# Apparent SVD decomposition of a $2 \times 2$ matrix

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.

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*}

• Oh I see! I didn't see that, thank you very much! Feb 5, 2023 at 2:42
• 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 $y=\tan(\theta/2)x$] Feb 5, 2023 at 19:20

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 $$-\sigma_1, -\sigma_2$$).
• Any comment ?... Feb 5, 2023 at 2:06
• 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 :) Feb 5, 2023 at 2:50
• The main difference between the 2 answers is that my explanation is based, as your title invites it, on the connection with the SVD. Feb 5, 2023 at 10:23