# Given positive definite $X\in\mathbb{R}^{4\times 4}$, find $Y\in\mathbb{R}^{4\times 2}$, such that $YY'\approx X$

Given positive definite $$X\in\mathbb{R}^{4\times 4}$$, I want to find $$Y\in\mathbb{R}^{4\times 2}$$, such that $$YY'\approx X$$.

My attempt:

Using SVD, $$X=U\Sigma U^*$$. Let $$U_i$$ be $$i'$$th column of $$U$$, then $$Y=[U_1 \quad U_2]\begin{bmatrix} \sigma_1 & \\ & \sigma_2 \end{bmatrix}.$$

I am not sure if this is a correct approach. Assumption: $$\sigma_1\geq \sigma_2\geq \sigma_3\geq \sigma_4$$. I think if $$\sigma_3$$ and $$\sigma_4$$ are very small compared to $$\sigma_1$$ and $$\sigma_2$$ then it might work???

You are almost correct. Instead, however, you should take $$Y=[U_1 \quad U_2]\begin{bmatrix} \sqrt{\sigma_1} & \\ & \sqrt{\sigma_2} \end{bmatrix}.$$ In particular, we find that $$YY' = U \pmatrix{\sigma_1 \\ & \sigma_2 \\ & & 0\\ &&& 0} U'$$ is a good approximation for $$X$$. In fact, the EYM theorem tells us that this $$YY'$$ is closer to $$X$$ than any other matrix of rank $$2$$.