# Algebra

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Q1. Obtain Smith normal form over the field of rational numbers and find invariant factors of the matrix
$\left[\begin{array}{ccc}-2& 0& 10\\ 0& -3& -4\\ 1& 2& -1\end{array}\right]$$\left[\begin{array}{ccc}-2& 0& 10\\ 0& -3& -4\\ 1& 2& -1\end{array}\right]$[[-2,0,10],[0,-3,-4],[1,2,-1]]\left[\begin{array}{ccc} -2 & 0 & 10 \\ 0 & -3 & -4 \\ 1 & 2 & -1 \end{array}\right]$\left[\begin{array}{ccc}-2& 0& 10\\ 0& -3& -4\\ 1& 2& -1\end{array}\right]$
To find the Smith normal form and the invariant factors of the given matrix over the field of rational numbers, we will follow these steps:
1. Introduce the Matrix: We are given the matrix
$A=\left[\begin{array}{ccc}-2& 0& 10\\ 0& -3& -4\\ 1& 2& -1\end{array}\right]$$A=\left[\begin{array}{ccc}-2& 0& 10\\ 0& -3& -4\\ 1& 2& -1\end{array}\right]$A=[[-2,0,10],[0,-3,-4],[1,2,-1]]A = \left[\begin{array}{ccc} -2 & 0 & 10 \\ 0 & -3 & -4 \\ 1 & 2 & -1 \end{array}\right]$A=\left[\begin{array}{ccc}-2& 0& 10\\ 0& -3& -4\\ 1& 2& -1\end{array}\right]$
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$AA$A$. After obtaining the Smith normal form, we will identify the invariant factors.

### Step 1: Initial Matrix

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

### Step 2: Making the First Pivot

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

### 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:
$\left[\begin{array}{ccc}1& 2& -1\\ 0& -3& -4\\ 0& 4& 8\end{array}\right]$$\left[\begin{array}{ccc}1& 2& -1\\ 0& -3& -4\\ 0& 4& 8\end{array}\right]$[[1,2,-1],[0,-3,-4],[0,4,8]]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & -3 & -4 \\ 0 & 4 & 8 \end{array}\right]$\left[\begin{array}{ccc}1& 2& -1\\ 0& -3& -4\\ 0& 4& 8\end{array}\right]$

### 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:
$\left[\begin{array}{ccc}1& 2& -1\\ 0& 4& 8\\ 0& -3& -4\end{array}\right]$$\left[\begin{array}{ccc}1& 2& -1\\ 0& 4& 8\\ 0& -3& -4\end{array}\right]$[[1,2,-1],[0,4,8],[0,-3,-4]]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & 4 & 8 \\ 0 & -3 & -4 \end{array}\right]$\left[\begin{array}{ccc}1& 2& -1\\ 0& 4& 8\\ 0& -3& -4\end{array}\right]$

### 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:
$\left[\begin{array}{ccc}1& 2& -1\\ 0& 4& 8\\ 0& 1& 4\end{array}\right]$$\left[\begin{array}{ccc}1& 2& -1\\ 0& 4& 8\\ 0& 1& 4\end{array}\right]$[[1,2,-1],[0,4,8],[0,1,4]]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & 4 & 8 \\ 0 & 1 & 4 \end{array}\right]$\left[\begin{array}{ccc}1& 2& -1\\ 0& 4& 8\\ 0& 1& 4\end{array}\right]$

### 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:
$\left[\begin{array}{ccc}1& 2& -1\\ 0& 1& 4\\ 0& 4& 8\end{array}\right]$$\left[\begin{array}{ccc}1& 2& -1\\ 0& 1& 4\\ 0& 4& 8\end{array}\right]$[[1,2,-1],[0,1,4],[0,4,8]]\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & 1 & 4 \\ 0 & 4 & 8 \end{array}\right]$\left[\begin{array}{ccc}1& 2& -1\\ 0& 1& 4\\ 0& 4& 8\end{array}\right]$

