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.

Step 1: Initial Matrix

Step 2: Making the First Pivot

Step 3: Clearing the First Column

Step 4: Working on the Second Pivot

Step 5: Clearing the Second Column

Step 6: Making the Third Pivot

Step 7: Clearing the Rest of the Matrix

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}$

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}$

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}$

1. Smith Normal Form: The Smith normal form of the matrix $A$$A$AA$A$ is given by
   $$\text{Smith Normal Form}(A)=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 8\end{array}\right]$$$$\text{Smith Normal Form}(A)=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 8\end{array}\right]$$“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]$$\text{Smith Normal Form}(A)=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 8\end{array}\right]$$
   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.

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$$v$vv$v$ (eigenvector) for which $T(v)=\lambda v$$T(v)=\lambda v$T(v)=lambda vT(v) = \lambda v$T(v)=\lambda v$.

Proof:

Conclusion:

Statement:

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:

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 \{v\}$$S^{\prime }=S\cup \{v\}$S^(‘)=S uu{v}S’ = S \cup \{v\}$S^{\prime }=S\cup \{v\}$.
   • Since $v$$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:

Free Module:

Example of a Non-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: