Show that a matrix is positive definite

Question

Let $U$ be a symmetric,positive definite matrix and let $Q \ne I$ be an orthogonal matrix s.t. $\det(Q)>0$. I want to show $$(Q-I)^tU(Q-I)$$ is positive definite

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I want to show that $$\langle (Q-I)^tU(Q-I) v,v\rangle \geq 0$$ for every $v \in \mathbb{R}^n$ and equality holds iff $v=0$.

So I wrote $$\langle (Q-I)^tU(Q-I) v,v\rangle = \langle U(Q-I) v,(Q-I)^ v\rangle $$

Let's set $w = (Q-I) v$, then I have

$$\langle Uw,w \rangle \geq 0$$

for every $w \in \mathbb{R^n}$ since $U$ is positive definite and equality holds iff $$w=0$$

Now, $w=0$ iff $Qv = v$

but this does not imply that $v=0$... I don't know how to move now

Answer 1

Take $$ Q = \begin{pmatrix} 0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 1 \end{pmatrix} $$ and $U=I$ as a counterexample.