# Cayley-Hamilton

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• Finite Integral Domain is a Field Show that any finite integral domain $R$ is a field.   Definition. A commutative ring $R$ with $1\neq 0$ is called an integral domain if it has no zero divisors. That is, if $ab=0$ for $a, b \in R$, then either $a=0$ or $b=0$. Proof. We give two proofs. Proof […]
• Linear Dependent/Independent Vectors of Polynomials Let $p_1(x), p_2(x), p_3(x), p_4(x)$ be (real) polynomials of degree at most $3$. Which (if any) of the following two conditions is sufficient for the conclusion that these polynomials are linearly dependent? (a) At $1$ each of the polynomials has the value $0$. Namely $p_i(1)=0$ […]
• Automorphism Group of $\Q(\sqrt[3]{2})$ Over $\Q$. Determine the automorphism group of $\Q(\sqrt[3]{2})$ over $\Q$. Proof. Let $\sigma \in \Aut(\Q(\sqrt[3]{2}/\Q)$ be an automorphism of $\Q(\sqrt[3]{2})$ over $\Q$. Then $\sigma$ is determined by the value $\sigma(\sqrt[3]{2})$ since any element $\alpha$ of $\Q(\sqrt[3]{2})$ […]
• Finite Group and Subgroup Criteria Let $G$ be a finite group and let $H$ be a subset of $G$ such that for any $a,b \in H$, $ab\in H$. Then show that $H$ is a subgroup of $G$.   Proof. Let $a \in H$. To show that $H$ is a subgroup of $G$, it suffices to show that the inverse $a^{-1}$ is in $H$. If […]
• The Union of Two Subspaces is Not a Subspace in a Vector Space Let $U$ and $V$ be subspaces of the vector space $\R^n$. If neither $U$ nor $V$ is a subset of the other, then prove that the union $U \cup V$ is not a subspace of $\R^n$.   Proof. Since $U$ is not contained in $V$, there exists a vector $\mathbf{u}\in U$ but […]
• Elements of Finite Order of an Abelian Group form a Subgroup Let $G$ be an abelian group and let $H$ be the subset of $G$ consisting of all elements of $G$ of finite order. That is, $H=\{ a\in G \mid \text{the order of a is finite}\}.$ Prove that $H$ is a subgroup of $G$.   Proof. Note that the identity element $e$ of […]
• Two Matrices with the Same Characteristic Polynomial. Diagonalize if Possible. Let $A=\begin{bmatrix} 1 & 3 & 3 \\ -3 &-5 &-3 \\ 3 & 3 & 1 \end{bmatrix} \text{ and } B=\begin{bmatrix} 2 & 4 & 3 \\ -4 &-6 &-3 \\ 3 & 3 & 1 \end{bmatrix}.$ For this problem, you may use the fact that both matrices have the same characteristic […]
• Column Rank = Row Rank. (The Rank of a Matrix is the Same as the Rank of its Transpose) Let $A$ be an $m\times n$ matrix. Prove that the rank of $A$ is the same as the rank of the transpose matrix $A^{\trans}$.   Hint. Recall that the rank of a matrix $A$ is the dimension of the range of $A$. The range of $A$ is spanned by the column vectors of the matrix […]