Exercise 3.22. Error in matrix example

>Dear Andy,
>In your homework 4 in ME 234, which is Exercise 3.22 in our book,
>you are supoosed to show that the smallest singular value of a certain
>lower triangular matrix is less than or equal to 2^{-n}
> But, as pointed out by Yi Cao, with Matlab we get for n=4:
>a =
> -1 1 1 1
> 0 -1 1 1
> 0 0 -1 1
> 0 0 0 -1
>>> svd(a)
>ans =
> 2.2631
> 1.5962
> 1.5157
> 0.1826
>and the smallest singular value is larger than 2^-4 = 0.0625
>What's wrong?
>Best regards, Sigurd

Hi Sigurd,

        I think the problem is wrong. close, but wrong. i
worked on it for a bit, and seems like you can show
that the minimum singular value is less than

        sqrt(3n) / [ 2^n sqrt(1 - (1/4)^n) ]

where n is the dimension. so, for large n, the minimmum
singular value is "less" than

        sqrt(n) / 2^n

Sorry. By the way, i derived this by considering the
input vector

        u = [1 ; 1/2 ; 1/4 ; 1/8 ; ... ; 1/(2^(n-1))]

which gives a very small output, and then taking the
ratio to get an upper bound on sigma_min.

Let me know if this works out for you. Sorry about the
error - thanks for pointing it out.

Andy

%From pack@erg.ME.Berkeley.EDU Thu Oct 24 20:38:09 1996

Me again,

The largest singular value is bigger than sqrt(n), so the condition
number is increasing like 2^n. maybe that was what i meant to
have in the problem...

Andy
Received on Thu Oct 24 20:33:51 1996