# Why is the first left and right singular vectos scale by the first singular values a good approximation of the original matrix

Conceptually, why is the first singular vector a good rank one approximation instead of something like the averaging of the total singular vectors?

If you have $$A = U\Sigma V^T$$

why isn't

$$\sqrt{\sigma_{avg}}u_{avg}v_{avg}^T$$ a good low rank approximation?



How about weighted average of the singular vectors?

Context of my Question:

An exam with $m$ questions is given to $n$ students. The instructor collects all the grades in a $n * m$ matrix $G$ with $G_{ij}$ the grade obtained by student $i$ on the question $j$. We would like to assign a difficulty score to each question based on the available data.

How would you compute a rank one approximation to $G$

Solution:

To approximate $G$ by a rank one vector we simply compute the SVD of $G$ and select the singular vectors corresponding to the largest singular value. Precisely, we set $s \sqrt{\sigma_1}u_1$ and $q =\sqrt{ \sigma_i}v_1$ where $u_1 \:\: v_1$ are the first columns of the matrices $U \:\: V$ in the SVD of $G = U\Sigma V^T$ and $\sigma_1$ is the largest singular value

The easiest way to see this is to use the Eckart–Young–Mirsky theorem, which states that the rank-k approximation $A_k$ to $A$ that minimizes Frobenius norm $|| A - A_k ||_F$ is $A_k = U_k \Sigma_k V_k^T$ for $\Sigma_k$ the top $k$ singular values and $U_k$ and $V_k$ the corresponding singular vectors. For $k=1$, this gives $A_1 = \sigma_1 u_1 v_1^T$.