of kernel
T plus dimension of image T so that we
have proved the first part where T is a
zero linear transformation in the second
cases the dimension of penalty is K it
is not zero now to prove this we have to
prove the dimensional image T is n minus
K for this let us say the basis for
penalty is alpha 1 alpha 2 alpha K and
kernel T being the subspace of V we can
prove that we have to prove that image T
is equal to alpha K plus 1 and for in so
this is what we are going to prove so in
HT is actually generated by the fit so
basis for V is let us consider it to be
alpha 1 alpha 2 alpha K alpha K plus 1 I
will find so we have M dimensional
vector space V and the basis alpha 1
alpha 2 alpha K this is the basis for
kernel T command T being the subspace of
V and we have some additional vectors so
this is for V is this let us consider
beta belonging to image T then there
must be some alpha in V so that beta is
equal to T of alpha so we write down
this alpha as a linear combination of
the vectors of basis of V that is alpha
is equal to C 1 alpha
plus c 2 alpha 2 plus C n L Phi n then
beta is equal to T of this vector so I
write down this vector as C 1 alpha 1
plus C 2 alpha 2 plus C K alpha K plus C
k plus 1 alpha K plus 1 up to C n alpha
and so that means I have divided this
into two parts this is the vector C 1
alpha 1 plus C 2 alpha 2 posickey alpha
K the linear combination of vectors
alpha 1 to alpha K that means this will
belong to parallel t and this is not to
0 so i use the property of new your
transformation and either they'll be
tires c 1 t alpha 1 plus c 2 t alpha 2
plus ck t alpha k on one side and ck
plus 1 T alpha k plus 1 plus up to CM t
alpha n as a second term so I have Putin
are representing beta as sum of two
terms this term and this term and that
means beta is equal to T of C 1 alpha 1
plus C 2 alpha 2 the sequel circuit plus
the second term and this is user penalty
is a subspace and therefore C 1 and Phi
1 plus C 2 alpha 2 plus C K alpha K
belongs to come L T and that is why TF
this is equal to theta so this means
beta is equal to simply this term so
beta is equal to CK plus 1
TK alpha k plus 1 plus the last term is
C n TL fine now every vector of n HT is
spanned by this set but you have to say
that if these vectors are linearly
independent then we have proved the
result so what we have just said that
every vector beta is a linear
combination of t alpha k plus 1 up to
alpha n so it is spanned by this set
what we have to see that they are
linearly independent and in that case B
1 is a basis for image T so to prove
this like
consider d1 TK plus 1 plus G 2 T of
alpha K plus to the N minus K T alpha n
is equal to theta so this linear
combination of M minus K vectors is 0
now this is 0 then T of D 1 alpha K plus
1 so linear property satisfied so T of D
1 alpha K plus 1 plus G 2 alpha K plus 2
plus DN minus K alpha is equal to 0 this
is equal to 2 times R added C 1 alpha 1
plus C 2 alpha 2 plus week alpha K added
this in this because the V of X this is
nothing to 0 so T of this is we also
added this e to this term so since there
is a basis so C 1 alpha 1 plus C 2 alpha
2 plus ek alpha K that's given alpha K
plus 1 therefore rest of the terms is
equal to 0 and say this a basis so this
is a combination is 0 means all the
linear all the scalar terms C 1 C 2 C K
D 1 D 2 D etcetera are 0 so C 1 C 2
etcetera they are all identically 0 and
that means alpha K plus 1 alpha K plus 2
alpha n are linearly independent because
these are 0 so these vectors are linear
these vectors are linearly independent
and that means the dimension of image T
is n minus K because we are having M
minus K vectors so what we have is
dimension of image T plus dimension of
penalty is dimension of V so we have
proved an important result for the
linear combination T rank of T is
defined to be the dimension of image T
this is the definition the drunk of T is
defined to be the dimension of image T
or rank of T is dimension image T
similarly nullity of T is defined to be
the dimension of kernel T so nullity T
is dimension of kernel T and in this
light we can say that a rank of T plus
nullity of T is equal to dimension
v so this is an important result and
we'll be using this at number of places
now now the result earlier
transformation is one-to-one if and only
if colonel t is equal to identity we
have only one element in Colonel T that
is identity such a transformation is
one-to-one now to prove this we have to
prove two parts one is the T is
one-to-one
then Colonel T is equal to theta and the
other part is if colonel T is equal to
theta then T is one-to-one so let alpha
belongs to V such that T alpha is Theta
