Let \(\phi\) be an \(n{\hbox{-}}\)form. If suffices to show these statements for \(n=2\).

\(\implies\): Suppose \(\phi\) is alternating, then \(\phi(b, b) = 0\) for all \(b\in B\).

Letting \(a,b \in B\) be arbitrary, we then have

\begin{align*}
0 &= \phi(a + b, a+b) \\
&= \phi(a, a+b) + \phi(b, a+b) \\
&= \phi(a, a) + \phi(a, b) + \phi(b, a) + \phi(b, b) \\
&= \phi(a, b) + \phi(b, a) \\
&\implies \phi(a,b) = -\phi(b, a)
,\end{align*}

which shows that \(\phi\) is skew-symmetric.

\(\impliedby\) Suppose \(\phi\) is skew-symmetric, so \(\phi(a,b) = -\phi(b, a)\) for all \(a, b\in B\). Then \(\phi(b, b) = - \phi(b, b)\) by transposing the terms, which says that \(\phi(b, b) = 0\) for all \(b\in B\) and thus \(\phi\) is alternating.

Problem 2

Let \(f(x) = \operatorname{det}(P + xQ) \in R[x]\), then \(f\) is a polynomial in \(x\) which is not identically zero.

To see that \(f \not \equiv 0\), we can use that fact that \(P\) is invertible to evaluate \(f(0) = \operatorname{det}(P) \neq 0\).

We can now note that \(f\) has finite degree, and thus finitely many zeroes in \(R\).

Problem 3

Letting \(k[x] \curvearrowright_\phi E\) to yield a \(k[x]{\hbox{-}}\)module structure on \(E\) and take an invariant factor decomposition, \begin{align*} E = E_1 \oplus E_2 \oplus \cdots \oplus E_t, \quad E_i = \frac{k[x]}{(q_i)}, \quad q_1 \divides q_2 \divides \cdots \divides q_t \end{align*}

where \(E_i = k[x] / (q_i)\). Then \(q_t = q\), the minimal polynomial of \(E\).

In particular, \(E_t\) is a \(\phi{\hbox{-}}\)invariant subspace of \(E\), and if \(\deg q_t = m\), then \(E_t\) is in fact an \(m{\hbox{-}}\)dimensional cyclic module with basis \(\left\{{\mathbf{v}, \phi(\mathbf{v}), \phi^2(\mathbf{v}), \cdots, \phi^{m-1}(\mathbf{v})}\right\}\) for some \(\mathbf{v} \in E_t\).

But since \(E_t \leq E\) is a subspace, we have \begin{align*} m = \deg q(x) = \deg q_t(x) = \dim E_t \leq \dim E. \end{align*}

Problem 4

\(\implies\): Suppose \(A \sim D\) where \(D\) is diagonal. Then \(JCF(A) = JCF(D) = D\), which means that every Jordan block of \(A\) has size exactly 1.

Since the elementary divisors of \(A\) are precisely the minimal polynomials of the Jordan blocks of \(A\), and the minimal polynomial of any \(1\times 1\) matrix \([a_{ij}]\) is given by the linear polynomial \(x - a_{ij}\), every elementary divisor of \(A\) must be linear.

\(\impliedby\): Suppose all of the elementary divisors of \(A\) are linear. Every elementary divisor is the minimal polynomial of a Jordan block of \(A\), and so if we write \(JCF(A) = \bigoplus M_i\), then the minimal polynomial of each \(M_i\) is linear.

Supposing that \(M_i\) has minimal polynomial \(p_i(x) = x - c\) for some scalar \(c\), we have \begin{align*} p_i(M_i) = 0 \implies M_i - c I_n = 0 \implies M_i = cI_n, \end{align*}

which shows that \(M_i\) is a diagonal matrix with only \(c\) on its diagonal.

But if every Jordan block of \(A\) is diagonal, then \(JCF(A) = D\) is diagonal and \(A \sim D\).

Problem 5

Part 1

We’ll use the fact that the minimal polynomial \(q\) is the invariant factor of highest degree, and so every other invariant factor must divide \(q\).

Moreover, \(RCF(A) = C_1 \oplus C_2 \oplus \cdots \oplus C_k\) where each \(C_i\) is the companion matrix of the \(i\)th invariant factor if we write \(V \cong \bigoplus_{i=1}^k k[x]/(a_i)\). So it suffices to determine all of the possible distinct combinations of invariant factors.

We can restrict this list by noting that the characteristic polynomial satisfies \(\chi_A(x) = \prod a_i\), and in particular, \(\deg \chi_A(x) = 6\). Noting that \(\deg q(x) = 3\), the degrees of the remaining invariant factors must sum to 3.

