Unit - 3

Vector Spaces

Q1) Are linearly independent?

A1) Independent if 

Has only the trivial solution 

Q2) Check whether the vectors are independent.

A2) Calculate the coefficients in which a linear combination of these vectors vector.

This vector equation can be written as a system of linear equations

Solve this system using the gauss method

From 2 row we subtract the 1-th row; from 3row we subtract the 1-th row:

From 1 row we subtract the 2row; for 3 row add 2 row:

This solution shows that the system has many solutions, i.e exist nonzero combination c such that the linear combination of is equal to the zero vector,

For Example, 

Means vectors are linearly dependent.

Answer: Vectors are linearly dependent.

Q3) Is the set S = {(1,1),(1,-1)} a basis for

1. Does S span       2. IS S linearly independent?

A3) Solve

This system is consistent for every x and y, therefore S spans

2.IS S linearly independent?

Solve

The system has a unique solution (trivial solution)

Therefore, S is linearly independent.

Consequently, S is a basis for

Q4)

A4) Is a basis for the vector space

1.+

Is equivalent to:

Which is consistent for every a,b,c and  d.

Therefore, S spans

2.+  is equivalent to:

The system has only the trivial solution is linearly independent.

Consequently, S is a basis for

Q5) Let L: V. Then ker L is a subspace of V.

A5) Proof: Notice that if L(v)=0 and L(u)=0, then for any constants c,d, L(cu+dv)=0.

Then by the subspace theorem, the kernel of L is a subspace of V.

This theorem has a nice interpretation in terms of the eigenspace of L. Suppose L has a zero eigenvalue. Then the associated eigenspace consists of all vectors v such that Lv=0v=0; in other words, the 0-eigenspace of L is exactly the kernel of L

Returning to the previous example, let L(x,y) =(x+y,x+2y,y). L is clearly not surjective, Since L sends to a plane in .

Notice that if x = L(v) and y = L(v), then for any constants c,d then cx+dy=L(cx+du). Then the subspace theorem strikes again, and we have the following theorem.

Q6) defined by is linear. Describe its kernel and range and give the dimension of each.

A6) It should be clear that  if, and only if, b= -a and d= -c. The kernel of T is therefore all matrices of the form

The two matrices and are not scalar multiples of each other, so they must be linearly independent. Therefore, the dimension of ker(T) is two.

Now suppose that we have any vector Clearly . So, the range of T of Thus the dimension of ran(T)is two.

Q7) defined by T(ax+b) = 2bx-a is linear. Describe its kernel and range and give the dimension of each.

A7) T(ax+b) =2bx-a=0 if, and only if, both a and b are zero. Therefore, the kernel of T is only the zero polynomial. By definition, the dimension of the subspace consisting of only the zero vector is zero, so ker(T) has dimension zero.

Suppose that we take a random polynomial cx+d in the codomain. If we consider the polynomial -dx+ in the domain we see that

This shows that the range of T is all of it has dimension two.

Q8) Is the given, matrix Invertible

A8)

A fail to be invertible, since rref(A)

Q9) Given 33 Rectangular Matrix 

A9) The augumented matrix is as follows

After applying the Gauss-Jordan elimination method:

The inverse of a matrix is as follows,

Q10) Find the inverse of

A10)

Step 1: Adjoin the identity matrix to the right side of A:

Step 2: Apply row operations to this matrix until the left side is reduced to I. The computations are:

R2  R2-R1  , R3R3-R1

R3 R3 + 2R2

R1R1 -3R3  , R2R2+3R3

R1R1-2R2

Step 3: Conclusion: The inverse matrix is:

Q11) Find the inverse of matrix A given below:

A11)   Let A = IA

Or

Thus, the inverse of matrix A is given by I =

Therefore,

Q12) Theorem: Let () be a basis of V and () an arbitrary list of vectors in W. Then there exists a unique linear map

A12) Proof: First we verify that there is at most one linear map T with . Take any Since () is a basis of V there are unique scalars such that . By linearity we must have

And hence T(v) is completely determined. To show existence, use (3) to define T. It remains to show that this T is linear and that. These two conditions are not hard to show and are left to the reader.

