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Suppose we have fields $\mathbb{K}_1, \mathbb{K}_2$ such that $$\mathbb{K}_1 \subset \mathbb{K}_2.$$

Question Let $A, B$ be square $n\times n$ matrices over $\mathbb{K}_1$. If there exists $P_2 \in \mathrm{GL}(n, \mathbb{K}_2)$ s.t.

$$ P_2^{-1}AP_2=B, $$

then there exists $P_1 \in \mathrm{GL}(n, \mathbb{K}_1)$ s.t. $$ P_1^{-1}AP_1=B. $$ Is it true?

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marked as duplicate by Jyrki Lahtonen abstract-algebra yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

up vote 5 down vote accepted

Here is a somewhat more elementary proof that works if the field $\mathbb{K}_1$ is sufficiently big (for example it works with infinite fields). Assume $A$ and $B$ are similar over $\mathbb{K}_2$, so you have $P \in \textrm{GL}_n(\mathbb{K}_2)$ such that

$$P A = B P$$

Now write $P = (p_{i,j})$, and pick a basis $(e_1, \ldots, e_r)$ of $\textrm{Vect}_{\mathbb{K}_1}(p_{i,j})$ as a $\mathbb{K}_1$-vector space. We can thus write

$$P = \sum_{i = 1}^r e_i P_i$$

with $P_i \in M_n(\mathbb{K}_1)$ for $i \in [\!|1,r|\!]$ (note that the $P_i$ need not be invertible in general). And because $(e_1, \ldots, e_r)$ is free, we get $P_i A = B P_i$ for all $i \in [\!|1,r|\!]$. Now consider the polynomial

$$f(X_1, \ldots, X_r) = \det \left(\sum_{i = 1}^r X_i P_i \right) \in \mathbb{K}_1[X_1, \ldots, X_r]$$

Since $P$ is invertible, we have $f(e_1, \ldots, e_r) \neq 0$, so $f$ is non zero. And if $|\mathbb{K}_1| > n$, then there exist $(\lambda_1, \ldots, \lambda_r) \in \mathbb{K}_1^r$ such that $f(\lambda_1, \ldots, \lambda_r) \neq 0$ (see lemma below for an explanation). So the matrix $P' = \sum_{i = 1}^r \lambda_i P_i$ is in $\textrm{GL}_n(\mathbb{K}_1)$ and satisfies $A = P'^{-1} B P'$.

Lemma : Let $f \in \mathbb{K}[X_1, \ldots, X_r]$ be a non zero polynomial. Assume

$$|\mathbb{K}| > \deg(f) = \max_{1 \le i \le r} (\deg_{i}(f))$$

Then there exist a point $(\lambda_1, \ldots, \lambda_r) \in \mathbb{K}^r$ such that $f(\lambda_1, \ldots, \lambda_r) \neq 0$.

Proof : We will prove this by induction on the number $r \ge 1$ of variables. The case $r = 1$ follows from the fact that a polynomial of degree $d$ has at most $d$ roots. If $r \ge 2$, write $d = \deg_{r}(f)$ and

$$f = \sum_{k = 0}^{d} a_k X_r^k$$

where the $a_i$ are polynomials in $r-1$ variables with $\deg(a_i) \le \deg(f)$. Now $a_d$ is non zero, so by induction, there is a point $(\lambda_1, \ldots, \lambda_{r-1}) \in \mathbb{K}_1^{r-1}$ such that $a_{d}(\lambda_1, \ldots, \lambda_{r-1}) \neq 0$. Finally, since $g(X) = f(\lambda_1, \ldots, \lambda_{r-1}, X)$ is a non zero polynomial with $\deg(g) \le \deg(f)$, you can find $\lambda_n$ such that $g(\lambda_n) \neq 0$. Which concludes.

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Sorry, I don't understand what $\text{Vect}_{\mathbb{K}_1}(p_{i,j})$ is supposed to be. – Qiaochu Yuan Jun 7 '11 at 16:28
@Qiaochu: I understand $\mathrm{Vect}_{\mathbb{K}_1}$ as the vector subspace of $\mathbb{K}_2$ (regarded as a $\mathbb{K}_1$-vector space) spanned by $(p_{1, 1}, p_{1, 2} \ldots p_{n, n})$. $$$$ @Joel: I like very much this proof. It couldn't be simpler than that. The only thing I can't understand is that "if $\mathbb{K}_1$ is sufficiently big...". Why? We have a polynomial $f$ and know that it is not null, because in an appropriate extension we have $f(e_1 \ldots e_r)\ne 0$. So we must have $f(\mathrm{something})\ne 0$ for $\mathrm{something} \in \mathbb{K}_1$... Where am I wrong? – Giuseppe Negro Jun 7 '11 at 16:50
@dissonance: take $\mathbb{K}_1 = \mathbb{F}_p$ and, for example, $f(x_1) = x_1^p - x_1$. For a homogeneous example take $f(x_1, x_2) = x_1^p x_2 - x_2^p x_1$. – Qiaochu Yuan Jun 7 '11 at 16:55
@Qiaochu: Ok, now I see. Thank you. Can we say that Joel's proof works only if $\mathrm{char}(\mathbb{K}_1)$ is greater than some lower bound? And, if affirmative, how big is this lower bound? – Giuseppe Negro Jun 7 '11 at 17:05
@dissonance : The assumption is not on the characteristic of the field, but rather on its cardinality. By sufficiently large, I mean that if $|\mathbb{K}_1| > n$ (where $n$ is the size of the matrices), the proof should work. I included an explanation in the answer above. – Joel Cohen Jun 7 '11 at 22:45

Yes. It is true. $A$ and $B$ are similar iff $\lambda I-A$ and $\lambda I-B$ are equivalent iff they have the same invariant factors or the same determinate factors. These are only depend on matrix itself.

see Similar matrix

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See also rational canonical form – Aaron Jun 7 '11 at 16:57

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