Part 1

Definition: An element \(r\in R\) is irreducible if whenever \(r=st\), then either \(s\) or \(t\) is a unit.

Definition: Two elements \(r,s\in R\) are associates if \(r = \ell s\) for some unit \(\ell\).

A ring \(R\) is a unique factorization domain iff for every \(r\in R\), there exists a set \(\left\{{p_i \mathrel{\Big|}1\leq i \leq n}\right\}\) such that \(r = u \prod_{i=1}^n p_i\) where \(u\) is a unit and each \(p_i\) is irreducible.

Moreover, this factorization is unique in the sense that if \(r = w \prod_{i=1}^n q_i\) for some \(w\) a unit and \(q_i\) irreducible elements, then each \(q_i\) is an associate of some \(p_i\).

Part 2

A ring \(R\) is a principal ideal domain iff whenever \(I {~\trianglelefteq~}R\) is an ideal of \(R\), there is a single element \(r_i \in R\) such that \(I = (r_i)\).

Part 3

An example of a UFD that is not a PID is given by \(R = k[x, y]\) for \(k\) a field.

That \(R\) is a UFD follows from the fact that if \(k\) is a field, then \(k\) has no prime elements since every non-zero element is a unit. So the factorization condition holds vacuously for \(k\), and \(k\) is a UFD. But then we can use the following result:

Theorem: If \(R\) is a UFD, then \(R[x]\) is a UFD.

Since \(k\) is a UFD, the theorem implies that \(k[x]\) is a UFD, from which it follows that \(k[x][y] = k[x, y]\) is also a UFD.

To see that \(R\) is not a PID, consider the ideal \(I = (x, y)\), and suppose \(I = (g)\) for some single \(g\in k[x, y]\).

Note that \(I \neq R\), since \(I\) contains no degree zero polynomials. Moreover, since \((x) \subset I = (g)\) (and similarly for \(y\)), we have \(g\divides x\) and \(g\divides y\), which forces \(\deg g = 0\).

So in fact \(g\in k\) and thus \(g\) is invertible, but then \((g) = g^{-1}(g) = (1) = k\), so this forces \(I = k {~\trianglelefteq~}k[x, y]\). However, \(x\not\in k\) (nor \(y\)), which is a contradiction.

Problem 2

Lemma 1

\(A\) has \(n\) distinct eigenvalues \(\iff m_A(x) = \chi_A(x)\).

Proof:

The eigenvalues are always root of both \(m_A(x)\) and \(\chi_A(x)\) (potentially with differing multiplicities), so we can write

\begin{align*}
m_A(x) = \prod_i (x-\lambda_i)^{p_i} \\
\chi_A(x) = \prod_i (x-\lambda_i)^{q_i} \\
\end{align*}

where \(1 \leq p_i \leq q_i\) for every \(i\).

\(\implies\): If \(A\) has \(n\) distinct eigenvalues, then \(\chi_A(x) = \prod_{i=1}^n (x-\lambda_i)\) in \(\overline{k} [x]\). Noting that every exponent is 1, we have \(q_i = 1\) for all \(i\), which forces \(p_i = 1\) and thus \(m_A(x) = \chi_A(x)\).

\(\impliedby\): If \(m_A(x) = \chi_A(x)\), then \(p_i = q_i\) for all \(i\). If we then consider \(JCF(A)\), we have

• The number of Jordan block \(J_{\lambda_i}\) is the dimension of the eigenspace \(E_{\lambda_i}\),
• \(q_i =\) the sum of the sizes of all Jordan blocks \(J_{\lambda_i}\), and
• \(p_i =\) the size of the largest Jordan block \(J_{\lambda_i}\).

Thus \(p_i = q_i\) for every \(i\) \(\iff\) there is one Jordan block for every \(\lambda_i\) \(\iff\) \(\dim E_{\lambda_i} = 1\) for every \(i\).

But \(\dim E_{\lambda_i}\) is precisely the multiplicity of \(\lambda_i\) in \(\chi_A(x)\), which means that \(\chi_A(x) = \prod_{i} (x-\lambda_i)\). Since \(\chi_A(x)\) is a degree \(n\) polynomial, this says that \(\chi_A\) has \(n\) distinct linear factors, corresponding to \(n\) distinct eigenvalues of \(A\).

