Vtyjkyuk

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...

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 $ntimes n$ symmetric matrix $A$ is positive-definite iff for every $n$-dimensional column vector $vneq 0$ it is true that
$$v^T! cdot A cdot v >0$$
(which makes sense, since this is a $1times 1$ matrix.)

Now, if $B$ is similar to $A$, being $A$ and $B$ both symmetric, there exists a matrix $Q$ such that
$$B=Q^Tcdot A cdot Q.quad (*)$$

So what can you say about
$$v^T! cdot B cdot v$$
for any $vneq0$?

Solution:

Since
$$v^T! cdot B cdot v=v^T! cdot Q^Tcdot Acdot Q cdot v=(Qv)^T! cdot Acdot (Qv),$$
and being $(Qv)$ a vector of dimension $n$, say $w$, because $A$ is positive-definite we have
$$v^T! cdot B cdot v=w^T!cdot Acdot w>0,$$
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
$$B=tilde P^T P A P^T tilde P=(P^Ttilde P)^T A (P^T tilde P),$$
and the desired result follows if we name $Q=P^Ttilde P$.