also T theta is theta because theta will
map to theta itself this we have proved
earlier so we have one more alpha which
will take which this now assumption t
will take two theta now this is one one
mapping so alpha has to be theta know
three different elements will go to same
element T type W so colonel T has to be
theta so by contradiction we have proved
that kernel T is equal to theta so this
part is proved the second is if penalty
is equal to theta then T will be 1 to 1
now to prove this result we consider
alpha 1 and alpha 2 belonging to V such
that T alpha 1 is equal to T of alpha 2
then T of alpha 1 minus T alpha 2 is
equal to theta and since the linear
transformation T of alpha 1 minus alpha
2 is will do theta but alpha 1 minus
alpha 2 will also belong to V because V
is a subspace so the linear combination
will also belong to V so T of alpha 1
minus alpha 2 is equal to theta but we
have said that only theta can map to
this
is equal to theta so that means alpha 1
minus alpha 2 is equal to theta and that
proves alpha 1 is equal to alpha 2 so
this means T is 1 1 so even if you have
started with two different values it
comes out to be that these two vectors
are the same and this proves that T is 1
to 1 so we have proved the result now I
will take some examples the first
example says that a transformation being
given to us from R 3 to R 2 T ABC is
equal to a vector in R 2 a plus 2b plus
C being the first component and minus a
plus C B plus C the second component now
this vehicle summation is being given to
us you have to find kernel T and its
basis and dimension so we will start
with the definition of kernel T if T ABC
is equal to theta then ABC will belong
to kernel T this kernel T will be a
member of r3 so that is why I consider a
three dimensional vector ABC so T of ABC
will be 0 then ABC will belong to kernel
T now to prove this we have to say that
a plus 2 B plus C is equal to 0 minus a
plus 3 B plus C equal to 0 this is to be
there this is because a kernel T will be
theta only so a the first component has
to be 0 and the second component has to
be 0 so if you simplify then C is equal
to minus 5 K and B is equal to 2 K you
can add the 2 and will get this result
that means penalty is equal to the a is
K B is equal to 2 K and C is equal to
minus 5 k k belonging to r that means
any vector of this form will satisfy
these equations so this vector will
belong to kernel t
so this is a penalty will be one 2-5 any
vector of this form can be generated
from this so this is the basis for
kernel T so kernel T will be having
vectors of this form which will be
generated by this vector so this forms a
basis of kernel T and since penalty
involves of the tail of this form so we
say that this is the basis having only
one vector so the dimension of kernel T
is 1 now we make use of rank T a melody
theorem it says the rank T plus nullity
T is equal to dimension V dimension V is
given V is given to be 3 right a nullity
is given to be 1 so Rho T is equal to
dimension n HT is equal to 3 minus 1 is
equal to 2 so dimension of image T is 2
in the second example we consider
standard basis for R 3 being II 1 e 2 e
3 and we consider linear transformation
from R 3 to R 3 define in this manner he
one will map to e1 plus e2 T of e 2 we
map 2 e 3 plus 2 e 3 to e 2 and T of e 3
will not to e1 minus e2 minus e 3 now
this is the map the transformation is
from R 3 to R 3 in my earlier result I
have proved that if you know how these
elements are being mapped you define the
mapping and the mapping can easily be
determined so we define the mapping in
terms of a mapping of e1 e2 and e3 so
this defines a mapping now you have to
show that the vectors T 1 T 2 and T 3
are not linearly independent so let us
consider V 1 as T of e 1 which is e1
plus e2 and veto s T
- which is 3 + 2 e - being defined here
V 3 is T 3 as e1 minus e2 - III these
are three vectors in r3 let us consider
the linear combination of these three
vectors that is C 1 te 1 plus C 2 te 2
plus C 3 te 3 as 0 if the C vector for
linearly independent and C 1 C 2 C 3
will come out to be 0 now we use the
definition of T 1 T 2 T 3 being given to
us so we write down C 1 e 1 plus e 2 s
te 1 plus C 2 te 2 s III plus 2 e 3 plus
C 3 times te 3 as e1 minus e2 - III 0 we
will combine the different different
terms
we'll have C 1 plus C 3 multiplied by e
1 plus C 1 plus 2 C 2 minus C 3
multiplied by e 2 plus C 2 minus C 3
multiplied by e 3 and this is equal to 0
since evil e 2 e 3 are linearly
independent therefore C 1 plus C 2 C 1
plus C 3 is equal to 0 C 1 because 2 C 2
minus C 3 equal to 0 this is the
coefficient of e 2 and coefficient of e