\(\hfill\blacksquare\)

Lemma 2

Let \(k[x] \curvearrowright V\) in the usual way with \(A\) to obtain an invariant factor decomposition \begin{align*} V = \frac{k[x]}{(f_1)} \oplus \frac{k[x]}{(f_2)} \oplus \cdots \oplus \frac{k[x]}{(f_k)}, \quad f_1 \divides f_2 \divides \cdots \divides f_k. \end{align*}

Then it is always the case that

• \(m_A(x) = f_k(x)\), i.e. the minimal polynomial is the invariant factor of largest degree,
• \(\chi_A(x) = \prod_{i=1}^k f_i(x)\), i.e. the characteristic polynomial is the product of all of the invariant factors.

\(\hfill\blacksquare\)

Main Result

\((1) \implies (2)\):

Suppose \begin{align*} V = \mathop{\mathrm{span}}_k\left\{{\mathbf{v}, A\mathbf{v}, A^2 \mathbf{v}, \cdots A^{n-1}\mathbf{v}}\right\} \coloneqq\mathop{\mathrm{span}}_k \mathcal B \end{align*} where \(\dim_k V = n\).

Then \(A^n \mathbf{v}\) is necessarily a linear combination of these basis elements, and in particular, there are coefficients \(c_i\) (not all zero) such that \begin{align*} A^n \mathbf{v} = \sum_{i=0}^{n-1} c_i A^i \mathbf{v}. \end{align*}

The consider computing the matrix of \(A\) in \(\mathcal B\) by considering the images of all basis elements under \(A\).

Letting \(\mathcal B = \left\{{\mathbf{w}_i \coloneqq A^i \mathbf{v} \mathrel{\Big|}0\leq i \leq n-1}\right\}\), we have

\begin{align*}
\mathbf{w}_0 \coloneqq\mathbf{v} &\mapsto A\mathbf{v} \coloneqq\mathbf{w}_1 \\
\mathbf{w}_1 \coloneqq A\mathbf{v} &\mapsto A^2\mathbf{v} \coloneqq\mathbf{w}_2 \\
\mathbf{w}_2 \coloneqq A^2\mathbf{v} &\mapsto A^3\mathbf{v} \coloneqq\mathbf{w}_3 \\
\vdots \quad & \quad \vdots \\
\mathbf{w}_{n-2} \coloneqq A^{n-2}\mathbf{v} &\mapsto A^{n-1}\mathbf{v} \coloneqq\mathbf{w}_{n-1} \\
\mathbf{w}_{n-1} \coloneqq A^{n-1}\mathbf{v} &\mapsto A^n\mathbf{v} = \sum_{i=0}^{n-1} c_i A^i \mathbf{v}_i \coloneqq\sum_{i=0}^{n-1} c_i \mathbf{w}_i
.\end{align*}

This means that with respect to the basis \(\mathcal B\), \(A\) has the following matrix representation:

\begin{align*}
[A]_{\mathcal B} = 
\left[\begin{array}{ccccc}{0} & {0} & {\dots} & {0} & {c_{0}} \\ {1} & {0} & {\dots} & {0} & {c_{1}} \\ {0} & {1} & {\dots} & {0} & {c_{2}} \\ {} & {} & {\ddots} & {} & {\vdots} \\ {0} & {0} & {\dots} & {1} & {c_{n-1}}\end{array}\right]
\end{align*}

But this is the companion matrix for \(p(x) = \sum_{i=0}^{n-1} c_i x^i\), which always satisfy the property that \(p(x)\) equals both their characteristic and their minimal polynomial.

Thus by lemma 1, the matrix \([A]_{\mathcal{B}}\) has distinct eigenvalues, and thus so does \(A\).

\((2) \implies (1)\):

Suppose \(A\) has distinct eigenvalues. By Lemma 1, \(\chi_A(x) = m_A(x)\), and so we have \begin{align*} \chi_A(x) = f_k(x) = \prod_{i=1}^k f_i(x) = m_A(x), \end{align*}

which can only happen if \(f_1(x) = f_2(x) = \cdots = f_{n-1}(x) = 1\), in which case there is only one nontrivial invariant factor.

