Study Materials for 1 st Year B.Tech Students Paper Name: Mathematics Paper Code : M201 Teacher Name: Rahuldeb Das Lecture 1 Definition DEFINITION (Vector Space) A vector space over denoted is a non-empty set, satisfying the following axioms: 1. VECTOR ADDITION: To every pair there corresponds a unique element in such that 1. (Commutative law). 2. (Associative law). 3. There is a unique element in (the zero vector) such that for every (called the additive identity). 4. For every there is a unique element such that (called the additive inverse). is called VECTOR ADDITION. 2. SCALAR MULTIPLICATION: For each and there corresponds a unique element in such that 1. for every and 2. for every where 3. DISTRIBUTIVE LAWS: RELATING VECTOR ADDITION WITH SCALAR MULTIPLICATION
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Study Materials for 1st Year B.Tech StudentsPaper Name: Mathematics
Paper Code : M201Teacher Name: Rahuldeb Das
Lecture 1
Definition
DEFINITION (Vector Space) A vector space over denoted is a non-empty set, satisfying the following axioms:
1. VECTOR ADDITION: To every pair there corresponds a unique element
in such that
1. (Commutative law).
2. (Associative law).
3. There is a unique element in (the zero vector) such that
for every (called the additive identity).
4. For every there is a unique element such that
(called the additive inverse).
is called VECTOR ADDITION.
2. SCALAR MULTIPLICATION: For each and there corresponds a
unique element in such that
1. for every and
2. for every where 3. DISTRIBUTIVE LAWS: RELATING VECTOR ADDITION WITH SCALAR
MULTIPLICATION
For any and the following distributive laws hold:
1.
2.
Note: the number 0 is the element of whereas is the zero vector.
Remark The elements of are called SCALARS, and that of are called VECTORS. If
the vector space is called a REAL VECTOR SPACE. If the vector space is called a complex vector space.
We may sometimes write for a vector space if is understood from the context.
THEOREM Let be a vector space over Then
1. implies
2. if and only if either is the zero vector or
3. for every
Proof. Proof of Part 1.
For by Axiom 1d there exists such that
Hence, is equivalent to
Proof of Part 2.
As , using the distributive law, we have
Thus, for any , the first part implies . In the same way,
Hence, using the first part, one has for any
Now suppose If then the proof is over. Therefore, let us assume
(note that is a real or complex number, hence exists and
Examples1. The set of real numbers, with the usual addition and multiplication (i.e.,
and ) forms a vector space over
2. Consider the set For and
define,
Then is a real vector space.
3. Let be the set of -tuples of real
numbers. For in and we define
(called component wise or coordinate wise operations). Then is a real vector space with addition and scalar multiplication defined as above. This vector space
is denoted by called the real vector space of -tuples.
4. Let (the set of positive real numbers). This is NOT A VECTOR SPACE under usual operations of addition and scalar multiplication (why?). We now define a new vector addition and scalar multiplication as
2. If S and T are two subspace of a vector space V then which one of the followings is a sub space of V also ?
a) S U T b) S ∩ T
c) S -T d) T- S
3. If a finite set has 5 elements, then its power set has
a) 5 ! b) 25
c) 10 d) None
Lecture 2
Linear Combinations
DEFINITION (Linear Span) Let be a vector space and let be a non-empty subset of The linear span of is the set defined by
If is an empty set we define
EXAMPLE
1. Is a linear combination of and
Solution: We want to find such that
(3.1.1)
2. Verify that is not a linear combination of the vectors and
?
3. The linear span of over is
4.
as , and if , take and
.
LEMMA (Linear Span is a subspace) Let be a vector space and let be a non-
empty subset of Then is a subspace of
Proof. By definition, and hence is non-empty subset of Let
Then, for there exist vectors and scalars such
that and Hence,
Thus, is a vector subspace of
Remark Let be a vector space and be a subspace. If then
is a subspace of as is a vector space in its own right.
THEOREM Let be a non-empty subset of a vector space Then is the smallest subspace of containing
Proof. For every and therefore, To show is the
smallest subspace of containing consider any subspace of containing Then
by Proposition, and hence the result follows.
