Notes - Rings and Moudles HT24, Rational and Jordan canonical forms
Flashcards
Suppose:
- $M$ is a finitely generated $k[t]$-module
We know that it is possible to identify $M$ with a vector space $V$ and an endomorphism $T : V \to V$. What relationship is there between the submodules of $M$ and the pair $(V, T)$?
Suppose we have a monic polynomial
\[f(t) = t^n + \sum^{n - 1}_
{i = 0} a_
i t^i \in \mathsf{k}[t]\]
Can you define the corresponding companion matrix of $f$, denoted $C(f)$?
Suppose:
- $V$ is a vector space
- $A : V \to V$ is a linear map
What does it mean for $v$ to be a cyclic vector of $A$?
spans $V$.
Suppose:
- $V$ is a vector space
- $A : V \to V$ is a linear map
- $v$ is a cyclic vector of $A$ (i.e. $\{v, Av, A^2 v, \cdots\}$ spans $V$)
Given this, what useful relationship between $m _ A$ and $\chi _ A$ is there?
Suppose we have the monic polynomial
\[f(t) = t^n + \sum^{n - 1}_{i = 0} a_i t^i \in \mathsf{k}[t]\]
with corresponding companion matrix
\[C(f) = \begin{pmatrix}
0 & \cdots & \cdots & 0 & -a_0 \\\\
1 & 0 & \ddots & \vdots & - a_1 \\\\
0 & 1 & \ddots & \vdots & \vdots \\\\
\vdots & \ddots & \ddots & 0 & -a_{n-2} \\\\
0 & \cdots & 0 & 1 & -a_{n - 1}
\end{pmatrix}\]
Can you state a theorem that relates this matrix to a particular $\mathbb F[t]$-module?
The $\mathbb F[t]$-module given by $\mathbb F[t] / \langle f \rangle$ has the basis
\[\\{t^i + \langle f \rangle : 0 \le i \le n-1\\}\]and the matrix for the action of $t$ with respect to this basis is given by $C(f)$.
Suppose we have the monic polynomial
\[f(t) = (t - \lambda)^n \in \mathbb F[t]\]
for some $\lambda \in \mathbb F$ and $n \in \mathbb Z _ {\ge 0}$, and the corresponding Jordan block $J _ n(\lambda)$ given by
\[J_n(\lambda) = \begin{pmatrix}
\lambda & 1 & \cdots & \cdots & 0 \\\\
0 & \lambda & 1 & \cdots & 0 \\\\
\vdots & \ddots & \ddots & \ddots & 0 \\\\
\vdots & \ddots & \ddots & \ddots & 1 \\\\
0 & \cdots & \cdots & 0 & \lambda
\end{pmatrix}\]
Can you state a theorem that relates this matrix to a particular $\mathbb F[t]$-module?
The $\mathbb F[t]$-module given by $\mathbb F[t] / \langle f \rangle$ has the basis
\[\\{(t - \lambda)^k + \langle f \rangle \mid 0 \le k \le n-1\\}\]and the matrix for the action of $t$ with respect to this basis is given by $J _ n(\lambda)$.
Suppose we have the monic polynomials
\[\begin{aligned}
&f(t) = t^n + \sum^{n - 1}_{i = 0} a_i t^i \in \mathbb{F}[t] \\\\
&g(t) = (t - \lambda)^n \in \mathbb F[t]
\end{aligned}\]
with the corresponding companion matrix
\[C(f) = \begin{pmatrix}
0 & \cdots & \cdots & 0 & -a_0 \\\\
1 & 0 & \ddots & \vdots & - a_1 \\\\
0 & 1 & \ddots & \vdots & \vdots \\\\
\vdots & \ddots & \ddots & 0 & -a_{n-2} \\\\
0 & \cdots & 0 & 1 & -a_{n - 1}
\end{pmatrix}\]
and the Jordan block
\[J_n(\lambda) = \begin{pmatrix}
\lambda & 1 & \cdots & \cdots & 0 \\\\
0 & \lambda & 1 & \cdots & 0 \\\\
\vdots & \ddots & \ddots & \ddots & 0 \\\\
\vdots & \ddots & \ddots & \ddots & 1 \\\\
0 & \cdots & \cdots & 0 & \lambda
\end{pmatrix}\]
Quickly prove the following statements:
- The $\mathbb F[t]$-modules $\mathbb F[t]/\langle f \rangle$ and $\mathbb F[t] / \langle g \rangle$ have respective bases
\[\\{t^i + \langle f \rangle \mid 0 \le i \le n-1\\}\]
\[\\{(t-\lambda)^k + \langle g \rangle \mid 0 \le k \le n-1\\} \text{ (ordered by decr. degree)}\]
- With respect to these bases the action of $t$ for each is given by $C(f)$ and $J _ n(\lambda)$.
