Proof that $M$ and $M^{T}$ are similar 
Triangular matrix $M \in \mathbb R^{n,n}$ and all elements on the diagonal are different. Proof that $M$ and $M^{T}$ are similar.

I know that the matrices are similar when the matrix similarity relation is the relation of equivalence so it is reflexive, symmetric  and transitive relation.Unfortunately I don't know what to use this information in my task.
 Can you get some tips?
 A: Presume that they are similar, and in particular $M = P^{-1}M^{T}P$. What can you say about $P$?
Then write the equation as $PM = M^{T}P$. Given what we know about $P$, this should suggest what we need for $M$ and $M^{T}$ to be similar (i.e. it should tell us what type of matrix $P$ is sufficient to make the equation true).
A: It turns out that any square matrix over any field is similar to its transpose. Here's why: First assume the field is algebraically closed. Write $M = AJA^{-1}$ where $J$ is the Jordan canonical form of $M$. Then $M^T = (A^T)^{-1} J^T A^T$. But it's not hard to show that $J$ and $J^T$ are similar (via permuting rows and columns within each Jordan block), so $M$ is similar to $M^T$.
If the field is not algebraically closed, apply the following theorem: If the field $F$ is the algebraic closure of the field $\Gamma$, and $M, N$ are square matrices over $\Gamma$ which are "similar over $F$" in the sense that there exists an invertible matrix $A$ over $F$ with $M = A N A^{-1}$, then $M$ and $N$ must also be "similar over $\Gamma$", i.e., there exists an invertible matrix $B$ over $\Gamma$ such that $M = B N B^{-1}$. For a proof via the rational canonical form, see Steven Roman's Advanced Linear Algebra, Third Edition, p. 182. For the special case $F = \mathbb C$ and $\Gamma = \mathbb R$, there's a nice alternate proof that avoids the rational canonical form: see Victor Prasolov's Problems and Theorems in Linear Algebra, first theorem in the section on the Jordan normal form.
By the way, the similarity $M \approx M^T$ has a nice interpretation in the more abstract framework of vector spaces and linear maps: For a finite-dimensional vector space $V$ with a fixed ordered basis, and its dual space $V^*$ with the dual basis, the matrices of a linear operator $\tau : V \to V$ and its operator adjoint $\tau^\times : V^* \to V^*$ are transposes of one another. So the theorem tells us that any linear operator looks just like its adjoint, up to a change of basis.
