Suppose square matrix $A$ is positive-definite and $B$ is similar to $A$. Is B positive-definite too? Suppose square matrix $A$ is positive-definite and $B$ is similar to $A$. Is B positive-definite too?  
I know a square matrix $A$ is positive-definite equal that $A$ is congruent to identity matrix $E$, i.e there exists an invertible matrix $C$ such that $C'AC=E$. So every matrix congruent to a positive-definite matrix is positive-definite. But what it will behavior when it's similar to a positive-definite matrix. I suppose it may not be positive-definite. But I can't get a counterexample. I tried from the definition...
 A: I assume $B$ is also symmetric, which is for most of the sources a precondition to talk about positive-definiteness, etc.
Try using the fact that an $n\times n$ symmetric matrix $A$ is positive-definite iff for every $n$-dimensional column vector $v\neq 0$ it is true that
$$v^T\! \cdot A \cdot v >0$$
(which makes sense, since this is a $1\times 1$ matrix.)
Now, if $B$ is similar to $A$, being $A$ and $B$ both symmetric, there exists a matrix $Q$ such that
$$B=Q^T\cdot A \cdot Q.\quad (*)$$
So what can you say about
$$v^T\! \cdot B \cdot v$$
for any $v\neq0$?

Solution:
Since
$$v^T\! \cdot B \cdot v=v^T\! \cdot Q^T\cdot A\cdot Q \cdot v=(Qv)^T\! \cdot A\cdot (Qv),$$
and being $(Qv)$ a vector of dimension $n$, say $w$, because $A$ is positive-definite we have
$$v^T\! \cdot B \cdot v=w^T\!\cdot A\cdot w>0,$$
which proves $B$ is also positive-definite.

(*) It is a theorem that if $A$ is a symmetric (real valued) matrix, then there exists an orthogonal matrix $P$ and a diagonal matrix $D$ such that
$$A=P^T D P.$$
Since the same is valid for $B$, say
$$B=\tilde P^T \tilde D \tilde P^,$$
and being $P^T=P^{-1}$ and also $D=\tilde D$ (because $A$ and $B$ are similar), we have
$$B=\tilde P^T P A P^T \tilde P=(P^T\tilde P)^T A (P^T \tilde P),$$
and the desired result follows if we name $Q=P^T\tilde P$.
