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


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



The matrix representing $V$ in the chosen basis is


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

I have spent a lot of time on this exercise and for some reason it does not look intuitive to me.I have a bounty on another question on this exercise, if you want to check out here is the link Orthogonal basis transformation matrix type


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.

Thanks in advance


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.

  • $\begingroup$ Thanks a lot for your insight. I find it very difficult to find an intuition behind the whole mechanism of this proof. $\endgroup$ – Pedro Gomes Aug 18 '17 at 20:42

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