# Identity with scalar product

In one proof of some theorem regarding normal distribution there is used I think an idenitity which states

$$(ABy,By)=(B^{-1}ABy,y)$$ where A is symetric and positive, B is orthogonal and y is a vector.

Can someone explain to me why is it so?

## 3 Answers

I assume $$(\cdot, \cdot)$$ denotes the scalar product. An orthogonal matrix $$B$$ preserves the scalar product by definition, i.e. $$(Bx,By)=(x,y).$$ Since $$B^{-1}$$ is also orthogonal, then $$(ABy, By) = (B^{-1}ABy,B^{-1}By)=(B^{-1}ABy,y).$$

It follows easily from
1. $$(u,v)=u^tv$$, where $$u^t$$ is the transpose of $$u$$,
2. $$(CD)^t=D^tC^t$$,
3. $$C$$ symmetric iff $$C=C^t$$, and
4. $$C$$ orthogonal iff $$C^t=C^{-1}$$.

• The fact that $A$ is symmetric is not needed. – Joppy Oct 6 '18 at 9:58

An equivalent definition of an orthogonal matrix is one for which $$(Bv, Bw) = (v, w)$$ for all vectors $$v$$ and $$w$$. You can check that if $$B$$ is orthogonal, then $$B^{-1}$$ exists and is orthogonal. Can you work it out from there?