### 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.
$\begin{array}{rl}& \left[\begin{array}{ccc}1& 2& -1\\ 0& 1& 4\\ 0& 4& 8\end{array}\right]\\ & {R}_{1}←{R}_{1}-2×{R}_{2}\\ & =\left[\begin{array}{ccc}1& 0& -9\\ 0& 1& 4\\ 0& 4& 8\end{array}\right]\\ & {R}_{3}←{R}_{3}-4×{R}_{2}\\ & =\left[\begin{array}{lll}1& 0& -9\\ 0& 1& 4\\ 0& 0& -8\end{array}\right]\end{array}$$\begin{array}{r}\left[\begin{array}{ccc}1& 2& -1\\ 0& 1& 4\\ 0& 4& 8\end{array}\right]\\ {R}_{1}←{R}_{1}-2×{R}_{2}\\ =\left[\begin{array}{ccc}1& 0& -9\\ 0& 1& 4\\ 0& 4& 8\end{array}\right]\\ {R}_{3}←{R}_{3}-4×{R}_{2}\\ =\left[\begin{array}{lll}1& 0& -9\\ 0& 1& 4\\ 0& 0& -8\end{array}\right]\end{array}${:[[[1,2,-1],[0,1,4],[0,4,8]]],[R_(1)larrR_(1)-2xxR_(2)],[=[[1,0,-9],[0,1,4],[0,4,8]]],[R_(3)larrR_(3)-4xxR_(2)],[=[[1,0,-9],[0,1,4],[0,0,-8]]]:}\begin{aligned} & {\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & 1 & 4 \\ 0 & 4 & 8 \end{array}\right]} \\ & R_1 \leftarrow R_1-2 \times R_2 \\ & =\left[\begin{array}{ccc} 1 & 0 & -9 \\ 0 & 1 & 4 \\ 0 & 4 & 8 \end{array}\right] \\ & R_3 \leftarrow R_3-4 \times R_2 \\ & =\left[\begin{array}{lll} 1 & 0 & -9 \\ 0 & 1 & 4 \\ 0 & 0 & -8 \end{array}\right] \\ \end{aligned}$\begin{array}{rl}& \left[\begin{array}{ccc}1& 2& -1\\ 0& 1& 4\\ 0& 4& 8\end{array}\right]\\ & {R}_{1}←{R}_{1}-2×{R}_{2}\\ & =\left[\begin{array}{ccc}1& 0& -9\\ 0& 1& 4\\ 0& 4& 8\end{array}\right]\\ & {R}_{3}←{R}_{3}-4×{R}_{2}\\ & =\left[\begin{array}{lll}1& 0& -9\\ 0& 1& 4\\ 0& 0& -8\end{array}\right]\end{array}$
$\left[\begin{array}{ccc}1& 0& -9\\ 0& 1& 4\\ 0& 0& -8\end{array}\right]$$\left[\begin{array}{ccc}1& 0& -9\\ 0& 1& 4\\ 0& 0& -8\end{array}\right]$[[1,0,-9],[0,1,4],[0,0,-8]]\left[\begin{array}{ccc} 1 & 0 & -9 \\ 0 & 1 & 4 \\ 0 & 0 & -8 \end{array}\right]$\left[\begin{array}{ccc}1& 0& -9\\ 0& 1& 4\\ 0& 0& -8\end{array}\right]$

### Operation 1: $R3\to \frac{R3}{-1}$$R3\to \frac{R3}{-1}$R3rarr(R3)/(-1)R3 \to \frac{R3}{-1}$R3\to \frac{R3}{-1}$

We divide the third row by $-1$$-1$-1-1$-1$. This operation changes the third row ($R3$$R3$R3R3$R3$):
$R3=\frac{R3}{-1}=\frac{1}{-1}\cdot \left[0,0,-8\right]=\left[0,0,8\right]$$R3=\frac{R3}{-1}=\frac{1}{-1}\cdot \left[0,0,-8\right]=\left[0,0,8\right]$R3=(R3)/(-1)=(1)/(-1)*[0,0,-8]=[0,0,8]R3 = \frac{R3}{-1} = \frac{1}{-1} \cdot \left[0, 0, -8\right] = \left[0, 0, 8\right]$R3=\frac{R3}{-1}=\frac{1}{-1}\cdot \left[0,0,-8\right]=\left[0,0,8\right]$
The matrix becomes:
$\left[\begin{array}{ccc}1& 0& -9\\ 0& 1& 4\\ 0& 0& 8\end{array}\right]$$\left[\begin{array}{ccc}1& 0& -9\\ 0& 1& 4\\ 0& 0& 8\end{array}\right]$[[1,0,-9],[0,1,4],[0,0,8]]\left[\begin{array}{ccc} 1 & 0 & -9 \\ 0 & 1 & 4 \\ 0 & 0 & 8 \end{array}\right]$\left[\begin{array}{ccc}1& 0& -9\\ 0& 1& 4\\ 0& 0& 8\end{array}\right]$

### Operation 2: $R1\to R1+\frac{9R3}{8}$$R1\to R1+\frac{9R3}{8}$R1rarr R1+(9R3)/(8)R1 \to R1 + \frac{9R3}{8}$R1\to R1+\frac{9R3}{8}$