The fact that these two sets are bases follows from the fact that any set containing one distinct polynomial of degree from $0$ to $n-1$ will form a basis.
For the second:
- Considering $f$. For $i < n -1$, we have $t(t^i + \langle f \rangle) = t^{i + 1} + \langle f \rangle$, and for $i = n-1$, $t(t^{n - 1} + \langle f \rangle) = t^n + \langle f \rangle = -\sum^{n - 1} _ {k = 0} a^k t^k + \langle f \rangle$
- Considering $g$. $t$ acts with matrix $J _ n(\lambda)$ if and only if $(t - \lambda)$ acts on the module with matrix $J _ n(0)$. Multiplying by $(t - \lambda)$ is equivalent to shifting to the next element of the basis, which is what the matrix shows (if we order the basis with decreasing degree).
Can you state the rational canonical form theorem?
Suppose:
- $V$ is a nonzero finite dimensional $\mathbb F$-vector space
- $T : V \to V$ is a linear map
Then:
There exists unique nonconstant monic polynomials $f _ 1, \cdots, f _ k \in \mathbb F[t]$ such that $f _ 1 \mid f _ 2 \mid \cdots \mid f _ k$ and a basis of $V$ with respect to which $T$ has the block diagonal form
\[\begin{pmatrix} C(f_1) & 0 & \cdots & 0 \\\\ 0 & C(f_2) & 0 & \vdots\\\\ \vdots & 0 & \ddots & \vdots \\\\ 0 & \cdots & 0 & C(f_k) \end{pmatrix}\]where each $C(f _ i)$ is the companion matrix, defined by
\[C(f_i) = \begin{pmatrix} 0 & \cdots & \cdots & 0 & -a_0 \\\\ 1 & 0 & \ddots & \vdots & - a_1 \\\\ 0 & 1 & \ddots & \vdots & \vdots \\\\ \vdots & \ddots & \ddots & 0 & -a_{n-2} \\\\ 0 & \cdots & 0 & 1 & -a_{n - 1} \end{pmatrix}\]and $a _ k$ are the corresponding coefficients of the polynomial.
Quickly prove the rational canonical form theorem, which states that if
- $V$ is a nonzero finite dimensional $\mathbb F$-vector space
- $T : V \to V$ is a linear map
then:
There exists unique nonconstant monic polynomials $f _ 1, \cdots, f _ r \in \mathbb F[t]$ such that $f _ 1 \mid f _ 2 \mid \cdots \mid f _ r$ and a basis of $V$ with respect to which $T$ has the block diagonal form
\[\begin{pmatrix}
C(f_1) & 0 & \cdots & 0 \\\\
0 & C(f_2) & 0 & \vdots\\\\
\vdots & 0 & \ddots & \vdots \\\\
0 & \cdots & 0 & C(f_r)
\end{pmatrix}\]
where each $C(f _ i)$ is the companion matrix, defined by
\[C(f_i) = \begin{pmatrix}
0 & \cdots & \cdots & 0 & -a_0 \\\\
1 & 0 & \ddots & \vdots & - a_1 \\\\
0 & 1 & \ddots & \vdots & \vdots \\\\
\vdots & \ddots & \ddots & 0 & -a_{n-2} \\\\
0 & \cdots & 0 & 1 & -a_{n - 1}
\end{pmatrix}\]
and $a _ k$ are the corresponding coefficients of the polynomial.
By the structure theorem, we know that there exists an isomorphism
\[\theta : V \to \left( \bigoplus^r_{i = 1} \frac{\mathbb F[t]}{\langle f_i\rangle} \right) \oplus \mathbb F[t]^s\]where $f _ 1 \mid f _ 2 \mid \cdots \mid f _ r$ and $f _ i$ are monic nonunits, which are unique. It must be the case that $s = 0$, since $\mathbb F[t]^s$ is infinite-dimensional as an $\mathbb F$-vector space (you would need a basis element for each monomial $t^k$) but $V$ is finite-dimensional as an $\mathbb F$-vector space since the left-hand side and right-hand sides are isomorphic.
