Algebra

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Q1. Obtain Smith normal form over the field of rational numbers and find invariant factors of the matrix
[ 2 0 10 0 3 4 1 2 1 ] 2 0 10 0 3 4 1 2 1 [[-2,0,10],[0,-3,-4],[1,2,-1]]\left[\begin{array}{ccc} -2 & 0 & 10 \\ 0 & -3 & -4 \\ 1 & 2 & -1 \end{array}\right][2010034121]
Answer:
To find the Smith normal form and the invariant factors of the given matrix over the field of rational numbers, we will follow these steps:
  1. Introduce the Matrix: We are given the matrix
    A = [ 2 0 10 0 3 4 1 2 1 ] A = 2 0 10 0 3 4 1 2 1 A=[[-2,0,10],[0,-3,-4],[1,2,-1]]A = \left[\begin{array}{ccc} -2 & 0 & 10 \\ 0 & -3 & -4 \\ 1 & 2 & -1 \end{array}\right]A=[2010034121]
    Our goal is to transform this matrix into its Smith normal form, which is a diagonal matrix where each diagonal element divides the next one.
  2. Compute the Smith Normal Form: We will use row and column operations to transform the matrix into its Smith normal form. These operations include adding multiples of one row to another, swapping rows, adding multiples of one column to another, and swapping columns. The operations are performed over the field of rational numbers.
  3. Find Invariant Factors: The diagonal elements of the Smith normal form are the invariant factors of the matrix. These factors give us information about the structure of modules over a principal ideal domain related to the matrix.
Let’s start by computing the Smith normal form of the matrix A A AAA. After obtaining the Smith normal form, we will identify the invariant factors.

Step 1: Initial Matrix

Start with the matrix A A AAA:
A = [ 2 0 10 0 3 4 1 2 1 ] A = 2 0 10 0 3 4 1 2 1 A=[[-2,0,10],[0,-3,-4],[1,2,-1]]A = \left[\begin{array}{ccc} -2 & 0 & 10 \\ 0 & -3 & -4 \\ 1 & 2 & -1 \end{array}\right]A=[2010034121]

Step 2: Making the First Pivot

We want a positive integer in the top-left corner. We can swap the first and third rows:
[ 1 2 1 0 3 4 2 0 10 ] 1 2 1 0 3 4 2 0 10 [[1,2,-1],[0,-3,-4],[-2,0,10]]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & -3 & -4 \\ -2 & 0 & 10 \end{array}\right][1210342010]

Step 3: Clearing the First Column

Next, we clear the first column below the pivot (1). We can add twice the first row to the third row:
[ 1 2 1 0 3 4 0 4 8 ] 1 2 1 0 3 4 0 4 8 [[1,2,-1],[0,-3,-4],[0,4,8]]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & -3 & -4 \\ 0 & 4 & 8 \end{array}\right][121034048]

Step 4: Working on the Second Pivot

We need to make the element in the second row and second column the next pivot. We can swap the second and third rows:
[ 1 2 1 0 4 8 0 3 4 ] 1 2 1 0 4 8 0 3 4 [[1,2,-1],[0,4,8],[0,-3,-4]]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & 4 & 8 \\ 0 & -3 & -4 \end{array}\right][121048034]

Step 5: Clearing the Second Column

We clear the second column below the pivot (4). We can add the second row to the third row:
[ 1 2 1 0 4 8 0 1 4 ] 1 2 1 0 4 8 0 1 4 [[1,2,-1],[0,4,8],[0,1,4]]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & 4 & 8 \\ 0 & 1 & 4 \end{array}\right][121048014]

Step 6: Making the Third Pivot

We need to make the element in the third row and third column the next pivot. We can swap the second and third rows:
[ 1 2 1 0 1 4 0 4 8 ] 1 2 1 0 1 4 0 4 8 [[1,2,-1],[0,1,4],[0,4,8]]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & 1 & 4 \\ 0 & 4 & 8 \end{array}\right][121014048]