We add $\frac{9}{8}$$\frac{9}{8}$(9)/(8)\frac{9}{8}$\frac{9}{8}$ times the third row to the first row. This operation changes the first row ($R1$$R1$R1R1$R1$):
$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]$$R1=R1+\frac{9}{8}\cdot R3=\left[1,0,-9\right]+\frac{9}{8}\cdot \left[0,0,8\right]=\left[1,0,-9+9\right]=\left[1,0,0\right]$R1=R1+(9)/(8)*R3=[1,0,-9]+(9)/(8)*[0,0,8]=[1,0,-9+9]=[1,0,0]R1 = R1 + \frac{9}{8} \cdot R3 = \left[1, 0, -9\right] + \frac{9}{8} \cdot \left[0, 0, 8\right] = \left[1, 0, -9 + 9\right] = \left[1, 0, 0\right]$R1=R1+\frac{9}{8}\cdot R3=\left[1,0,-9\right]+\frac{9}{8}\cdot \left[0,0,8\right]=\left[1,0,-9+9\right]=\left[1,0,0\right]$
The matrix becomes:
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 4\\ 0& 0& 8\end{array}\right]$$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 4\\ 0& 0& 8\end{array}\right]$[[1,0,0],[0,1,4],[0,0,8]]\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 8 \end{array}\right]$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 4\\ 0& 0& 8\end{array}\right]$

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

We subtract $\frac{1}{2}$$\frac{1}{2}$(1)/(2)\frac{1}{2}$\frac{1}{2}$ times the third row from the second row. This operation changes the second row ($R2$$R2$R2R2$R2$):
$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]$$R2=R2-\frac{1}{2}\cdot R3=\left[0,1,4\right]-\frac{1}{2}\cdot \left[0,0,8\right]=\left[0,1,4-4\right]=\left[0,1,0\right]$R2=R2-(1)/(2)*R3=[0,1,4]-(1)/(2)*[0,0,8]=[0,1,4-4]=[0,1,0]R2 = R2 – \frac{1}{2} \cdot R3 = \left[0, 1, 4\right] – \frac{1}{2} \cdot \left[0, 0, 8\right] = \left[0, 1, 4 – 4\right] = \left[0, 1, 0\right]$R2=R2-\frac{1}{2}\cdot R3=\left[0,1,4\right]-\frac{1}{2}\cdot \left[0,0,8\right]=\left[0,1,4-4\right]=\left[0,1,0\right]$
The final matrix after all operations is:
$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 8\end{array}\right]$$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 8\end{array}\right]$[[1,0,0],[0,1,0],[0,0,8]]\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 8 \end{array}\right]$\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 8\end{array}\right]$
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$$A$AA$A$ is given by
$\text{Smith Normal Form}\left(A\right)=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 8\end{array}\right]$$\text{Smith Normal Form}\left(A\right)=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 8\end{array}\right]$“Smith Normal Form”(A)=[[1,0,0],[0,1,0],[0,0,8]]\text{Smith Normal Form}(A) = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 8 \end{array}\right]$\text{Smith Normal Form}\left(A\right)=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 8\end{array}\right]$
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$$A$AA$A$, the invariant factors are $1$$1$11$1$, $1$$1$11$1$, and $8$$8$88$8$. 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$$A$AA$A$ is a diagonal matrix with elements $1$$1$11$1$, $1$$1$11$1$, and $8$$8$88$8$, and these are also the invariant factors of the matrix.
Q2. Prove that all the characteristic values of a nilpotent operator are zero.
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$$T$TT$T$ on a vector space is said to be nilpotent if there exists some positive integer $k$$k$kk$k$ such that ${T}^{k}=0$${T}^{k}=0$T^(k)=0T^k = 0${T}^{k}=0$, where $0$$0$00$0$ is the zero operator. This means that when you apply $T$$T$TT$T$ repeatedly $k$$k$kk$k$ times to any vector in the space, you end up with the zero vector.
2. Eigenvalues (Characteristic Values): An eigenvalue of an operator $T$$T$TT$T$ is a scalar $\lambda$$\lambda$lambda\lambda$\lambda$ such that there exists a non-zero vector $v$$v$vv$v$ (eigenvector) for which $T\left(v\right)=\lambda v$$T\left(v\right)=\lambda v$T(v)=lambda vT(v) = \lambda v$T\left(v\right)=\lambda v$.