So we have \begin{align*} V \cong \frac{k[x]}{(f_k)}, \quad \operatorname{Ann}(V) = (f_k), \quad \deg f_k = n. \end{align*}

If we now take the Rational Canonical Form of \(A\), it follows that \(RCF(A)\) has only a single block in a suitable ordered basis \(\mathcal B = \left\{{\mathbf{w}_0, \cdots, \mathbf{w}_{n-1}}\right\}\).

So write \(f_k(x) = \sum_{i=0}^n c_i x^i\); then \([A]_{\mathcal B}\) is the companion matrix to \(f_k(x)\) in the basis \(\mathcal B\), which by construction satisfies \begin{align*} A = \left[\begin{array}{ccccc}{0} & {0} & {\dots} & {0} & {c_{0}} \\ {1} & {0} & {\dots} & {0} & {c_{1}} \\ {0} & {1} & {\dots} & {0} & {c_{2}} \\ {} & {} & {\ddots} & {} & {\vdots} \\ {0} & {0} & {\dots} & {1} & {c_{n-1}}\end{array}\right] \implies A \mathbf{w}_i = \begin{cases} \mathbf{w}_{i+1} & 0 \leq i < n-2 \\ \sum_{i=0}^{n-1} c_i \mathbf{w}_i & i = n-1, \end{cases} \end{align*}

and thus we have

\begin{align*} V \cong \mathop{\mathrm{span}}_k \mathcal{B} = \mathop{\mathrm{span}}_k\left\{{\mathbf{w}_0, \cdots \mathbf{w}_{n-1}}\right\} \cong \mathop{\mathrm{span}}_k\left\{{\mathbf{w}_0, A\mathbf{w}_0, A^2\mathbf{w}_0, \cdots, A^{n-1} \mathbf{w}_0}\right\}. \end{align*}

\(\hfill\blacksquare\)

Problem 3

Part 1

Let \(\mathbf{v} = [0,1,0]^t\), We compute

\begin{align*}
M\mathbf{v} =
\left[\begin{array}{ccc}
1 & 0 & x \\
0 & 1 & 0 \\
y & 0 & 1
\end{array}\right]
\left[\begin{array}{c} 0  \\ 1  \\ 0 \end{array}\right] = 
\left[\begin{array}{cc} 1(0) + 0(1) + x(0)  \\ 0(0) + 1(1) + 0(0) \\ y(0) + 0(1) + 1(0) \end{array}\right] = 
\left[\begin{array}{cc} 0 \\ 1 \\ 0 \end{array}\right] = 
1 \left[\begin{array}{cc} 0 \\ 1 \\ 0 \end{array}\right]
,\end{align*}

which shows that \(\mathbf{v}\) is an eigenvector of \(M\) with eigenvalue \(\lambda = 1\).

Part 2

Noting that the rank is the dimension of the column space, we find that

• \(\operatorname{rank}(M) \geq 1\), since it is not the zero matrix,
• \(\operatorname{rank}(M) \geq 2\), since neither \([1,0,y]^t\) or \([x,0,1]^t\) can be in the span of \([0,1,0]^t\), and
• \(\operatorname{rank}(M) = 3 \iff \operatorname{det}(M) \neq 0\).

So we compute

\begin{align*}
\operatorname{det}_M(x, y) = \left|\begin{array}{ccc}
1 & 0 & x \\
0 & 1 & 0 \\
y & 0 & 1
\end{array}\right| = 1(1-0) - 0(1-xy) + x(-y) = 1 - xy
,\end{align*}

and so \(\operatorname{det}_M(x, y) = 0 \iff xy = 1\). Thus

\begin{align*}
\operatorname{rank}(M) = \begin{cases} 3 & xy = 1 \\ 2 & \text{else.} \end{cases}
\end{align*}

Part 3

Since \(M\) is diagonalizable \(\iff M\) is full rank, which in this case means \(\operatorname{rank}(M) = 3\), we have \begin{align*} S = \left\{{(x, y) \in {\mathbf{R}}^2 \mathrel{\Big|}M \text{ is diagonalizable }}\right\} = \left\{{ \left( x, \frac 1 x \right) \mathrel{\Big|}x\in {\mathbf{R}}\setminus\left\{{0}\right\} }\right\} \subset {\mathbf{R}}^2. \end{align*}