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


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?

Thanks in advance!

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

1 Answer 1


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 $$ \langle v_1,v_1\rangle=\langle Av_1,Av_1\rangle=\langle av_1+bv_2,av_1+bv_2\rangle=a^2\langle v_1,v_1\rangle+b^2\langle v_2,v_2\rangle. $$

  • $\begingroup$ 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. $\endgroup$ Aug 17, 2017 at 11:37
  • $\begingroup$ 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? $\endgroup$ Aug 18, 2017 at 18:19
  • $\begingroup$ $\langle v_1,v_1\rangle=d^2\langle v_1,v_1\rangle\ b^2\langle v_2,v_2\rangle\implies (d^2-1)\langle v_1,v_1\rangle{ b}^2\langle v_2,v_2\rangle=0$, 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 $c^2=b^2.$ $\endgroup$ Aug 18, 2017 at 18:19
  • 1
    $\begingroup$ It follows from $\langle v_1,v_1\rangle=\langle Av_1,Av_1\rangle=\langle av_1+bv_2,av_1+bv_2\rangle=a^2\langle v_1,v_1\rangle+b^2\langle v_2,v_2\rangle$. $\endgroup$
    – san
    Aug 19, 2017 at 19:51
  • 1
    $\begingroup$ You start from $\begin{pmatrix} a&c\\b&d\end{pmatrix}$, and obtain $c=-b$ and $d=a$. $\endgroup$
    – san
    Aug 20, 2017 at 15:49

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