### Proof:

We need to show that if $T$$T$TT$T$ is a nilpotent operator, then any eigenvalue $\lambda$$\lambda$lambda\lambda$\lambda$ of $T$$T$TT$T$ must be zero.
Let $\lambda$$\lambda$lambda\lambda$\lambda$ be an eigenvalue of $T$$T$TT$T$ and $v$$v$vv$v$ be a corresponding non-zero eigenvector. This means $T\left(v\right)=\lambda v$$T\left(v\right)=\lambda v$T(v)=lambda vT(v) = \lambda v$T\left(v\right)=\lambda v$.
Now, consider ${T}^{k}$${T}^{k}$T^(k)T^k${T}^{k}$ where $k$$k$kk$k$ is the smallest positive integer such that ${T}^{k}=0$${T}^{k}=0$T^(k)=0T^k = 0${T}^{k}=0$ (the definition of a nilpotent operator).
Apply ${T}^{k}$${T}^{k}$T^(k)T^k${T}^{k}$ to $v$$v$vv$v$:
${T}^{k}\left(v\right)=0$${T}^{k}\left(v\right)=0$T^(k)(v)=0T^k(v) = 0${T}^{k}\left(v\right)=0$
Since $T\left(v\right)=\lambda v$$T\left(v\right)=\lambda v$T(v)=lambda vT(v) = \lambda v$T\left(v\right)=\lambda v$, we can write:
${T}^{k}\left(v\right)={T}^{k-1}\left(T\left(v\right)\right)={T}^{k-1}\left(\lambda v\right)=\lambda {T}^{k-1}\left(v\right)$${T}^{k}\left(v\right)={T}^{k-1}\left(T\left(v\right)\right)={T}^{k-1}\left(\lambda v\right)=\lambda {T}^{k-1}\left(v\right)$T^(k)(v)=T^(k-1)(T(v))=T^(k-1)(lambda v)=lambdaT^(k-1)(v)T^k(v) = T^{k-1}(T(v)) = T^{k-1}(\lambda v) = \lambda T^{k-1}(v)${T}^{k}\left(v\right)={T}^{k-1}\left(T\left(v\right)\right)={T}^{k-1}\left(\lambda v\right)=\lambda {T}^{k-1}\left(v\right)$
Continuing this process, we apply $T$$T$TT$T$ repeatedly:
$={\lambda }^{2}{T}^{k-2}\left(v\right)=\dots ={\lambda }^{k}v$$={\lambda }^{2}{T}^{k-2}\left(v\right)=\dots ={\lambda }^{k}v$=lambda^(2)T^(k-2)(v)=dots=lambda ^(k)v= \lambda^2 T^{k-2}(v) = \ldots = \lambda^k v$={\lambda }^{2}{T}^{k-2}\left(v\right)=\dots ={\lambda }^{k}v$
Since ${T}^{k}=0$${T}^{k}=0$T^(k)=0T^k = 0${T}^{k}=0$, we have:
${\lambda }^{k}v=0$${\lambda }^{k}v=0$lambda ^(k)v=0\lambda^k v = 0${\lambda }^{k}v=0$
But $v$$v$vv$v$ is a non-zero vector. Therefore, for this equation to hold, it must be that ${\lambda }^{k}=0$${\lambda }^{k}=0$lambda ^(k)=0\lambda^k = 0${\lambda }^{k}=0$. Since $\lambda$$\lambda$lambda\lambda$\lambda$ is a scalar, the only way for ${\lambda }^{k}$${\lambda }^{k}$lambda ^(k)\lambda^k${\lambda }^{k}$ to be zero for some positive integer $k$$k$kk$k$ is for $\lambda$$\lambda$lambda\lambda$\lambda$ itself to be zero.

### Conclusion:

Thus, we have shown that for a nilpotent operator $T$$T$TT$T$, any eigenvalue $\lambda$$\lambda$lambda\lambda$\lambda$ 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 $\mathrm{V}$$\mathrm{V}$V\mathrm{V}$\mathrm{V}$.
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$VV$V$.”