DEFINITION Let be an matrix with real entries. Then using the rows
and columns we define
1.
2.
3. denoted as
4. denoted
Note that the ``column space" of a matrix consists of all such that has a
solution. Hence,
LEMMA Let be a real matrix. Suppose for some elementary matrix
Then
THEOREM Let be an matrix with real entries. Then
1. is a subspace of ; 2. the non-zero row vectors of a matrix in row-reduced form, forms a basis for the
row-space. Hence
Proof. Part can be easily proved. Let be an matrix. For part let be the
row-reduced form of with non-zero rows Then
for some elementary matrices Then, a repeated
application of Lemma implies That is, if the rows
of the matrix are then
Hence the required result follows.
Linear Independence
DEFINITION (Linear Independence and Dependence) Let be
any non-empty subset of If there exist some non-zero 's such that
then the set is called a linearly dependent set. Otherwise, the set is called linearly independent.
EXAMPLE
1. Let Then check that
Since and
is a solution of, so the set is a linearly dependent subset of
2. Let Suppose there exists such that
Then check that in this case we
necessarily have which shows that the set
is a linearly independent subset of
In other words, if is a non-empty subset of a vector space then to check whether the set is linearly dependent or independent, one needs to consider the equation
(1)
In case is THE ONLY SOLUTION of (1), the set becomes a linearly independent subset of Otherwise, the set becomes a linearly dependent subset of
PROPOSITION Let be a vector space. 1. Then the zero-vector cannot belong to a linearly independent set.
2. If is a linearly independent subset of then every subset of is also linearly independent.
3. If is a linearly dependent subset of then every set containing is also linearly dependent.
Proof. We give the proof of the first part. The reader is required to supply the proof of other parts.
Let be a set consisting of the zero vector. Then for any
Hence, for the system
we have a non-zero solution and
Therefore, the set is linearly dependent.
THEOREM Let be a linearly independent subset of a vector space
Suppose there exists a vector such that the set is
linearly dependent, then is a linear combination of
Proof. Since the set is linearly dependent, there exist scalars
NOT ALL ZERO such that
(2)
CLAIM:
Let if possible Then equation (2) gives with
not all zero. Hence, by the definition of linear independence, the set
is linearly dependent which is contradictory to our hypothesis. Thus,
and we get
Note that for every and hence for Hence the result follows.
9. Consider the vector space Let Find all choices for the vector
such that the set is linear independent subset of Does there exist
choices for vectors and such that the set is linearly independent subset of ?
10. If none of the elements appearing along the principal diagonal of a lower triangular matrix is zero, show that the row vectors are linearly independent in
The same is true for column vectors.
11. Let Determine whether or not
the vector
12. Show that is linearly dependent in
13. Show that is a linearly independent set in In
general if is a linearly independent set then
is also a linearly independent set.
14. In give an example of vectors and such that is linearly
dependent but any set of vectors from is linearly independent. 15. What is the maximum number of linearly independent vectors in
Multiple Choice Questions
1. If α, β are two vectors of a vector space then indicate which one of the following is not a liner combination of α and β
a) 3α-4 β b) α
c) None d) α.β
2. The maximum number of independent vectors among the four (1,2,3,4), (0,4,3,1), (0,0,7,1) and (0,0,0,0) is
a) 4 b) 3
c) 2 d) 1
3. The maximum number of independent vectors in
V = { (x1,x2,x3,x4,x5) : xi Є R} is
a) 4 b) 3
c) 5 d) None
Lecture 3
Bases DEFINITION (Basis of a Vector Space)
1. A non-empty subset of a vector space is called a basis of if 1. is a linearly independent set, and
2. i.e., every vector in can be expressed as a linear combination of the elements of
2. A vector in is called a basis vector.
Remark Let be a basis of a vector space Then any is a
unique linear combination of the basis vectors,
Observe that if there exists a such that and
then
But then the set is linearly independent and therefore the scalars
for must all be equal to zero. Hence, for and we have the uniqueness.
