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I am reading a paper. There is an upper bound for largest singular value of matrix $A$. Suppose its largest singular value is $d$ and $d\leq p$. Now we have a row vector $U$. It is said that in the paper $\|UA\|\leq \|U\|p$ but really I don't understand it. Could you please help me? Thanks in advance.

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Assuming you are using $2$-norm, then it has the sub-multiplicative property.

Hence $$\|UA\| \le \|U\|\|A\|=\|U\|d \le \|U\|p$$

Siong Thye Goh
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    Yes it is 2-norm. Why the first ineqaulity is true? What is definition of norm $A$? –  Feb 20 '18 at 17:21
  • refer to the wikipedia page. It is an induced norm. First we define the vector norm and use it to define a matrix norm as follows: $|A|2 = \max{x \neq 0} \frac{|Ax|}{|x|}$. It is equal to the largest singular value. Refer here for a proof of submultipllicative property of the matrix norm. – Siong Thye Goh Feb 20 '18 at 17:30