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Recently, I read a proposition in a lecture notes that says the following:

Proposition. Let $a,b,c,d \in \mathbb{R}$, then there exist $\alpha, \beta, r, s \in \mathbb{R}$ such that

$$ \begin{bmatrix}{a}&{b}\\{c}&{d}\end{bmatrix}= \begin{bmatrix}{cos(\beta)}&{-sin(\beta)}\\{sin(\beta)}&{cos(\beta)}\end{bmatrix} \begin{bmatrix}{r}&{0}\\{0}&{s}\end{bmatrix} \begin{bmatrix}{cos(\alpha)}&{-sin(\alpha)}\\{sin(\alpha)}&{cos(\alpha)}\end{bmatrix} $$

At the outset, I didn't know how to proceed. However, I read the following theorem in Sheldon Axler's book, namely, "Linear algebra done right":

Theorem. Let $V$ be an $n$-dimensional inner product vector space. Suppose $T$ is an endomorphism of $V$ with singular values $s_1, \ldots, s_n$. Then, there exist orthonormal bases $e_1, \ldots, e_n$ and $f_1,\ldots,f_n$ of $V$such that $$ Tv=s_1 \langle v, e_1 \rangle f_1 + \ldots + s_n \langle v, e_n \rangle f_n $$ for every $v \in V$.

This led me to the following attempt: Let's denote by $T$ the endomorphism associated with the left matrix of the proposition. Then, if $s_1$ and $s_2$ are the singular values of $T$, by the above theorem we know that there exist orthonormal bases $\beta=\left\{{e_1,e_2}\right\}$ and $\gamma=\left\{{f_1,f_2}\right\}$ such that $$ [T]_{\beta}^{\gamma}=\begin{bmatrix}{s_1}&{0}\\{0}&{s_2}\end{bmatrix} $$ where $[T]_{\beta}^{\gamma}$ is matrix of $T$ with respect to the bases $\beta$ and $\gamma$, that is, $T(e_1)=s_1f_1$ and $T(e_2)=s_2f_2$. Let $\alpha$ be the canonical basis of $\mathbb{R}^2$, $P$ the change-of-basis matrix from $\beta$ to $\alpha$ and $Q$ the change-of-basis matrix from $\gamma$ to $\alpha$. Then, one has that $$ \begin{bmatrix}{a}&{b}\\{c}&{d}\end{bmatrix}=[T]_{\alpha}^{\alpha}=Q^{-1} [T]_{\beta}^{\gamma}P $$ Now, since $\beta$ and $\gamma$ are orthonormal bases, it should then be the case that $P$ and $Q$ are orthogonal, and therefore they have the form $$ \begin{bmatrix}{cos(\theta)}&{(-1)^{k+1}sin(\theta)}\\{sin(\theta)}&{(-1)^kcos(\theta)}\end{bmatrix} $$ for some integer $k$. The proposition claims that we can assure that $k$ is even, why? Is not clear at all for me.

In advance thank you.

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2 Answers 2

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For the purpose in this question, even if $P$ "has odd $k$":

$$\begin{align*} P&=\begin{bmatrix}\cos\theta&\sin\theta\\ \sin\theta&-\cos\theta\end{bmatrix}\\ &=\begin{bmatrix}1&0\\ 0&-1\end{bmatrix}\begin{bmatrix}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{bmatrix}\\ &=\begin{bmatrix}1&0\\ 0&-1\end{bmatrix}\begin{bmatrix}\cos(-\theta)&-\sin(-\theta)\\ \sin(-\theta)&\cos(-\theta)\end{bmatrix}\\ \begin{bmatrix}s_1&0\\0&s_2\end{bmatrix} P &=\begin{bmatrix}s_1&0\\0&-s_2\end{bmatrix} \begin{bmatrix}\cos(-\theta)&-\sin(-\theta)\\ \sin(-\theta)&\cos(-\theta)\end{bmatrix}\\ \end{align*}$$

So by choosing a different $s_2$ and $\theta$, the new orthogonal matrix replacing $P$ can "have even $k$" and be a rotation matrix.