By convention, the linear span of an empty set is Hence, the empty set is a basis of
the vector space
EXAMPLE
1. Check that if then or
or or are bases of
2. For let Then, the set
forms a basis of This set is called the standard basis of
That is, if then the set forms an standard basis of
3. Let be a vector subspace of
Then It can be easily verified that the
vector and
Then by Remark, cannot be a basis of
A basis of can be obtained by the following method:
The condition is equivalent to we replace the value of
with to get
Hence, forms a basis of
4. Let and That is, is a complex vector space.
Note that any element can be written as Hence, a
basis of is
5. Let and That is, is a real vector space. Any
element is expressible as Hence a basis of is
Observe that is a vector in Also, and hence is not defined.
6. Recall the vector space the vector space of all polynomials with real coefficients. A basis of this vector space is the set
This basis has infinite number of vectors as the degree of the polynomial can be any positive integer.
DEFINITION (Finite Dimensional Vector Space) A vector space is said to be finite dimensional if there exists a basis consisting of finite number of elements. Otherwise, the vector space is called infinite dimensional.
Remark We can use the above results to obtain a basis of any finite dimensional vector space as follows:
Step 1: Choose a non-zero vector, say, Then the set is linearly independent.
Step 2: If we have got a basis of Else there exists a vector, say,
such that Then by Corollary, the set is linearly independent.
Step 3: If then is a basis of Else there exists a vector,
say, such that So, by Corollary, the set is linearly independent.
At the step, either or
In the first case, we have as a basis of
In the second case, . So, we choose a vector, say,
such that Therefore, by Corollary, the set
is linearly independent.
This process will finally end as is a finite dimensional vector space.
Assignment
1. Let be a subset of a vector space Suppose
but is not a linearly independent set. Then prove that each vector in can be expressed in more than one way as a linear combination of vectors from
2. Show that the set is a basis of 3. Let be a matrix of rank Then show that the non-zero rows in the row-
reduced echelon form of are linearly independent and they form a basis of the row space of
Multiple Choice Questions
1. The value of k for which the two vectors (k,6) and (2,k) form a basis of V2 is
a) 2√ 3 b) - 2√ 3
c) any value d) any value except ±2√3
2. The vectyors (1,2,3) and (4,-2,7) are
a) linearly independent b) linearly dependent
c) form a basis of V3 d) None
3. The dimension of the sub space {(x1,x2,x3) : xi Є R and 2x1 + x2 – x3= 0} is
a) 2 b) 1
c) 3 d) None
Lecture 4
Basic Properties
DEFINITION (Linear Transformation) Let and be vector spaces over A map is called a linear transformation if
We now give a few examples of linear transformations.
EXAMPLE
1. Define by for all Then is a linear transformation as
2. Verify that the maps given below from to are linear transformations. Let
1. Define
2. For any define
3. For a fixed vector define
Note that examples and can be obtained by assigning particular values for the vector
3. Define by
Then is a linear transformation with and
DEFINITION (Zero Transformation) Let be a vector space and let be the map defined by
Then is a linear transformation. Such a linear transformation is called the zero transformation and is denoted by DEFINITION (Identity Transformation) Let be a vector space and let be the map defined by
Then is a linear transformation. Such a linear transformation is called the Identity transformation and is denoted by
THEOREM Let be a linear transformation and be an ordered basis of Then the linear transformation is a linear combination of the
vectors
In other words, is determined by
Proof. Since is a basis of for any there exist scalars such that
So, by the definition of a linear transformation
Observe that, given we know the scalars Therefore, to know
we just need to know the vectors in
That is, for every is determined by the coordinates of
with respect to the ordered basis and the vectors
DEFINITION (Inverse Linear Transformation) Let be a linear transformation. If the map is one-one and onto, then the map defined by
is called the inverse of the linear transformation
EXAMPLE
1. Define by Then is defined by
Note that
Hence, the identity transformation. Verify that Thus,
the map is indeed the inverse of the linear transformation
2. Recall the vector space and the linear transformation
defined by
for Then is defined as
for Verify that Hence, conclude that the map is indeed the inverse of the linear transformation
Assignment 1. Which of the following are linear transformations Justify your
answers.
