Rings and Modules 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
\[f(t) = t^n + \sum^{n - 1} _
{i = 0} a _
i t^i \in \mathsf{k}[t]\]
@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
\[C(f) = \begin{pmatrix}
0 & \cdots & \cdots & 0 & -a _ 0 \\
1 & 0 & \ddots & \vdots & - a _ 1 \\
0 & 1 & \ddots & \vdots & \vdots \\
\vdots & \ddots & \ddots & 0 & -a _ {n-2} \\
0 & \cdots & 0 & 1 & -a _ {n - 1}
\end{pmatrix}\]
@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
\[J _ n(\lambda) = \begin{pmatrix}
\lambda & 1 & \cdots & \cdots & 0 \\
0 & \lambda & 1 & \cdots & 0 \\
\vdots & \ddots & \ddots & \ddots & 0 \\
\vdots & \ddots & \ddots & \ddots & 1 \\
0 & \cdots & \cdots & 0 & \lambda
\end{pmatrix}\]
@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).
@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 $x \cdot v = Av$
- $A$ 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 $\mathrm{Im}\, \phi = \langle v _ {\lambda _ 1}\rangle$ (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 $\mathbb F[x]$-module $M$ where the action of $f(x) \in \mathbb F[x]$ 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 characteristic 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 $d _ i$