# 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?

• $M$ is diagonalizable, that is $P^{-1}MP = D$ Feb 2, 2019 at 22:40

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).
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.