If $A,B,C$ and $D$ are $n\times n$ matrices. Then if $A$ and $B$ are similar and also $C$ ad $D$, is it true that $A+C$ and $B+D$ are similar? Is it true that $AC$ and $BD$ are similar?
Here $A$ and $B$ are similar iff there is some invertible $P$ such that $A=PBP^{-1}$.
My attempt was just about working directly with the definition and try to show $A+C$ as $M(B+D)M^{-1}$ etc... but it doesn't work because I couldn't link different "change of basis matrix" which makes the similarity between two different pair of matrices...
Thinking on the matrices as matrices of linear transformations and $P$ as a change of basis matrix I can see that $A+C$ and $B+D$ could not be similar and $AC$ and $BD$ maybe... in the sense that if $A=[T]^{\alpha}_{\alpha}$ for some $T$ linear and the same for $B,C$ and $D$ with different basis, I can see that there is no reason for the two sums being two representations for the same linear transformation in different basis.
E.g.:
$
A=\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}
$,
$
C= \begin{pmatrix} -1 & 1 \\ 0 & 1\end{pmatrix}
$,
$ D=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $.
Then $A$ and $A$ are trivially similar, and so $C$ and $D$. But $\begin{pmatrix}2&0\\0&2\end{pmatrix}$ and $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ clearly doesn't represent the same linear transformation.
At a first glance I thought that $AC$ and $BD$ would be similar because that was just a composition of transformations. But I couldn't prove it... and I found a problem too... Can I say that any matrix can be seen as a $[T]^{\beta}_{\alpha}$ for $\alpha\not =\beta$? Because if I can't, the argument on this part of the problem becomes a little harder...
Anyone can help me?
Unfortunately your $C(=AC)$ is similar to your $D(=BD)$ over any field of characteristic $\ne2$, because both of them have the same spectrum and the two eigenvalues in the spectrum are different. In fact, $$ \pmatrix{0&\frac12\\ 1&-\frac12}C\pmatrix{1&1\\ 2&0}=\pmatrix{0&\frac12\\ 1&-\frac12}\pmatrix{-1&1\\ 0&1}\pmatrix{1&1\\ 2&0}=\pmatrix{1&0\\ 0&-1}=D. $$
For a counterexample that works over any field, let $A=B=\pmatrix{0&1\\ 1&1},\ C=\pmatrix{1&0\\ 0&0}$ and $D=\pmatrix{0&0\\ 0&1}$. Then $A$ is similar to $B$ and $C$ is similar to $D$, but the singular matrix $A+C=\pmatrix{1&1\\ 1&1}$ is not similar to the non-singular matrix $B+D=\pmatrix{0&1\\ 1&2}$ and the traceless matrix $AC=\pmatrix{0&0\\ 1&0}$ is not similar to the matrix $BD=\pmatrix{0&1\\ 0&1}$ with trace $1$.