This post will prove the uniqueness of Jordan form, Smith normal form and Rational canonical form in an equivalence class of similar matrices. A criterion to check similarlity of matrices using these forms is given, and a nice lemma of similarity being irrelevant to field extension is shown using rational canonical form.

This post will prove the uniqueness of Jordan form, Smith normal form and Rational canonical form in an equivalence class of similar matrices. This is why they are called canonical.

Theorem 1 Let $A$ be an $n \times n$ matrix (over an algebraically closed field $k$). Then there exists a unique matrix among all the similar matrices of $A$, that is in Jordan form. (up to permutation of Jordan blocks)

Proof

We have seen in a previous post that how $\mathrm{dim} \, \mathrm{Ker} \, (A - \lambda I)^k$ determines the Jordan blocks. This determines the uniqueness.

Alternatively, notice that if we regard $A$ to be the matrix of a linear transformation $T: V \rightarrow V$, then a Jordan form corresopnds to the primary decomposition of finitely generated modules over PID, which is unique up to permutation. $\Box$

Theorem 2 Let $A$ be an $n \times n$ matrix (over a PID). Then there exists a unique matrix among all the similar matrices of $A$, that is in Smith normal form. (up to associatedness of elementary divisors)

Proof

Define $\Delta_{A,j}$ to be the greatest common divisor of all minors of $A$ of order $j$ for $j = 1, 2, \cdots , n$.

Once we have put the matrix $A$ in its Smith normal form $A'$ = $diag( a_1, \cdots , a_k, 0, \cdots , 0)$, it is clear that

$\Delta_{A',j} = a_1 \cdots a_j$

(if we set $a_j = 0$ if $j > k$) Therefore the theorem is proved once we have

Lemma 1 (Invariance of $\Delta_j$) Let $B = PAQ$, where $P,Q$ are invertible matrices. Then for any $j$, $\Delta_{B,j} = \Delta_{A,j}$.

Proof Clear from Binet-Cauchy formula. $\Box$

Theorem 3 Let $A$ be an $n \times n$ matrix (over a field $k$). Then there exists a unique matrix among all the similar matrices of $A$, that is in rational canonical form.

Proof

If we regard $A$ to be the matrix of a linear transformation $T: V \rightarrow V$, then rational canonical form corresopnds to the decomposition of finitely generated modules over PID, which is unique. $\Box$

This gives us some methods to check when two matrices are similar.

Corollary 1 Two $n \times n$ matrices $A$ and $B$ over a field $k$ iff they have the same Jordan form/rational canonical form.

Smith normal form also helps determine the similarity of matrices.

Corollary 2 Two $n \times n$ matrices $A$ and $B$ over a field $k$ are similar iff $xI - A$ and $xI - B$ has the same Smith normal form.

Proof

Regard $A,B$ to be the matrices of linear transformations $S, T: V \rightarrow V$. Remember that the Smith normal form of $xI - A$ ($xI - B$) represents exactly the decomposition of $k[S]$-module. ($k[T]$-module) Therefore it suffices to show that $A$, $B$ are similar iff the $k[S]$-module structure of $V$ is the same as its $k[T]$-module structure, which is obvious. $\Box$

Similarity and underlying field

As Jordan form needs the field to be extended to its algebraic closure, it raises a natural question: Let $L \subset K$ are two fields, and $A,B$ are two $n \times n$ matrices with entries in $L$. If $A,B$ are similar over $K$, are they similar over $L$? The answer is yes.

Theorem 4 Let $L \subset K$ are two fields, and $A,B$ are two $n \times n$ matrices with entries in $L$. If $A,B$ are similar over $K$, then they are similar over $L$.

Proof for infinite field case

$B = PAP^{-1}$ is equivalent to $BP - PA$ = 0. Thus we are asking if $BX - XA = 0$ is solvable over $K$, can it be solved over $L$.

Using Kronecker product, rewrite the equation as

$\left(B^T \otimes I - I \otimes A\right)X = 0$

This shows that the solution space of $X$ (over either $K$ or $L$) can be spanned by matrices with entries in $L$. Let $C_1, \cdots, C_k$ be one such basis, and consider the multinomial

$p(x_1,\cdots, x_k) = det\left(x_1C_1 + \cdots + x_kC_k \right)$

$p$ is not identically 0, because over $K$, there exists some $(x_1, \cdots, x_k) \in K^n$ such that $p(x_1,\cdots, x_k) \neq 0$. If the field is infinite, then this implies that for some $(x_1, \cdots, x_k) \in L^n$, $p(x_1,\cdots,x_k) \neq 0$, meaning that $BX - XA = 0$ is solvable over $L$. $\Box$

There is a swft argument for the general case, as indicated by loup blanc in this post.

Proof 2

1. $A$, $B$ are similar if and only if they have the same rational canonical form, as proved above.

2. Notice furthermore that the rational canonical form of $A$ in $L$ is the same as that in $K$. Reason:

Suppose that we can find invertible $P,Q \in M_n(L[x])$ such that $P(xI - A^t)Q$ is in Smith normal form. By uniqueness of normal form, and that $M_n(L[x]) \subset M_n(K[x])$, the Smith normal form in $M_n(K[x])$ is the same. This implies that the rational canonical form are the same.

If $A, B$ are similar over $K$, they have the same rational canonical form over $K$. But their rational canonical form over $L$ are the same, so they are similar over $L$.

Quick summary

1. Jordan form/rational canonical form of a linear map $T: V \rightarrow V$ is merely the matrix form of the structure of $V$ as a $k[T]$-module.

2. The similarity problem is the same as whether the $k[T]$-module structure of $V$ are the same.

3. $xI-T$ is exactly the relations matrix/kernel of the natural map $k[T]^n \rightarrow V$. Thus it is significant in rational canonical form/Jordan form.