# Orthogonal basis transformation matrix type

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

Questions:

1) Since the basis are assumed orthogonal and not orthonormal. How can the author possible derive $$(a^2-1)\langle v_1,v_1\rangle + b^2\langle v_2,v_2\rangle=0$$? How does he knows the last equality? What is the intuition?

2) $$dw_1-bw_2=(ad-bc)v_1=v_1$$ How does the author know $$(ad-cb)=1$$ if we are not working with orthonormal basis?

• Some authors use "orthogonal" for bases, even though each vector may be unit length. Check the author's conventions/definition of orthogonal basis. Aug 13, 2017 at 21:24
• Also, since you assumed $A$ to have determinant 1, its representing matrix will have determinant 1, so of course $ad-bc=1$. Aug 13, 2017 at 21:26
• @Randall I have checked and the author uses the term orthonormal and orthogonal basis, for unitary and non-unitary orthogonal basis. Thanks for the reply Aug 13, 2017 at 21:26
• @Randall Yes that is right! Thanks for the insight! Aug 13, 2017 at 21:27
• @Randall What do you think about the first question? How did the author got that expression if the basis are not assumed normal? Aug 13, 2017 at 21:28

If you only assume that the basis $V=\{v_1,v_2\}$ is orthogonal but not ortonormal, then the result is false. Assume for example that the matrix of $A$ in the canonical basis is $\begin{pmatrix} 0&1\\ -1&0\end{pmatrix}$, and that the basis is $v_1=\binom{1}{0}$ and $v_2=\binom{0}{2}$. Then matrix of $A$ relative to the orthogonal basis $V=\{v_1,v_2\}$ is the matrix $\begin{pmatrix} 0&2\\ -\frac 12&0\end{pmatrix}$, which is not of the required form.

On the other hand if the basis is orthonormal, the result is straightforward using $$\langle Av_i,Av_j\rangle=\langle v_i,v_j\rangle=\delta_{ij}$$

In fact, from these equations we obtain $$ac+bd=0$$ $$a^2+b^2=1$$ $$c^2+d^2=1$$ Now, if $a=0$, then $b^2=1$ and $d=0$, so $c^2=1$. From $det(A)=1$ we get $c=-b$, as desired.

On the other hand, if $a\ne 0$, then $c=-bd/a$ and so $$1=c^2+d^2=d^2+b^2d^2/a^2\quad \Rightarrow\quad a^2=(a^2+b^2)d^2=d^2$$ and so $d=\pm a$, i.e., $a=\varepsilon d$, with $\varepsilon=\pm 1$. Then $c=-\varepsilon b$, and from $$1=\det(A)=\det\begin{pmatrix}a&-\varepsilon b\\ b& \varepsilon a\end{pmatrix}=\varepsilon a^2+\varepsilon b^2=\varepsilon$$ it follows that $\varepsilon =1$ and so $A=\begin{pmatrix}a&- b\\ b& a\end{pmatrix}$.

${\bf Edit:}$

Now let's break down the solution in the OP. There are some typos, so I write what I think should be written, and point out the error (there has to be an error, since the result is false).

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

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

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

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

$\langle v_1,v_1\rangle=\langle A(dv_1-\color{red}bv_2),A(dv_1-\color{red}{b}v_2)\rangle=d^2\langle v_1,v_1\rangle\color{red}{+ b}^2\langle v_2,v_2\rangle$,

thus implies $a^2=d^2$ and $\color{red}{b^2=c^2.}$ $\color{red}{Here\ is\ the\ error}$.

You have $a^2=d^2$, but from the similar equation $\langle v_2,v_2\rangle= a^2\langle v_2,v_2\rangle+c^2\langle v_1,v_1\rangle$ you obtain again $a^2=d^2$, but not the equality $b^2=c^2$. In the example I gave above clearly $a^2=d^2$ but $b^2\ne c^2$.

So the proof is wrong, but $(ad-bc)=\det(A)=1$ is correct, since the determinant does not depend on the chosen basis.

The equality $(a^2-1)\langle v_1,v_1\rangle+ b^2\langle v_2,v_2\rangle=0$ can be derived from $$\langle v_1,v_1\rangle=\langle Av_1,Av_1\rangle=\langle av_1+bv_2,av_1+bv_2\rangle=a^2\langle v_1,v_1\rangle+b^2\langle v_2,v_2\rangle.$$

• That was not my proof but it was copied by me from the book: Solutions Manual for Lang´s Linear Algebra, by Rami Sharkarchi. The problem rises from the fact that the author is assuming the basis to be only orthogonal, not orthonormal as you suggest. I want to know if you consider the author´s answer wrong. Aug 17, 2017 at 11:37
• I am so sorry but I copied the solution wrongly. I have corrected the pointed mistakes. You assume in your answer the base is orthonormal however the book states that it is only orthogonal. Is the latest edited answer wrong? Aug 18, 2017 at 18:19
• $\langle v_1,v_1\rangle=d^2\langle v_1,v_1\rangle\ b^2\langle v_2,v_2\rangle\implies (d^2-1)\langle v_1,v_1\rangle{ b}^2\langle v_2,v_2\rangle=0$, therefore $(d^2-1)\langle v_1,v_1\rangle b^2\langle v_2,v_2\rangle=(a^2-1)\langle v_1,v_1\rangle + b^2\langle v_2,v_2\rangle$ and $(d^2-1)\langle v_1,v_1\rangle b^2\langle v_2,v_2\rangle=(c^2)\langle v_1,v_1\rangle + (d^2-1)\langle v_2,v_2\rangle$, hence $a^2=d^2$ and $c^2=b^2.$ Aug 18, 2017 at 18:19
• It follows from $\langle v_1,v_1\rangle=\langle Av_1,Av_1\rangle=\langle av_1+bv_2,av_1+bv_2\rangle=a^2\langle v_1,v_1\rangle+b^2\langle v_2,v_2\rangle$.
– san
Aug 19, 2017 at 19:51
• You start from $\begin{pmatrix} a&c\\b&d\end{pmatrix}$, and obtain $c=-b$ and $d=a$.
– san
Aug 20, 2017 at 15:49
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