Unit 2C | unit 2c vector spaces

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Unit - 2C

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

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 , R3 R3-R1

R3 R3 + 2R2

R1R1 -3R3 , R2 R2+3R3

R1R1-2R2

Step 3: Conclusion: The inverse matrix is:

A-1 =

Q11) Find the inverse of matrix A given below:

A11) Let A = IA

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