Show that if $$\sigma _{1}$$ represents the largest singular value of a matrix $$A=(a _{ij})$$ that $$\sigma _{1}\ge|a _{ij}|_{\max}$$

Question

anonymous · Answer

I'm wondering if I could use some property of the SVD in order to answer the question.

zarkon · Answer

yes...using the SVD you can prove what you posted

anonymous · Answer

@Zarkon Could you help me work it out?  I'm not sure how to go about this.

phi · Answer

Here are some ideas

the $ a_{ij} $ of matrix A will be the ith row of U times Σ times the jth column of $V^T$
expand this out
factor out $\sigma_1$ (largest value)

now consider
we know that both the row of U and the column of $V^T$ have unit length
we know Cauchy-Schwarz   
$$ | X \cdot Y| ≤ |X| |Y| $$

anonymous · Answer

I ended up using the Frobenius Norm, though I'm not sure I did it right.  Unfortunately, it's past the due time now (I did turn in an answer though), but on the plus side next week I can get a copy of the solutions.

phi · Answer

I should have posted earlier...

anonymous · Answer

Do not worry about it.