Step 7: Clearing the Rest of the Matrix

We clear the rest of the matrix around the pivots. This involves making the non-diagonal elements zero, while ensuring the diagonal elements divide each other.
[ 1 2 1 0 1 4 0 4 8 ] R 1 R 1 2 × R 2 = [ 1 0 9 0 1 4 0 4 8 ] R 3 R 3 4 × R 2 = [ 1 0 9 0 1 4 0 0 8 ] 1 2 1 0 1 4 0 4 8 R 1 R 1 2 × R 2 = 1 0 9 0 1 4 0 4 8 R 3 R 3 4 × R 2 = 1 0 9 0 1 4 0 0 8 {:[[[1,2,-1],[0,1,4],[0,4,8]]],[R_(1)larrR_(1)-2xxR_(2)],[=[[1,0,-9],[0,1,4],[0,4,8]]],[R_(3)larrR_(3)-4xxR_(2)],[=[[1,0,-9],[0,1,4],[0,0,-8]]]:}\begin{aligned} & {\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & 1 & 4 \\ 0 & 4 & 8 \end{array}\right]} \\ & R_1 \leftarrow R_1-2 \times R_2 \\ & =\left[\begin{array}{ccc} 1 & 0 & -9 \\ 0 & 1 & 4 \\ 0 & 4 & 8 \end{array}\right] \\ & R_3 \leftarrow R_3-4 \times R_2 \\ & =\left[\begin{array}{lll} 1 & 0 & -9 \\ 0 & 1 & 4 \\ 0 & 0 & -8 \end{array}\right] \\ \end{aligned}[121014048]R1R12×R2=[109014048]R3R34×R2=[109014008]
[ 1 0 9 0 1 4 0 0 8 ] 1 0 9 0 1 4 0 0 8 [[1,0,-9],[0,1,4],[0,0,-8]]\left[\begin{array}{ccc} 1 & 0 & -9 \\ 0 & 1 & 4 \\ 0 & 0 & -8 \end{array}\right][109014008]

Operation 1: R 3 R 3 1 R 3 R 3 1 R3rarr(R3)/(-1)R3 \to \frac{R3}{-1}R3R31

We divide the third row by 1 1 -1-11. This operation changes the third row ( R 3 R 3 R3R3R3):
R 3 = R 3 1 = 1 1 [ 0 , 0 , 8 ] = [ 0 , 0 , 8 ] R 3 = R 3 1 = 1 1 0 , 0 , 8 = 0 , 0 , 8 R3=(R3)/(-1)=(1)/(-1)*[0,0,-8]=[0,0,8]R3 = \frac{R3}{-1} = \frac{1}{-1} \cdot \left[0, 0, -8\right] = \left[0, 0, 8\right]R3=R31=11[0,0,8]=[0,0,8]
The matrix becomes:
[ 1 0 9 0 1 4 0 0 8 ] 1 0 9 0 1 4 0 0 8 [[1,0,-9],[0,1,4],[0,0,8]]\left[\begin{array}{ccc} 1 & 0 & -9 \\ 0 & 1 & 4 \\ 0 & 0 & 8 \end{array}\right][109014008]

Operation 2: R 1 R 1 + 9 R 3 8 R 1 R 1 + 9 R 3 8 R1rarr R1+(9R3)/(8)R1 \to R1 + \frac{9R3}{8}R1R1+9R38

