$A$ is similar to $B$ if $A\oplus A$ is similar to $B\oplus B$ Question:
If the matrix $\begin{pmatrix} A & 0 \\ 0& A \end{pmatrix}$ is similar to $\begin{pmatrix} B & 0 \\ 0 & B \end{pmatrix}$
show that:
the matrix $A$ is similar the matrix $B$
My try:
since  the matrix diag $(A,A)$  is  similar  matrix diag $(B,B)$;
and  I want use this elementary divisor,But there is not in The plural number field
then I can't.Thank you someone help me,Thank you  very much!
 A: This solution uses the notion of elementary divisors from the theory of canonical forms: see e.g. $\S$ 5 of these notes for the definition and properties.   I would be interested to know if there is a more elementary argument.
Use the fact that the elementary divisors of $\operatorname{diag}(A,B)$ are obtained by combining the elementary divisors of $A$ and $B$ (i.e., add the multiplicities) together with the fact that two matrices are similar if and only if they have the same elementary divisors (Theorem 7.2 of loc. cit.).
A: Please bare with me. It would be nice if someone would help me write this in a simpler way. Anyways, here is my attempt.
Proof:
Let's write the rational canonical form of $A$ as $A_{RCF}$ and the RCF of $B$ as $B_{RCF}$.
We proceed by contradiction. Suppose that $A$ is not similar to $B$. Then $A$ and $B$ have a different rational canonical forms. That is there exists $h(x)$ such that $a:=$ number of $h(x)$'s in the list of elementary divisors of $A$ is not equal to $b:=$ number of $h(x)$'s in the list of elementary divisors of $B$. 
Case 1.
WLOG, if $b=0$, then $h(x)$ is not an elementary divisor of $B$ but is an elementary divisor of $A$ since $a \neq b$. Since the rational canonical form of $A \oplus A$ is $A_{RCF} \oplus A_{RCF}$, then $h(x)$ is also an elementary divisor $A \oplus A$. However, $h(x)$ is not an elementary divisor of $B \oplus B$.
Case 2. Let's assume $a,b \neq 0$.
Then, $A_{RCF} = \bigoplus_a C(h(x)) \oplus A_1$ where $A_1 = \bigoplus_{i=1}^{k_1} C(f_{i}(x))$ for some $k_1$ and such that for all $i, f_i(x) \neq h(x)$. Also, $B_{RCF}= \bigoplus_b C(h(x)) \oplus A_2$ where $A_2 = \bigoplus_{i=1}^{k_2} C(g_{i}(x))$ for some $k_2$ and such that for all $i, g_i(x) \neq h(x)$. Hence, the RCF of $A \oplus A$ is $\bigoplus_{2a} C(h(x)) \oplus A_2$ where $A_2 = \bigoplus_{i=1}^{l_1} C(g_i(x))$ for some $l_1$ and such that for all $i, r_i(x) \neq h(x)$. And the RCF of $B \oplus B$ is $\bigoplus_{2b} C(h(x)) \oplus B_2$ where $B_2 = \bigoplus_{i=1}^{l_2} C(s_i(x))$ for some $1_2$ and such that for all $i, s_i(x) \neq h(x)$. Therefore, the number of $h(x)$'s in the list of elementary divisor s of $A \oplus A$ is $2a$ and the number of $h(x)$'s in the list of elementary divisor s of $B \oplus B$ is $2b$. Note that $2a \neq 2b$.
In both cases, we find a contradiction since $A \oplus A$ is similar to $B \oplus B$.
Thus, $A$ is similar to $B$.