So the possibilities are:

\begin{align*}
R_1: a_1 = (x-2), && a_2 = (x-2)^2,    && a_3 = q(x), \\
R_2: a_1 = (x-2), && a_2 = (x-2)(x+3), && a_3 = q(x), \\
R_3: a_1 = (x+3), && a_2 = (x-2)(x+3), && a_3 = q(x), \\
R_4: a_1 = (x-2), && a_2 = (x-2),      && a_3 = (x-2) && a_4 = q(x), \\
R_5: a_1 = (x+3), && a_2 = (x+3),      && a_3 = (x+3) && a_4 = q(x).
\end{align*}

This exhausts all possibilities, because the degrees of \(a_i\) must be a weakly increasing integer partitions of 3, namely \((1,2)\) or \((1, 1, 1)\). A \((1,2)\) partition can only yield a quadratic factor for \(a_2\), and since \(a_2 \divides a_3\) there are only two choices. If a repeated factor is chosen like \((x-2)^2\), then \(a_1 \divides a_2\) forces \(a_1 = x-2\) ,yielding \(R_1\). Otherwise, we can pick either distinct factor of \(a_2\) as a choice for \(a_1\), yielding \(R_2, R_3\). Any \((1,1,1)\) partition can only be a repeated linear factor, since we must have \(a_1 \divides a_2 \divides a_3\), and there are only two choices. This yields \(R_4, R_5\).

Noting that

\begin{align*}
(x-2)^2 &= x^2-4x+4 \\
(x-2)(x+3) &= x^2+x-6 \\
q(x) &= x^3-x^2-8x+12
,\end{align*}

these choices correspond to the matrices

\begin{align*}
&R_1 = \left[\begin{array}{c|cc|ccc} 
2 & 0 & 0 & 0 & 0 & 0 \\ \hline 
0 & 0 & -4 & 0 & 0 & 0 \\ 
0 & 1 & 4 & 0 & 0 & 0 \\ \hline 
0 & 0 & 0 & 0 & 0 & -12 \\ 
0 & 0 & 0 & 1 & 0 & 8 \\ 
0 & 0 & 0 & 0 & 1 & 1 \\ 
\end{array}\right] ,
&R_2 = \left[\begin{array}{c|cc|ccc} 
2 & 0 & 0 & 0 & 0 & 0 \\ \hline 
0 & 0 & 6 & 0 & 0 & 0 \\ 
0 & 1 & -1 & 0 & 0 & 0 \\ \hline 
0 & 0 & 0 & 0 & 0 & -12 \\ 
0 & 0 & 0 & 1 & 0 & 8 \\ 
0 & 0 & 0 & 0 & 1 & 1 \\ 
\end{array}\right],
&R_3 = \left[\begin{array}{c|cc|ccc} 
3 & 0 & 0 & 0 & 0 & 0 \\ \hline 
0 & 0 & 6 & 0 & 0 & 0 \\ 
0 & 1 & -1 & 0 & 0 & 0 \\ \hline 
0 & 0 & 0 & 0 & 0 & -12 \\ 
0 & 0 & 0 & 1 & 0 & 8 \\ 
0 & 0 & 0 & 0 & 1 & 1 \\ 
\end{array}\right] \\ \\
&R_4 = 
\left[\begin{array}{c|c|c|ccc}
2 & 0 & 0 & 0 & 0 & 0 \\ \hline
0 & 2 & 0 & 0 & 0 & 0 \\ \hline
0 & 0 & 2 & 0 & 0 & 0 \\ \hline
0 & 0 & 0 & 0 & 0 & -12 \\ 
0 & 0 & 0 & 1 & 0 & 8 \\ 
0 & 0 & 0 & 0 & 1 & 1 \\ 
\end{array}\right]
&R_5 = 
\left[\begin{array}{c|c|c|ccc}
-3 & 0 & 0 & 0 & 0 & 0 \\ \hline
0 & -3 & 0 & 0 & 0 & 0 \\ \hline
0 & 0 & -3 & 0 & 0 & 0 \\ \hline
0 & 0 & 0 & 0 & 0 & -12 \\ 
0 & 0 & 0 & 1 & 0 & 8 \\ 
0 & 0 & 0 & 0 & 1 & 1 \\ 
\end{array}\right]
&
.\end{align*}

Note: these are perhaps transposed from Hungerford’s notation.

Since none of the associated polynomials were irreducible over \({\mathbf{Q}}\), \(RCF(A)\) takes these forms over \({\mathbf{C}}\) as well.