The set of linear maps is itself a vector space. For S,T addition is defined as

(S+T)v = S v + Tv

For a F an T scalar multiplication is defined as

(aT)(v) = a(Tv) 

We should verify that S+T and aT are indeed linear maps again and that all properties of a vector space is satisfied.

I.e., associavity, identity and distributive property. Hence by the above properties we conclude that the given transformation is linear transformation.

Q13) Consider a linear map Tv: which rotates vectors be an angle 0 . From a known statement we have the corresponding matrix for this linear transformation is given below, solve for linear mapping

A13) the composition of two rotations b angles are given by v and v’ that is

TV’TV is clearly the rotation TV+V’ hence we must have

Av+v’ = Av’Av

The left hand side of this equation is given by

And the right hand-side is,

=

Comparing corresponding entries we find the claim that Tv’Tv = Tv+v’ is equivalent to well-known addition formulas

Cos(v+v’) = cosv. Cosv’ – sinv. Sinv’

Sin(v+v’) = cosv’. Sinv +cosv. Sinv’.

Q14) Let T L be defined by T(p(x)) = 2xp’(x), and let B1 = be a basis of (R). Determine

A14) First we will calculate the images of the basis vectors in B1 under the linear transformation T.

T(1) = 2x(1)’ = 2x(0) = 0(1) + 0(x) + 0(x2) + 0(x3) + 0(x4)

T(x) = 2x(x)’ = 2x(1) = 0(1) +0(x) + 0(x2) + 0(x3) + 0(x4)

T(x2) = 2x(x2)’ = 2x(2x) = 0(1) +0(x) + 0(x2) + 0(x3) + 0(x4)

T(x3) = 2x(x3)’ = 2x(3x3) = 0(1) +0(x) + 0(x2) + 0(x3) + 0(x4)

T(x4) = 2x(x4)’ = 2x (4x3) = 0(1) +0(x) + 0(x2) + 0(x3) + 0(x4)

Therefore, we can now construct our matrix as follows.

The first column of corresponds to the co-efficients determinend by T(1).

The second column corresponds to the co-efficients determined by T(x) and so fourth

  =

Q15) Let T L(M22, M22) be defined by T = for a,b,c,d R and let B1 = be a basis of M22.Determine

A15) we will first calculate the images of our basic vectors under T.

T = = 0 + 0 + 0 + 1

T = = 0 -1 + 0 + 0

T = = 0 + 0 -1 + 0

T = = 1 + 0 + 0 +0

Therefore, we construct

=

Q16) Prove that

A16)

We have

We can write

So that,

Now V is an abelian group with respect to addition.

So that by right cancellation law, we get-

Q17) The necessary and sufficient conditions for a non-empty sub-set W of a vector space V(F) to be a subspace of V are-

1.

2.

A17)

Necessary conditions-

W is an abelian group with respect to vector addition If W is a subspace of V.

So that

Here W must be closed under a scalar multiplication so that the second condition is also necessary.

Sufficient conditions-

Let W is a non-empty subset of V satisfying the two given conditions.

From first condition-

So that we can say that zero vector of V belongs to W. It is a zero vector of W as well.

Now

So that the additive inverse of each element of W is also in W.

So that-

Thus W is closed with respect to vector addition.

Q18) If the set S = of vector V(F) is linearly independent, then none of the vectors can be zero vector.

A18)

Let be equal to zero vector where

Then

For any in F.

Here therefore from this relation we can say that S is linearly dependent. This is the contradiction because it is given that S is linearly independent.

Hence none of the vectors can be zero.

We can conclude that a set of vectors which contains the zero vector is necessarily linear dependent.