We add 9 8 9 8 (9)/(8)\frac{9}{8}98 times the third row to the first row. This operation changes the first row ( R 1 R 1 R1R1R1):
R 1 = R 1 + 9 8 R 3 = [ 1 , 0 , 9 ] + 9 8 [ 0 , 0 , 8 ] = [ 1 , 0 , 9 + 9 ] = [ 1 , 0 , 0 ] R 1 = R 1 + 9 8 R 3 = 1 , 0 , 9 + 9 8 0 , 0 , 8 = 1 , 0 , 9 + 9 = 1 , 0 , 0 R1=R1+(9)/(8)*R3=[1,0,-9]+(9)/(8)*[0,0,8]=[1,0,-9+9]=[1,0,0]R1 = R1 + \frac{9}{8} \cdot R3 = \left[1, 0, -9\right] + \frac{9}{8} \cdot \left[0, 0, 8\right] = \left[1, 0, -9 + 9\right] = \left[1, 0, 0\right]R1=R1+98R3=[1,0,9]+98[0,0,8]=[1,0,9+9]=[1,0,0]
The matrix becomes:
[ 1 0 0 0 1 4 0 0 8 ] 1 0 0 0 1 4 0 0 8 [[1,0,0],[0,1,4],[0,0,8]]\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 8 \end{array}\right][100014008]

Operation 3: R 2 R 2 R 3 2 R 2 R 2 R 3 2 R2rarr R2-(R3)/(2)R2 \to R2 – \frac{R3}{2}R2R2R32

We subtract 1 2 1 2 (1)/(2)\frac{1}{2}12 times the third row from the second row. This operation changes the second row ( R 2 R 2 R2R2R2):
R 2 = R 2 1 2 R 3 = [ 0 , 1 , 4 ] 1 2 [ 0 , 0 , 8 ] = [ 0 , 1 , 4 4 ] = [ 0 , 1 , 0 ] R 2 = R 2 1 2 R 3 = 0 , 1 , 4 1 2 0 , 0 , 8 = 0 , 1 , 4 4 = 0 , 1 , 0 R2=R2-(1)/(2)*R3=[0,1,4]-(1)/(2)*[0,0,8]=[0,1,4-4]=[0,1,0]R2 = R2 – \frac{1}{2} \cdot R3 = \left[0, 1, 4\right] – \frac{1}{2} \cdot \left[0, 0, 8\right] = \left[0, 1, 4 – 4\right] = \left[0, 1, 0\right]R2=R212R3=[0,1,4]12[0,0,8]=[0,1,44]=[0,1,0]
The final matrix after all operations is:
[ 1 0 0 0 1 0 0 0 8 ] 1 0 0 0 1 0 0 0 8 [[1,0,0],[0,1,0],[0,0,8]]\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 8 \end{array}\right][100010008]
This matrix is now in Smith normal form.
The Smith normal form of the given matrix is obtained as follows:
  1. Smith Normal Form: The Smith normal form of the matrix A A AAA is given by
    Smith Normal Form ( A ) = [ 1 0 0 0 1 0 0 0 8 ] Smith Normal Form ( A ) = 1 0 0 0 1 0 0 0 8 “Smith Normal Form”(A)=[[1,0,0],[0,1,0],[0,0,8]]\text{Smith Normal Form}(A) = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 8 \end{array}\right]Smith Normal Form(A)=[100010008]
    This is a diagonal matrix where each diagonal element divides the next one.
  2. Invariant Factors: The diagonal elements of the Smith normal form are the invariant factors of the matrix. For our matrix A A AAA, the invariant factors are 1 1 111, 1 1 111, and 8 8 888. These factors provide insights into the structure of modules related to the matrix over a principal ideal domain.
In summary, the Smith normal form of the matrix A A AAA is a diagonal matrix with elements 1 1 111, 1 1 111, and 8 8 888, and these are also the invariant factors of the matrix.
Q2. Prove that all the characteristic values of a nilpotent operator are zero.
Answer:
To prove that all the characteristic values (eigenvalues) of a nilpotent operator are zero, we need to understand the definitions and properties of nilpotent operators and eigenvalues.

Definitions:

  1. Nilpotent Operator: An operator T T TTT on a vector space is said to be nilpotent if there exists some positive integer k k kkk such that T k = 0 T k = 0 T^(k)=0T^k = 0Tk=0, where 0 0 000 is the zero operator. This means that when you apply T T TTT repeatedly k k kkk times to any vector in the space, you end up with the zero vector.
  2. Eigenvalues (Characteristic Values): An eigenvalue of an operator T T TTT is a scalar λ λ lambda\lambdaλ such that there exists a non-zero vector v v vvv (eigenvector) for which T ( v ) = λ v T ( v ) = λ v T(v)=lambda vT(v) = \lambda vT(v)=λv.