To obtain the possible Jordan Canonical forms, we’ll instead need to consider elementary divisors. These can be obtained from the invariant factors above, yielding the possibilities:

\begin{align*}
R_1: &(x-2), ~~(x-2), ~(x-2)^2 ~~(x+3)  \\
R_2: &(x-2), ~~(x-2), ~~(x-2)^2, ~~(x+3), ~~ (x+3) \\
R_3: &(x-2), ~~(x-2)^2, ~~(x+3), ~~(x+3), ~~ (x+3) \\
R_4: &(x-2), ~~(x-2), ~~(x-2), ~~(x-2)^2, ~~ (x+3) \\
R_5: &(x+3), ~~(x+3), ~~(x+3), ~~(x+3), ~~ (x-2)^2 \\
\end{align*}

For the sake of notation, write \(J_\lambda^k\) for a \(k\times k\) Jordan block with \(\lambda\) on the diagonal and \(0_k\) for the \(k\times k\) zero matrix. We then have

\begin{align*}
R_1: &0_2 \oplus J_2^1 \oplus J_2^1 \oplus J_2^2 \oplus J_3^1 \\ \\
R_2: &J_2^1 \oplus J_2^1 \oplus J_2^2 \oplus J_3^1 \oplus J_3^1 \\ \\
R_3: &J_2^1 \oplus J_2^2 \oplus J_3^1 \oplus J_3^1 \oplus J_3^1 \\ \\ 
R_4: &J_2^1 \oplus J_2^1 \oplus J_2^1 \oplus J_2^2 \oplus J_3 \\ \\
R_5: &J_2^2 \oplus J_3^1 \oplus J_3^1 \oplus J_3^1 \oplus J_3^1 \\
.\end{align*}

Part 2

We’ll first exhibit the possibilities over \({\mathbf{C}}\), then show what subset can be obtained over \({\mathbf{Q}}\).

Over \({\mathbf{C}}\), we have \(x^2 + 1 = (x-i)(x+i)\). By the same argument used in Part 1, we know that \(q(x)\) is the largest invariant factor, and since \(\deg q = 3\), the degrees of the remaining factors must sum to 4 (since the degree \(\chi_A\) will be 7, and it’s the product of these factors).

We also know that the degrees must forma weakly decreasing partition of \(4\), which are

• \((1,1,1,1)\)
  • This can only be \(a_1=a_2=a_3=a_4\), a repeated linear factor, so there are 3 possibilities
• \((1,1,2)\)
  • This must satisfy \(a_1 = a_2\), so there are 3 possibilities for \(a_1=a_2\) and 2 for \(a_3\), for 6 total.
• \((2, 2)\)
  • This also must satisfy \(a_1 = a_2\), so there are \({3 \choose 2}/2 = 3\) possibilities

The possibilities are thus

\begin{align*}
R_1: a_1 = (x-i) && a_2 = (x-i) && a_3 = (x-i) && a_4 = (x-i) && a_5 = q(x) \\
R_2: a_1 = (x+i) && a_2 = (x+i) && a_3 = (x+i) && a_4 = (x+i) && a_5 = q(x) \\
R_3: a_1 = (x-7) && a_2 = (x-7) && a_3 = (x-7) && a_4 = (x-7) && a_5 = q(x) \\
\\
R_4: a_1 = (x+i) && a_2 = (x+i) && a_3 = (x+i)(x-i) && a_4 = q(x) \\
R_5: a_1  = (x+i) && a_2 = (x+i) && a_3 = (x+i)(x-7) && a_4 = q(x) \\
R_6: a_1  = (x-i) && a_2 = (x-i) && a_3 = (x-i)(x+i) && a_4 = q(x) \\
R_7: a_1  = (x-i) && a_2 = (x-i) && a_3 = (x-i)(x-7) && a_4 = q(x) \\
R_8: a_1  = (x-7) && a_2 = (x-7) && a_3 = (x-7)(x+i) && a_4 = q(x) \\
R_9: a_1  = (x-7) && a_2 = (x-7) && a_3 = (x-7)(x-i) && a_4 = q(x) \\
\\
R_{10}: a_1 = (x+i)(x-i) && a_2 = (x+i)(x-i) && a_3 = q(x) && \\
R_{11}: a_1 = (x+i)(x-7) && a_2 = (x+i)(x-7) && a_3 = q(x) && \\
R_{12}: a_1 = (x-i)(x-7) && a_2 = (x-i)(x-7) && a_3 = q(x) && \\
.\end{align*}

The corresponding Rational Canonical Forms for each \(R_j\) can be obtained by writing the companion matrix for the blocks \(a_i\) and taking their direct sum.

It is then easy to see that if \(A\) is taken over \({\mathbf{Q}}\) instead, only form \(R_3\) is possible (since \(x^2+1\) does not split over \({\mathbf{Q}}\)).

