# 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.

## 2 Answers

See proposition 1 of Chi-Kwong Li and Edward Poon, Additive Decomposition of Real Matrices, Linear and Multilinear Algebra, 50(4):321-326, 2002.

My memory is a bit hazy, but if I remember correctly, the least upper bound of the number of orthogonal matrices needed was an open problem two decades ago. The paper by Li and Poon above shows that at most four are needed, but it wasn't known if three are enough.

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.