Proof:

We need to show that if T T TTT is a nilpotent operator, then any eigenvalue λ λ lambda\lambdaλ of T T TTT must be zero.
Let λ λ lambda\lambdaλ be an eigenvalue of T T TTT and v v vvv be a corresponding non-zero eigenvector. This means T ( v ) = λ v T ( v ) = λ v T(v)=lambda vT(v) = \lambda vT(v)=λv.
Now, consider T k T k T^(k)T^kTk where k k kkk is the smallest positive integer such that T k = 0 T k = 0 T^(k)=0T^k = 0Tk=0 (the definition of a nilpotent operator).
Apply T k T k T^(k)T^kTk to v v vvv:
T k ( v ) = 0 T k ( v ) = 0 T^(k)(v)=0T^k(v) = 0Tk(v)=0
Since T ( v ) = λ v T ( v ) = λ v T(v)=lambda vT(v) = \lambda vT(v)=λv, we can write:
T k ( v ) = T k 1 ( T ( v ) ) = T k 1 ( λ v ) = λ T k 1 ( v ) T k ( v ) = T k 1 ( T ( v ) ) = T k 1 ( λ v ) = λ T k 1 ( v ) T^(k)(v)=T^(k-1)(T(v))=T^(k-1)(lambda v)=lambdaT^(k-1)(v)T^k(v) = T^{k-1}(T(v)) = T^{k-1}(\lambda v) = \lambda T^{k-1}(v)Tk(v)=Tk1(T(v))=Tk1(λv)=λTk1(v)
Continuing this process, we apply T T TTT repeatedly:
= λ 2 T k 2 ( v ) = = λ k v = λ 2 T k 2 ( v ) = = λ k v =lambda^(2)T^(k-2)(v)=dots=lambda ^(k)v= \lambda^2 T^{k-2}(v) = \ldots = \lambda^k v=λ2Tk2(v)==λkv
Since T k = 0 T k = 0 T^(k)=0T^k = 0Tk=0, we have:
λ k v = 0 λ k v = 0 lambda ^(k)v=0\lambda^k v = 0λkv=0
But v v vvv is a non-zero vector. Therefore, for this equation to hold, it must be that λ k = 0 λ k = 0 lambda ^(k)=0\lambda^k = 0λk=0. Since λ λ lambda\lambdaλ is a scalar, the only way for λ k λ k lambda ^(k)\lambda^kλk to be zero for some positive integer k k kkk is for λ λ lambda\lambdaλ itself to be zero.

Conclusion:

Thus, we have shown that for a nilpotent operator T T TTT, any eigenvalue λ λ lambda\lambdaλ must be zero. This completes the proof that all characteristic values of a nilpotent operator are zero.
Q3. Prove or disprove: Every linearly independent subset of a finite dimensional vector space can be extended to a basis of V V V\mathrm{V}V.
Answer:
To address this statement, we need to understand the concepts of linear independence, basis, and finite-dimensional vector spaces.

Statement:

“Every linearly independent subset of a finite-dimensional vector space can be extended to a basis of V V VVV.”

Definitions:

  1. Linearly Independent Set: A subset S S SSS of a vector space V V VVV is linearly independent if no vector in S S SSS can be written as a linear combination of the others.
  2. Basis of a Vector Space: A basis of a vector space V V VVV is a linearly independent set of vectors in V V VVV that spans V V VVV. This means every vector in V V VVV can be expressed as a linear combination of the vectors in the basis.
  3. Finite-Dimensional Vector Space: A vector space is finite-dimensional if it has a basis consisting of a finite number of vectors.