Let \(n J_\lambda^k\) denote \(J_\lambda^k \oplus J_\lambda^k \oplus \cdots \oplus J_\lambda^k\), where \(n\) copies appear in the direct sum corresponding to \(n\) Jordan blocks. We can immediately obtain the corresponding Jordan forms:

\begin{align*}
R_1: 5 J_i^1 \oplus J_{-i}^1 \oplus J_7^1 \\
R_2: 5 J_{-i}^1 \oplus J_{i}^1 \oplus J_7^1\\
R_3: 5J_7^1 \oplus J_i^1 \oplus J_{-i}^1 \\ \\
R_4: 4J_{-i}^1 \oplus 2J_i^1 \oplus J_7^1 \\
R_5:  4J_{-i}^1 \oplus J_{i}^1 \oplus 2J_7^1 \\
R_6: 4J_i^1 \oplus 2J_{-i}^1 \oplus J_7^1 \\
R_7:  4J_i^1 \oplus J_{-i}^1 \oplus 2J_7^1\\
R_8:  2J_{-i}^1 \oplus J_i^1 \oplus 2J_7^1 \\
R_9: 2J_i^1 \oplus J_{-i}^1 \oplus 4J_7^1 \\ \\
R_{10}:  3J_i^1 \oplus 3J_{-i}^1 \oplus J_7^1 \\
R_{11}:  J_i^1 \oplus 3J_{-i}^1 \oplus 3J_7^1 \\
R_{12}:  3J_i^1 \oplus J_{-i}^1 \oplus 3J_7^1
.\end{align*}

Problem 6

Let \(\phi \in \endo(V)\), then following a different proof than what is suggested in Hungerford, define an action

\begin{align*}
k[x] &\curvearrowright V \\
p(x)\curvearrowright\mathbf{v} &= p(\phi)(\mathbf{v})
,\end{align*}

which induces an invariant factor decomposition \begin{align*} V \cong \bigoplus_{i=1}^n \frac{k[x]}{(f_i)},\quad f_i \in k[x],\quad f_1 \divides f_2 \divides \cdots \divides f_n. \end{align*}

Then \(f_n(x)\) is the minimal polynomial of \(\phi\), and the characteristic polynomial is given by \(p_\phi(x) = \prod_{i=1}^n f_i(x)\). In particular, \(f_n(x) \divides p_\phi(x)\) and \(f_n(\phi) = 0\) by definition, so \(p_\phi(\phi) = 0\) as well. \(\hfill\blacksquare\)

Problem 7

Part 1

Suppose \(\phi \psi = \psi \phi\) and both \(\phi, \psi\) have bases of eigenvectors.

Letting \(\lambda_i\) denote the eigenvalues of \(\phi\), write \begin{align*} V = \bigoplus_i V_{\lambda_i}. \end{align*}

Now let \(\mathbf{v}\) be an eigenvector corresponding to \(\lambda_i\). We have \(\phi(\mathbf{v}) = \lambda_i \mathbf{v}\), and \begin{align*} \phi \psi(\mathbf{v}) = \psi \phi(\mathbf{v}) = \psi(\lambda_i \mathbf{v}) = \lambda_i \psi(\mathbf{v}), \end{align*}

which demonstrates that \(\psi(\mathbf{v})\) is also an eigenvector for \(\phi\), and moreover \(\psi(V_{\lambda_i}) \subseteq V_{\lambda_i}\), so it only sends \(\lambda_i\) eigenvectors to other \(\lambda_i\) eigenvectors.

Now consider \({\left.{{\psi}} \right|_{{V_{\lambda_i}}} }\), the restriction of \(\psi\) this eigenspace. Since \(\psi\) had an eigenbasis on \(V\), this restricts to an eigenbasis \(\mathcal{B}_i = \left\{{\mathbf{w}_i}\right\}\) of \(V_{\lambda_i}\). But then every element of \(\mathbf{w}_i\) is an eigenvector of \(\psi\) by definition, and we also have \(\mathbf{w}_i \in V_{\lambda_i}\), so the \(\mathbf{w}_i\) are also eigenvectors for \(\phi\).

Doing this for every \(i\), we obtain \(\mathcal{B} = {\textstyle\coprod}_i \mathcal{B}_i\) where \(\mathop{\mathrm{span}}(\mathcal B) = E\), which yields a simultaneous eigenbasis of \(E\) for both \(\psi\) and \(\phi\).

Part 2

Writing \(\mathcal{B} = \left\{{\mathbf{v}_i \mathrel{\Big|}1\leq i \leq n}\right\}\), this means we can form an invertible matrix \(P = [\mathbf{v}_1^t, \cdots, \mathbf{v}_n^t]\). Then if \(A\) is the matrix of \(\phi\) in the standard basis and \(B\) is the matrix of \(\psi\), we have \begin{align*} PAP^{-1}= D_1 \quad \text{and} \quad PBP^{-1}= D_2 \end{align*}

where \(D_1, D_2\) are diagonal. In other words, \(P\) simultaneously diagonalizes both \(A\) and \(B\).