$\langle Av_1,Av_2\rangle=ac\langle v_1,v_1\rangle+bd\langle v_2,v_2\rangle$?

Define a rotation of $$V$$ to be a real unitary map $$A$$ of $$V$$ whose determinant is 1. Show that the matrix of $$A$$ relative to an orthogonal basis of $$V$$ is of type

$$\begin{bmatrix}a&-b\\b&a\end{bmatrix}$$

for some real numbers $$a,b$$ such that $$a^2+b^2=1$$.

SOLUTION. Let $$\{v_1,v_2\}$$ be an orthogonal basis for $$V$$. Let $$w_i=Av_i$$ and

$$w_1=av_1+bv_2$$

$$w_2=cv_1+dv_2$$

The matrix representing $$V$$ in the chosen basis is

$$\begin{bmatrix}a&c\\b&d\end{bmatrix}$$.

Then, since $$\langle Av_i,Av_i\rangle=\langle v_i,v_i\rangle$$ we have

$$(a^2-1)\langle v_1,v_1\rangle + b^2\langle v_2,v_2\rangle=0$$

$$(c^2)\langle v_1,v_1\rangle + (d^2-1)\langle v_2,v_2\rangle=0$$

But $$dw_1-bw_2=(ad-bc)v_1=v_1$$,so

$$\langle v_1,v_1\rangle=\langle A(dv_1-dv_2),A(dv_1-dv_2)\rangle=d^2\langle v_1,v_1\rangle + b^2\langle v_2,v_2\rangle$$,

thus implies $$a^2=d^2$$ and $$b^2=c^2$$. Moreover,

$$0=\langle v_1,v_2\rangle=\langle Av_1,Av_2\rangle=ac\langle v_1,v_1\rangle+bd\langle v_2,v_2\rangle$$,

so $$ac$$ and $$bd$$ are of opposite signs and therefore the matrix $$A$$ has the desired form.Solutions Manual for Lang´s Linear Algebra, Rami Sharcharchi

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Question:

How can I derive this expression $$\langle Av_1,Av_2\rangle=ac\langle v_1,v_1\rangle+bd\langle v_2,v_2\rangle$$? What step did the author give? I tried to multiply the matrix $$A^tA$$ by the basis $$\{v_1,v_2\}$$ and unsuccessfully I got nothing that looks like the expression given.

Note that the scalar product is bilinear and: $$\langle Av_1,Av_2\rangle =\langle w_1,w_2 \rangle =\langle av_1+bv_2,cv_1+dv_2 \rangle =\langle av_1,cv_1+dv_2\rangle+\langle bv_2,cv_1+dv_2\rangle = ac\langle v_1,v_1\rangle +ad\langle v_1,v_2\rangle+bc\langle v_2,v_1\rangle +bd\langle v_2,v_2\rangle.$$ Since $\langle v_1,v_2\rangle=\langle v_2,v_1\rangle=0$, the result follows.