1. Let and with
2. Let with
3. Let with 4. Let and with
Lecture 5
Rank-Nullity Theorem
DEFINITION (Range and Null Space) Let be finite dimensional vector spaces over the same set of scalars and be a linear transformation. We define
1. and
2.
We now prove some results associated with the above definitions.
PROPOSITION Let and be finite dimensional vector spaces and let
be a linear transformation. Suppose that is an ordered basis of Then
1.
1. is a subspace of
2.
3.2.
1. is a subspace of
2.
3. is one-one is the zero subspace of
is a basis of
4. if and only if
Proof. The results about and can be easily proved. We thus leave the proof for the readers.
We now assume that is one-one. We need to show that
Let Then by definition, Also for any linear transformation,
Thus So, is one-one implies That is,
Let We need to show that is one-one. So, let us assume that for some
Then, by linearity of This implies,
This in turn implies Hence, is one-one.
The other parts can be similarly proved.
Remark
1. The space is called the RANGE SPACE of and is called the NULL SPACE of
2. We write and
3. is called the rank of the linear transformation and is called the nullity of
EXAMPLE Determine the range and null space of the linear transformation
Solution: By Definition We therefore have
Also, by definition
THEOREM (Rank Nullity Theorem) Let be a linear transformation and be a finite dimensional vector space. Then
or equivalently
Proof. Let and Suppose is a basis of
Since is a linearly independent set in we can extend it to form
a basis of . So, there exist vectors such that
is a basis of Therefore,
We now prove that the set is linearly independent. Suppose the set is not linearly independent. Then, there exists scalars,
not all zero such that
That is,
So, by definition of
Hence, there exists scalars such that
That is,
But the set is a basis of and so linearly independent. Thus by definition of linear independence
In other words, we have shown that is a basis of Hence,
COROLLARY Let be a linear transformation on a finite dimensional vector space Then
Proof. By Proposition, is one-one if and only if By the rank-nullity
Theorem is equivalent to the condition Or equivalently is onto.
By definition, is invertible if is one-one and onto. But we have shown that is one-one if and only if is onto. Thus, we have the last equivalent condition.
Remark Let be a finite dimensional vector space and let be a linear transformation. If either is one-one or is onto, then is invertible.
The following are some of the consequences of the rank-nullity theorem. The proof is left as an exercise for the reader.
COROLLARY The following are equivalent for an real matrix
1.2. There exist exactly rows of that are linearly independent. 3. There exist exactly columns of that are linearly independent.
4. There is a sub matrix of with non-zero determinant and every
sub matrix of has zero determinant. 5. The dimension of the range space of is 6. There is a subset of consisting of exactly linearly independent vectors
such that the system for is consistent.
7. The dimension of the null space of
Assignment
1. Let be a linear transformation and let be
linearly independent in Prove that is linearly independent.
2. Let be defined by
Then the vectors and are linearly independent whereas and
are linearly dependent.
3. Is there a linear transformation
4. Let be defined by
1. Find for
2. Find and Also calculate and 3. Show that and find the matrix of the linear transformation with
respect to the standard basis.
5. Let be a linear transformation.
1. If is finite dimensional then show that the null space and the range space of are also finite dimensional.
2. If and are both finite dimensional then show that
1. if then is onto.
2. if then is not one-one.
6. Let be an real matrix. Then
3. if then the system has infinitely many solutions,
4. if then there exists a non-zero vector such that the system does not have any solution.
7. Let be a vector space of dimension and let be an
ordered basis of . Suppose and let
. Put . Then prove that is a basis of if and only if the matrix is invertible.
8. Let be an matrix. Prove that
Lecture 6
Definition
In , given two vectors , we know the inner product
. Note that for any and , this inner product satisfies the conditions
and if and only if . Thus, we are motivated to define an inner product on an arbitrary vector space.
DEFINITION (Inner Product) Let be a vector space over An inner product
over denoted by is a map,
such that for and
1.