The direct sum
\[\bigoplus^r_{i = 1} \frac{\mathbb F[t]}{\langle f_i\rangle}\]Has a basis $\mathcal B$ given by the union of bases for each summand. Then $\theta^{-1}(\mathcal B)$ is a basis for $V$.
Using the basis $\{t^j + \langle f \rangle \mid 0 \le j \le n-1\}$, for each $\mathbb F[t]/\langle f _ i \rangle$, the action of $t$ on $V$ is given by $C(f _ i)$ (since for $j < n -1$, we have $t(t^j + \langle f \rangle) = t^{j + 1} + \langle f \rangle$, and for $j = n-1$, $t(t^{n - 1} + \langle f \rangle) = t^n + \langle f \rangle = -\sum^{n - 1} _ {k = 0} a^k t^k + \langle f \rangle$).
Suppose:
- $M = \mathbb R^3$ and is being viewed as an $R[x]$ module via the action $x \cdot v = Ax$
- $A$ is a linear transformation $M \to M$
- We have the decomposition of $V$ into eigenspaces of $A$ given by $V = \langle v _ {\lambda _ 1} \rangle \oplus \langle v _ {\lambda _ 2} \rangle \oplus \langle v _ {\lambda _ 3} \rangle$
Via an isomorphism, how can you rewrite each eigenspace, and what decomposition does this give you for $A$?
Consider $v _ {\lambda _ 1}$. Then
\[\langle v_{\lambda_1} \rangle \cong \frac{\mathbb R[x]}{\langle x - \lambda_1\rangle}\]This is because for $v \in \langle v _ {\lambda _ 1} \rangle$, we have $x \cdot v = Av = \lambda v$. Hence for any polynomial $f(x) \in \mathbb R[x]$ and $v \in \langle v _ {\lambda _ 1} \rangle$, we have
\[f(x) \cdot v = f(A) v = f(\lambda) v\]so via the map $\phi : \mathbb R[x] \to \langle v _ {\lambda _ 1} \rangle$ where $f \mapsto f(\lambda) v _ {\lambda _ 1}$, we can use the 1st isomorphism theorem since $\ker \phi = \langle x - \lambda _ 1 \rangle$ and $\text{Im } \phi = \langle v _ {\lambda _ 1}\rangle$ (note we are relying on the fact that $M = \mathbb R^3$ and there are three eigenspaces).
Then
\[V = \frac{\mathbb R[x]}{\langle x - \lambda_1\rangle} \oplus \frac{\mathbb R[x]}{\langle x - \lambda_2\rangle} \oplus \frac{\mathbb R[x]}{\langle x - \lambda_3\rangle}\]Suppose we have a $\mathbb F[t]$-module $M$ where the action of $f(x) \in \mathbb F[t]$ corresponds to $f(A) v$. After forming the presentation matrix $xI - A$, we put it into Smith Normal Form and get the invariant factors $d _ 1(x) \mid d _ 2(x) \mid \cdots \mid d _ k(x)$.
How can you deduce the minimal and characteristic polynomial from these factors, and why does this work?
- Minimal polynomial is $d _ k(x)$. If you consider the rational canonical form of $A$, each invariant factor corresponds to a companion matrix $C(d _ i)$ in a big block matrix. So the minimal polynomial of $A$ must be the lowest common multiple of all of the minimal polynomials for each $C(d _ i)$. Since each $C(d _ i)$ has minimal polynomial $d _ i$ (this follows from the definition of the companion matrix), the lowest common multiple is $d _ k(x)$. So the minimal polynomial of $A$ is $d _ k(x)$.
- Characteristic polynomial is $d _ 1(x) \cdots d _ k(x)$. Again considering the rational canonical form of $A$, each invariant factor corresponds to a companion matrix with characteristic polynomial $d _ i(x)$ in the big block matrix. Since the characteristic polynomial of the block matrix is a product of the characteristic polynomials of the blocks, the characteristic polynomial is $d _ 1(x) \cdots d _ k(x)$.
In other words:
- The minimal polynomial of a block matrix is a the lowest common multiple of the minimal polynomials of each of the blocks
- The characteristic polynomial of a block matrix is the product of the minimal polynomials in each block
- The rational canonical form of $A$ means that there is a direct correspondence between each invariant factor $d _ i$ and a block matrix with minimal and characteristic polynomial $d _ i$