### Definitions:

1. Linearly Independent Set: A subset $S$$S$SS$S$ of a vector space $V$$V$VV$V$ is linearly independent if no vector in $S$$S$SS$S$ can be written as a linear combination of the others.
2. Basis of a Vector Space: A basis of a vector space $V$$V$VV$V$ is a linearly independent set of vectors in $V$$V$VV$V$ that spans $V$$V$VV$V$. This means every vector in $V$$V$VV$V$ 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$$V$VV$V$ can be extended to form a basis of $V$$V$VV$V$.
Let $S$$S$SS$S$ be a linearly independent subset of a finite-dimensional vector space $V$$V$VV$V$. We want to show that $S$$S$SS$S$ can be extended to a basis of $V$$V$VV$V$.
1. Case 1: $S$$S$SS$S$ Spans $V$$V$VV$V$:
• If $S$$S$SS$S$ already spans $V$$V$VV$V$, then $S$$S$SS$S$ is itself a basis of $V$$V$VV$V$, and there is nothing to extend.
2. Case 2: $S$$S$SS$S$ Does Not Span $V$$V$VV$V$:
• If $S$$S$SS$S$ does not span $V$$V$VV$V$, then there exists at least one vector $v\in V$$v\in V$v in Vv \in V$v\in V$ that cannot be expressed as a linear combination of vectors in $S$$S$SS$S$.
• Add $v$$v$vv$v$ to $S$$S$SS$S$ to form a new set ${S}^{\prime }=S\cup \left\{v\right\}$${S}^{\prime }=S\cup \left\{v\right\}$S^(‘)=S uu{v}S’ = S \cup \{v\}${S}^{\prime }=S\cup \left\{v\right\}$.
• Since $v$$v$vv$v$ is not a linear combination of vectors in $S$$S$SS$S$, the set ${S}^{\prime }$${S}^{\prime }$S^(‘)S’${S}^{\prime }$ is still linearly independent.
• Repeat this process: If ${S}^{\prime }$${S}^{\prime }$S^(‘)S’${S}^{\prime }$ does not span $V$$V$VV$V$, find another vector ${v}^{\prime }\in V$${v}^{\prime }\in V$v^(‘)in Vv’ \in V${v}^{\prime }\in V$ not in the span of ${S}^{\prime }$${S}^{\prime }$S^(‘)S’${S}^{\prime }$ and add it to ${S}^{\prime }$${S}^{\prime }$S^(‘)S’${S}^{\prime }$.
3. Termination of the Process:
• Since $V$$V$VV$V$ is finite-dimensional, this process must terminate after a finite number of steps. We cannot keep adding vectors indefinitely without eventually spanning $V$$V$VV$V$ or violating linear independence.
• The process stops when we have a set that spans $V$$V$VV$V$ and is linearly independent, which is a basis of $V$$V$VV$V$.

### Conclusion:

Therefore, every linearly independent subset of a finite-dimensional vector space $V$$V$VV$V$ can indeed be extended to a basis of $V$$V$VV$V$. This proves the statement.
Q4. Give an example of a module over a commutative ring $\mathrm{R}$$\mathrm{R}$R\mathrm{R}$\mathrm{R}$ which is not free.
To provide an example of a module over a commutative ring $R$$R$RR$R$ 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$$M$MM$M$ over a ring $R$$R$RR$R$ is called a free module if it has a basis. This means there exists a set of elements in $M$$M$MM$M$ such that every element of $M$$M$MM$M$ can be uniquely expressed as a linear combination of these basis elements, with coefficients in $R$$R$RR$R$. 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=\mathbb{Z}$$R=\mathbb{Z}$R=ZR = \mathbb{Z}$R=\mathbb{Z}$ (the ring of integers) and the module $M=\mathbb{Z}/2\mathbb{Z}$$M=\mathbb{Z}/2\mathbb{Z}$M=Z//2ZM = \mathbb{Z}/2\mathbb{Z}$M=\mathbb{Z}/2\mathbb{Z}$ (the integers modulo 2). This module $M$$M$MM$M$ is actually an $R$$R$RR$R$-module because we can multiply any element of $M$$M$MM$M$ by any integer, and the result (modulo 2) is still in $M$$M$MM$M$.
Now, let’s see why $M$$M$MM$M$ is not a free module:
1. No Basis: In $M$$M$MM$M$, every element is either 0 or 1. There is no subset of $M$$M$MM$M$ that can serve as a basis for $M$$M$MM$M$ over $R$$R$RR$R$. This is because in a free module, the basis elements must be able to generate every module element through linear combinations with coefficients in $R$$R$RR$R$. However, in $M$$M$MM$M$, you cannot generate both 0 and 1 using integer coefficients without either being redundant or failing to generate all elements. For example, if you choose 1 as a “basis element,” then multiplying it by any integer always gives you either 0 or 1 (modulo 2), which means you can’t generate a “new” element that isn’t already in $M$$M$MM$M$.
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$$M$MM$M$, any element multiplied by 2 (which is a valid element of $R$$R$RR$R$) gives 0 (in $M$$M$MM$M$). 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, $\mathbb{Z}/2\mathbb{Z}$$\mathbb{Z}/2\mathbb{Z}$Z//2Z\mathbb{Z}/2\mathbb{Z}$\mathbb{Z}/2\mathbb{Z}$ as a module over the ring $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{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 $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{Z}$, and it does not satisfy the condition of linear independence required for a free module.
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