2. the complex conjugate of and
3. for all and equality holds if and only if
DEFINITION (Inner Product Space) Let be a vector space with an inner product
Then is called an inner product space, in short denoted by IPS.
EXAMPLE The first two examples given below are called the STANDARD INNER
PRODUCT or the DOT PRODUCT on and respectively.. 1. Let be the real vector space of dimension Given two vectors
and of we define
Verify is an inner product.
2. Let be a complex vector space of dimension Then for
and in check that
is an inner product.
3. Let and let Define Check that is
an inner product. Hint: Note that
4. let Show that
is an inner
product in 5. Consider the real vector space . In this example, we define three products that
satisfy two conditions out of the three conditions for an inner product. Hence the three products are not inner products.
1. Define Then it is easy to verify that the third condition is not valid whereas the first two conditions are valid.
2. Define Then it is easy to verify that the first condition is not valid whereas the second and third conditions are valid.
3. Define Then it is easy to verify that the second condition is not valid whereas the first and third conditions are valid.
DEFINITION (Length/Norm of a Vector) For we define the length (norm) of
denoted by the positive square root.
A very useful and a fundamental inequality concerning the inner product is due to Cauchy and Schwartz. The next theorem gives the statement and a proof of this inequality.
THEOREM (Cauchy-Schwartz inequality) Let be an inner product space. Then
for any
The equality holds if and only if the vectors and are linearly dependent. Further, if
, then DEFINITION 5.1.8 (Angle between two vectors) Let be a real vector space. Then
for every by the Cauchy-Schwartz inequality, we have
We know that is an one-one and onto function. Therefore, for
every real number there exists a unique such that
1. The real number with and satisfying is called the angle between the two vectors and in
2. The vectors and in are said to be orthogonal if
3. A set of vectors is called mutually orthogonal if
for all
DEFINITION (Orthogonal Complement) Let be a subspace of a vector space
with inner product . Then the subspace
is called the orthogonal complement of in
THEOREM Let be an inner product space. Let be a set of non-zero, mutually orthogonal vectors of
1. Then the set is linearly independent.
2.
3. Let and also let for Then for any
In particular, for all if and only if
Therefore, we have obtained the required result.
DEFINITION (Orthonormal Set) Let be an inner product space. A set of non-zero,
mutually orthogonal vectors in is called an orthonormal set if
for
If the set is also a basis of then the set of vectors is called an orthonormal basis of
EXAMPLE 1. Consider the vector space with the standard inner product. Then the standard
ordered basis is an orthonormal set. Also, the basis
is an orthonormal set.
Assignment
1. Recall the following inner product on for and
1. Find the angle between the vectors and
2. Let Find such that
3. Find two vectors such that and 2. Find an inner product in such that the following conditions hold:
[Hint: Consider a symmetric matrix Define and
solve a system of equations for the unknowns .]
3. Let Find with respect to the standard inner product.
4. Let be a subspace of a finite dimensional inner product space . Prove that
5. Let be the real vector space of all continuous functions with domain
That is, Then show that is an inner product space with inner
product
For different values of and find the angle between the functions
and
6. Let be an inner product space. Prove that
This inequality is called the TRIANGLE INEQUALITY.
7. Let Use the Cauchy-Schwartz inequality to prove that
When does the equality hold?
Lecture 7
Definitions
In this chapter, the linear transformations are from a given finite dimensional vector space to itself. Observe that in this case, the matrix of the linear transformation is a square matrix. So, in this chapter, all the matrices are square matrices and a vector
means for some positive integer
Let be a matrix of order In general, we ask the question:
For what values of there exist a non-zero vector such that
(1)
Here, stands for either the vector space over or over Equation (1) is equivalent to the equation
This system of linear equations has a non-zero solution, if
So, to solve (1), we are forced to choose those values of for which
Observe that is a polynomial in of degree We are therefore lead to the following definition.
DEFINITION (Characteristic Polynomial) Let be a matrix of order The
polynomial is called the characteristic polynomial of and is denoted by
The equation is called the characteristic equation of If is a
solution of the characteristic equation then is called a characteristic value of
Some books use the term EIGEN VALUE in place of characteristic value.
