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???


1 Answer 1


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 $2$.

  • $\begingroup$ thanks, good to know this theorem $\endgroup$
    – Lee
    Commented Jun 23, 2020 at 10:31

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