Proof:

We need to prove that any linearly independent set in a finite-dimensional vector space V V VVV can be extended to form a basis of V V VVV.
Let S S SSS be a linearly independent subset of a finite-dimensional vector space V V VVV. We want to show that S S SSS can be extended to a basis of V V VVV.
  1. Case 1: S S SSS Spans V V VVV:
    • If S S SSS already spans V V VVV, then S S SSS is itself a basis of V V VVV, and there is nothing to extend.
  2. Case 2: S S SSS Does Not Span V V VVV:
    • If S S SSS does not span V V VVV, then there exists at least one vector v V v V v in Vv \in VvV that cannot be expressed as a linear combination of vectors in S S SSS.
    • Add v v vvv to S S SSS to form a new set S = S { v } S = S { v } S^(‘)=S uu{v}S’ = S \cup \{v\}S=S{v}.
    • Since v v vvv is not a linear combination of vectors in S S SSS, the set S S S^(‘)S’S is still linearly independent.
    • Repeat this process: If S S S^(‘)S’S does not span V V VVV, find another vector v V v V v^(‘)in Vv’ \in VvV not in the span of S S S^(‘)S’S and add it to S S S^(‘)S’S.
  3. Termination of the Process:
    • Since V V VVV is finite-dimensional, this process must terminate after a finite number of steps. We cannot keep adding vectors indefinitely without eventually spanning V V VVV or violating linear independence.
    • The process stops when we have a set that spans V V VVV and is linearly independent, which is a basis of V V VVV.

Conclusion:

Therefore, every linearly independent subset of a finite-dimensional vector space V V VVV can indeed be extended to a basis of V V VVV. This proves the statement.
Q4. Give an example of a module over a commutative ring R R R\mathrm{R}R which is not free.
Answer:
To provide an example of a module over a commutative ring R R RRR that is not free, let’s first understand what a free module is and then consider a module that does not meet these criteria.

Free Module:

A module M M MMM over a ring R R RRR is called a free module if it has a basis. This means there exists a set of elements in M M MMM such that every element of M M MMM can be uniquely expressed as a linear combination of these basis elements, with coefficients in R R RRR. In simpler terms, a free module is analogous to a vector space in linear algebra, where the basis vectors can be used to generate the entire space.

Example of a Non-Free Module:

Consider the ring R = Z R = Z R=ZR = \mathbb{Z}R=Z (the ring of integers) and the module M = Z / 2 Z M = Z / 2 Z M=Z//2ZM = \mathbb{Z}/2\mathbb{Z}M=Z/2Z (the integers modulo 2). This module M M MMM is actually an R R RRR-module because we can multiply any element of M M MMM by any integer, and the result (modulo 2) is still in M M MMM.
Now, let’s see why M M MMM is not a free module:
  1. No Basis: In M M MMM, every element is either 0 or 1. There is no subset of M M MMM that can serve as a basis for M M MMM over R R RRR. This is because in a free module, the basis elements must be able to generate every module element through linear combinations with coefficients in R R RRR. However, in M M MMM, you cannot generate both 0 and 1 using integer coefficients without either being redundant or failing to generate all elements. For example, if you choose 1 as a “basis element,” then multiplying it by any integer always gives you either 0 or 1 (modulo 2), which means you can’t generate a “new” element that isn’t already in M M MMM.
  2. Linear Independence: Another way to see this is through the concept of linear independence. In a free module, the basis elements must be linearly independent. However, in M M MMM, any element multiplied by 2 (which is a valid element of R R RRR) gives 0 (in M M MMM). This violates the condition of linear independence, as a non-trivial linear combination of module elements (in this case, just the element 1 multiplied by 2) results in the zero element of the module.

Conclusion:

Thus, Z / 2 Z Z / 2 Z Z//2Z\mathbb{Z}/2\mathbb{Z}Z/2Z as a module over the ring Z Z Z\mathbb{Z}Z is an example of a module that is not free. It lacks a basis that can generate the module through linear combinations with coefficients in Z Z Z\mathbb{Z}Z, and it does not satisfy the condition of linear independence required for a free module.
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