3 is C 2 minus C 3 equal to 0 now we
have three equations to solve C 1 C 2
and C 3 from the first equation C 1 is
equal to minus C 3 and from second C 2
is equal to C 3 so if you substitute C 1
plus C 3 is equal to 0 here then we'll
have C 1 plus C 2 equal to 0 and that
means C 1 is equal to minus K C 2 is
equal to K and C 3 is equal to K is a
possible solution for this and cade need
not be 0 so we have obtained a solution
for this equation which is a nonzero
solution and that means te 1t
- t3 are linearly dependent they cannot
be linearly independent because we have
got this nonzero solution now we can put
a limiter the set of nearly independent
vectors may have linearly dependent
images and their earlier transformation
so this is what has happened in the
earlier example we have started with a
linearly independent set of vectors e1
e2 e3 we have a Lu a transformation T
but what we have TT of e1 T of e 2 T of
e 3 they are not linearly dependent they
are not linearly independent but they
are linearly dependent and that is the
set of linearly independent vectors may
have linearly dependent image under a
linear transformation now this is a
theorem we say that if we have a linear
transformation u into V
if it is one-one linear transformation
then the set of vectors u 1 u 2 UN
belonging to you are linearly
independent vectors then its images tu 1
tu tu tu n are also linearly independent
vectors so basically this theorem
provides the condition under which
linearly independent vectors will map to
linearly independent vectors and the
result is that the transformation has to
be 1 1 so here is the proof we can start
with a linear combination of vectors to
u 1 to u 2p u N and since T is a linear
transformation so we'll have T of C you
see 1 u 1 plus C 2 u 2 plus C n u n
equal to 0 and U will belong to you
because U is a vector space the linear
combination law so belong to you
now T is 1 1 or celerity is equal to 0
so we have been given that T is one one
transformation and we have either if I
may prove that kernel T is equal to 0 if
T is one-one that means T of U is equal
to 0 so this vector U has to be 0
because the MLT is equal to zero and
that means C value alpha C
you 2 + 3 mu n is equal to 0 but we have
you are--oh UN has to be 0 that means C
1 C 2 C n has to be 0 so that means we
have another combination which is 0 and
these officials come out comes out to be
0 and that means tu 1 tu tu tu and are
linearly independent so under this
condition will have a dealer but this
will go to linear linear independent
vectors will go to linearly independent
set of vectors now what happened in the
earlier example which was TE 1 is equal
to e1 plus e2 te 2 was e 3 plus 2 e 2 te
three is e1 minus e2 minus e 3 in this
example the vector even minus e 2 minus
e 3 is actually linear combination as te
1 minus T 2 minus PE 3 and this is
actually 0 and that means the vector e1
minus e2 minus e 3 belongs to n of T and
dimension of NT is equal to 1 in that
case so according to regulative theorem
the rank of T is dimension of image T
which is 2 and in this case they are
actually not mapping to the three
vectors cannot be linearly independent
now I have one more dosa one more simple
result that if I have a linear
transformation from our four to our six
linear transformation and if dimension
of penalty is given to be told
then dimension of rate of T or image of
T can be computed as 4 minus 2 it is 2
you have to make it clear that the
dimension of ways to be considered in
very nullity theorem nor the dimension
of W so that is result here is 2
similarly if dimension of range T is 3
then the emission of penalty is 1 so 3
plus 1 is equal to 4 so this rank of
negative rank and melty theory
can be used to obtain and the dimension
of penalty if other two things are given
can be obtained for the emission of Rach
T if dimension of V is given and
dimension of Colonel being given to us
viewers with this we have come to the
end of this lecture to summarize what we
have done today I have started with the
definition of linear transformation I
have given some examples I have
illustrated with the help of examples
what do you mean by linear
transformation what we mean by additive
property what do you mean by homogeneous
homogeneous property and then we have
this discussed several results related
with linear transformations I have
introduced the concept of kernel and
nullity and we have finally established
theorem relating range dimension of
range and dimension of kernel with the
dimension of the vector space V we have
we continue with this will discuss more
conservator to this in my next lecture
thank you