The case for $Q$ is similar,

$$\begin{align*} Q&=\begin{bmatrix}1&0\\0&-1\end{bmatrix} \begin{bmatrix}\cos(-\theta)&-\sin(-\theta)\\ \sin(-\theta)&\cos(-\theta)\end{bmatrix}\\ Q^{-1}&= \begin{bmatrix}\cos(-\theta)&-\sin(-\theta)\\ \sin(-\theta)&\cos(-\theta)\end{bmatrix}^{-1} \begin{bmatrix}1&0\\0&-1\end{bmatrix}\\ Q^{-1}\begin{bmatrix}s_1&0\\0&s_2\end{bmatrix} &= \begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix} \begin{bmatrix}s_1&0\\0&-s_2\end{bmatrix}\\ \end{align*}$$

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  • $\begingroup$ Oh I see! I didn't see that, thank you very much! $\endgroup$
    – ferolimen
    Feb 5, 2023 at 2:42
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    $\begingroup$ I think it is interesting to mention that $\begin{bmatrix}\cos\theta&\sin\theta\\ \sin\theta&-\cos\theta\end{bmatrix}$ is a symmetry matrix [with respect to the straight line with equation $y=\tan(\theta/2)x$] $\endgroup$
    – Jean Marie
    Feb 5, 2023 at 19:20
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I will consider it in a somewhat different way:

There are two types of orthogonal $2 \times 2$ matrices :

$$\begin{cases}\text{Rotation matrices (type "R") :}& \begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}\\ \text{Symmetry matrices (type "S") : }& \begin{bmatrix}\cos\theta&\sin\theta\\ \sin\theta&-\cos\theta\end{bmatrix} \end{cases}\tag{1}$$

Remark : please note that $R^T$ is still a rotation matrix (by angle $-\theta$) and $S^T$ is still a symmetry matrix (with $S^T=S$).

One can "convert" type "S" into type "R" by right-multiplying it in this way :

$$\underbrace{\begin{bmatrix}\cos\theta&\sin\theta\\ \sin\theta&-\cos\theta\end{bmatrix}}_{S}\underbrace{\begin{bmatrix}1&0\\ 0&-1 \end{bmatrix}}_D=\underbrace{\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}}_R \tag{2}$$

Besides, the SVD decomposition of a $2 \times 2$ matrix is :

$$A=U \Sigma V^T \ \text{where} \ \Sigma=\begin{bmatrix}\sigma_1&0\\ 0&\sigma_2\end{bmatrix}\tag{3}$$

where $U,V$ are orthogonal matrices and $\sigma_1,\sigma_2 >0$ are the singular values of matrix $A$.

Therefore 4 cases can occur in (3):

$$R\Sigma R, S\Sigma R, R\Sigma S, S \Sigma S \tag{4}$$

(we have dropped the transposition sign $^T$ due to remark above).

  • in the first case: nothing has to be done.

  • in the second case $A=S\Sigma R$, which can be written, using (2) :

$$A = S I \Sigma R = S (D D)\Sigma R=\underbrace{(SD)}_{R_1} \underbrace{(D \Sigma)}_{\Sigma_1} R$$

boiling down to a change of $\sigma_2$ into $-\sigma_2$.

  • The two other cases of (4) can be treated in the same way (changing $\sigma_1$ into $-\sigma_1$, or changing both $\sigma_1, \sigma_2$ into $-\sigma_1, -\sigma_2$).
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  • $\begingroup$ Any comment ?... $\endgroup$
    – Jean Marie
    Feb 5, 2023 at 2:06
  • $\begingroup$ I see, the approach is more or less similar to the other answer, thank you very much for your clear and complete explanation! :) Now I have a better understanding of this stuff :) $\endgroup$
    – ferolimen
    Feb 5, 2023 at 2:50
  • $\begingroup$ The main difference between the 2 answers is that my explanation is based, as your title invites it, on the connection with the SVD. $\endgroup$
    – Jean Marie
    Feb 5, 2023 at 10:23

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