THEOREM Let Suppose is a root of
the characteristic equation. Then there exists a non-zero such that
Proof. Since is a root of the characteristic equation, This shows
that the matrix is singular and therefore by Theorem the linear system
Remark Observe that the linear system has a solution for every
So, we consider only those that are non-zero and are solutions of the linear system
DEFINITION (Eigen value and Eigenvector) If the linear system has a non-
zero solution for some then
1. is called an eigen value of
2. is called an eigenvector corresponding to the eigen value of and
3. the tuple is called an eigen pair.
Remark To understand the difference between a characteristic value and an eigen value, we give the following example.
Consider the matrix Then the characteristic polynomial of is
Given the matrix recall the linear transformation defined by
1. If that is, if is considered a COMPLEX matrix, then the roots of
in are So, has and as eigen pairs.
2. If that is, if is considered a REAL matrix, then has no solution
in Therefore, if then has no eigen value but it has as characteristic values.
Remark Note that if is an eigen pair for an matrix then for any non-zero
is also an eigen pair for Similarly, if are
eigenvectors of corresponding to the eigen value then for any non-zero
it is easily seen that if , then is also an eigenvector of corresponding to the eigen value Hence, when we talk of
eigenvectors corresponding to an eigen value we mean linearly independent eigenvectors.
Suppose is a root of the characteristic equation Then
is singular and Suppose Then
by Corollary, the linear system has linearly independent
solutions. That is, has linearly independent eigenvectors corresponding to the
eigen value whenever
EXAMPLE
1. Let Then Hence, the characteristic
equation has roots That is is a repeated eigen value. Now check that the
equation for is equivalent to the equation
And this has the solution Hence, from the above remark, is a representative for the eigenvector. Therefore, here we have two eigen values mathend000# but only one eigenvector.
2. Let Then The characteristic equation
has roots Here, the matrix that we have is and we know that for
every and we can choose any two linearly independent vectors
from to get and as the two eigen pairs.
In general, if are linearly independent vectors in then
are eigen pairs for the identity matrix,
3. Let Then The characteristic
equation has roots Now check that the eigen pairs are and
In this case, we have two distinct eigen values and the corresponding eigenvectors are also linearly independent. The reader is required to prove the linear independence of the two eigenvectors.
4. Let Then The characteristic
equation has roots Hence, over the matrix has no eigen value.
Over the reader is required to show that the eigen pairs are and
Assignment1. Find the eigen values of a triangular matrix.
2. Find eigen pairs over for each of the following matrices:
and
3. Prove that the matrices and have the same set of eigen values. Construct a
matrix such that the eigenvectors of and are different. 4. Let be a matrix such that ( is called an idempotent matrix). Then
prove that its eigen values are either 0 or or both.
Lecture 8
THEOREM 6.1.11 Let be an matrix with eigen values
not necessarily distinct. Then and
EXERCISE
1. Let be a skew symmetric matrix of order Then prove that 0 is an eigen value of
2. Let be a orthogonal matrix .If , then prove that
there exists a non-zero vector such that
Let be an matrix. Then in the proof of the above theorem, we observed that the
characteristic equation is a polynomial equation of degree in Also,
for some numbers it has the form
Note that, in the expression is an element of Thus, we can only substitute by elements of
It turns out that the expression
holds true as a matrix identity. This is a celebrated theorem called the Cayley Hamilton Theorem. We state this theorem without proof and give some implications.
THEOREM (Cayley Hamilton Theorem) Let be a square matrix of order Then satisfies its characteristic equation. That is,
holds true as a matrix identity.
Some of the implications of Cayley Hamilton Theorem are as follows.
Remark
1. Let Then its characteristic polynomial is Also, for
the function, and This shows that the
condition for each eigen value of does not imply that 2. Suppose we are given a square matrix of order and we are interested in
calculating where is large compared to Then we can use the division
algorithm to find numbers and a polynomial such that
3.Hence, by the Cayley Hamilton Theorem,
4.
5. Let be a non-singular matrix of order Then note that and
This matrix identity can be used to calculate the inverse.
