positive definiteness similarity Does the concept of matrix similarity apply to the condition < Xv,v >? 
In other words, if a real square non-symmetric matrix X is similar to a symmetric positive definite matrix, do we have < Xv,v > > 0 for all nonzero vector v? 
I feel this is a trivial question, but I am a bit confused with the concept of matrix similarity. 
Thank you very much.
 A: Nope. If a matrix $A$ is similar to an SPD matrix $B$, this does not generally mean that $x^TAx>0$ for nonzero all $x\neq 0$ (well, obviously yes for some, namely the eigenvectors).
Consider a simple example:
$$
A=\begin{bmatrix}1 & 0 \\ 5 & 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 2\end{bmatrix}\begin{bmatrix}1 & 0 \\ 5 & 1\end{bmatrix}^{-1}
=\begin{bmatrix}
1 & 0 \\ -5 & 2
\end{bmatrix}.
$$
The matrix $A$ is clearly similar to an SPD matrix (a diagonal one with positive diagonal entries), but
$$
x^TAx=-2<0 \quad\text{for}\quad x=[1,1]^T.
$$
A: Matrix similarity is an equivalence relation between matrices in the same space.
A similarity transformation (one way to see it) transforms a vector $\vec{x}$ represented in the basis vectors of matrix $A$ to a vector $\vec{y}$ represented to the basis vectors of matrix $B$ in such a way so that the inner product (or norm) remains equal.
i.e $$<\vec{x}, A\vec{x}> = <B\vec{y}, \vec{y}>$$ (for a similarity transformation by an orthogonal matrix)
Let Q be an orthogonal matrix (i.e $Q^{-1} = Q^T$) and let $\vec{x} = Q\vec{y}$, then:
$$ x^TAx  = (Qy)^T A (Qy) = y^T Q^TAQy = y^TQ^{-1}AQy = y^TBy$$
What this means intuitively is that in general for any similarity transform that is an isometry both maintain the inner product of the space (in a normed space).
For an investigation and results on general (similarity) transformations that preserve positive-definiteness check this arxiv paper
A: The answer by Nikos M. can be generalized as follows:
Let $A\geq 0$, i.e. $\langle x|A|x\rangle \geq 0$ for all $|x\rangle$, and let $B$ be arbitrary. Then we have $B^\dagger A B\geq 0$:
Let $|x\rangle$ be arbitrary, and define $|y\rangle = B|x\rangle$. Then
$$\langle x| B^\dagger A B |x\rangle = (B|x\rangle)^\dagger A (B|x\rangle) = \langle y|A|y\rangle \geq 0$$
by the assumption that $A\geq 0$.
In particular, if $A$ is positive and $H$ is Hermitian, then $HAH$ is positive.
(Hope you're fine with bra-ket notation...)
