Every real matrix $A$ is the linear combination of $4$ orthogonal matrices Question:

Prove that every matrix $A\in M_n(\mathbb R)$ is the linear combination of $4$ orthogonal matrices $X, Y, Z, W$ , i.e. $A=aX+bY+cZ+dW$ for some $a,b,c,d\in\mathbb R$. 

This problem is taken from a forum and this is my paraphrase. It is not obviously true. But I think the proof must invoke the singular-value decomposition (SVD) of a real matrix, but it's unclear to me what the next step is. Any idea is appreciated. Many thanks.

In view of SVD, we may assume that $A\neq 0$ and  $A = \operatorname{diag}(\lambda_1, \cdots, \lambda_n)$ such that $\lambda_i$'s are non-negative, and $\lambda_n$ is the largest among $\lambda_i$'s.
Under this assumption, we have $\lambda_n > 0$. Now write $\lambda_{2i-1} = a_i - b_i$ and $\lambda_{2i} = a_i + b_i$. In case $n$ is odd, we also set $a_{(n+1)/2} = 2\lambda_n$ and $b_{(n+1)/2} = \lambda_n$. This gives
$$ A = \operatorname{diag}(a_1, a_1, a_2, a_2, \cdots) - \operatorname{diag}(b_1, -b_1, b_2, -b_2, \cdots). $$
Now we notice that $|a_i| \leq 2\lambda_n$ and $|b_i| \leq \lambda_n$, aod so, there exist $\alpha_i$ and $\beta_i$ such that $a_i = 2\lambda_n \cos\alpha_i$ and $b_i = \lambda_n \cos\beta_i$. Now the trick is to consider matrices
$$ R_{\theta} = \begin{pmatrix} \cos \theta & -\sin\theta \\ \sin\theta & \cos \theta \end{pmatrix} \qquad \text{and} \qquad S_{\theta} = \begin{pmatrix} \cos \theta & \sin\theta \\ \sin\theta & -\cos \theta \end{pmatrix}. $$
Then both $R_\theta$ and $S_\theta$ are orthogonal. Moreover, we have
$$R_\theta + R_{-\theta} = (2 \cos\theta) \operatorname{diag}(1, 1) \qquad \text{and} \qquad S_\theta + S_{-\theta} = (2 \cos\theta) \operatorname{diag}(1, -1).$$
Now using this, we may write
$$ \operatorname{diag}(a_1, a_1, a_2, a_2, \cdots) = \lambda_n \begin{pmatrix} R_{\alpha_1} & & \\ & R_{\alpha_2} & \\ & & \ddots \end{pmatrix} + \lambda_n \begin{pmatrix} R_{-\alpha_1} & & \\ & R_{-\alpha_2} & \\ & & \ddots \end{pmatrix} $$
and
$$ \operatorname{diag}(b_1, -b_1, b_2, -b_2, \cdots) = \frac{\lambda_n}{2} \begin{pmatrix} S_{\beta_1} & & \\ & S_{\beta_2} & \\ & & \ddots \end{pmatrix} + \frac{\lambda_n}{2} \begin{pmatrix} S_{-\beta_1} & & \\ & S_{-\beta_2} & \\ & & \ddots \end{pmatrix}. $$
When $n$ is odd, the last block in the diagonal in each block matrix is the $1\times 1$ matrix with the entry $1$. Now all these four block matrices are orthogonal, and therefore the desired claim follows.