Note that the vector (as an element of the vector space of all matrices) is a linear
combination of the vectors
EXERCISE Find inverse of the following matrices by using the Cayley Hamilton Theorem
THEOREM If are distinct eigen values of a matrix with
corresponding eigenvectors then the set is linearly independent.
Proof. The proof is by induction on the number of eigen values. The result is obviously true if as the corresponding eigenvector is non-zero and we know that any set containing exactly one non-zero vector is linearly independent.
Let the result be true for We prove the result for We consider the equation
(1)
for the unknowns We have
(2)
From Equations (1) and (2), we get
This is an equation in eigenvectors. So, by the induction hypothesis, we have
But the eigen values are distinct implies for We therefore
get for Also, and therefore (1) gives
Thus, we have the required result.
We are thus lead to the following important corollary.
COROLLARY The eigenvectors corresponding to distinct eigen values of an matrix are linearly independent.
Assignment
1. For an matrix prove the following.
1. and have the same set of eigen values.
2. If is an eigen value of an invertible matrix then is an eigen value of
3. If is an eigen value of then is an eigen value of for any positive integer
4. If and are matrices with nonsingular then and have the same set of eigen values.
In each case, what can you say about the eigenvectors?
2. Let and be matrices for which and 1. Do and have the same set of eigen values? 2. Give examples to show that the matrices and need not be similar.
3. Let be an eigen pair for a matrix and let be an eigen pair for another matrix
1. Then prove that is an eigen pair for the matrix
2. Give an example to show that if are respectively the eigen values of
and then need not be an eigen value of
Lecture 9
Diagonalisation
DEFINITION (Matrix Diagonalisation) A matrix is said to be diagonalisable if there exists a non-singular matrix such that is a diagonal matrix.
Remark Let be an diagonalisable matrix with eigen values By
definition, is similar to a diagonal matrix Observe that as similar matrices have the same set of eigen values and the eigen values of a diagonal matrix are its diagonal entries.
EXAMPLE Let Then we have the following: 1. Let Then has no real eigen value and hence doesn't have
eigenvectors that are vectors in Hence, there does not exist any non-singular
real matrix such that is a diagonal matrix.
2. In case, the two complex eigen values of are and the
corresponding eigenvectors are and respectively. Also, and
can be taken as a basis of Define a complex matrix by
Then
THEOREM let be an matrix. Then is diagonalisable if and only if has linearly independent eigenvectors.
COROLLARY 6.2.5 let be an matrix. Suppose that the eigen values of are distinct. Then is diagonalisable.
EXAMPLE
1. Let Then Hence, has
eigen values It is easily seen that and are the only eigen pairs. That is, the matrix has exactly one eigenvector
corresponding to the repeated eigen value Hence, by Theorem the matrix is not diagonalisable.
2. Let Then Hence, has
eigen values It can be easily verified that and
correspond to the eigen value and corresponds to the eigen value
Note that the set consisting of eigenvectors corresponding to the eigen value are not orthogonal. This set can be replaced by
the orthogonal set which still consists of eigenvectors
corresponding to the eigen value as . Also,
the set forms a basis of So, by Theorem , the
matrix is diagonalisable. Also, if is the
corresponding unitary matrix then
Observe that the matrix is a symmetric matrix. In this case, the eigenvectors are
mutually orthogonal. In general, for any real symmetric matrix there always exist eigenvectors and they are mutually orthogonal. This result will be proved later.
Assignment1. By finding the eigen values of the following matrices, justify whether or not
for some real non-singular matrix and a real diagonal matrix
for any with
2. Are the two matrices and diagonalisable?
3. Find the eigen values and eigenvectors of , where if and otherwise.
4. Let be an matrix and an matrix. Suppose Then show that is diagonalisable if and only if both and are diagonalisable.
5. Let be a linear transformation with and
Then
1. determine the eigen values of 2. find the number of linearly independent eigenvectors corresponding to
each eigen value? 3. is diagonalisable? Justify your answer.
6. Let be a non-zero square matrix such that Show that cannot be diagonalised.