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Pacopag
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Homework Statement
I am trying to show that
(1) [tex]||A||_1 = \max_j \sum_i |a_{ij}|[/tex]
(2) [tex]||A||_2 = \sqrt{max\_eigval\_ of\_{ } A^* A}[/tex] where A* is the conjugate transpose
(3) [tex]||A||_\infty = \max_i \sum_i |a_{ij}|[/tex]
Homework Equations
In general,
[tex]||A||_p = max_{x\neq 0} {{||Ax||_p}\over{||x||_p}}[/tex]
where for a vector we have
[tex]||x||_p = \left( \sum_i |x_i|^p \right)^{1\over p}[/tex]
The Attempt at a Solution
First of all, write
[tex](Ax)_i = \sum_j a_{ij} x_j[/tex].
(1) Here we have
[tex]||Ax||_1 = max_{x} {{\sum_i \left|\sum_j a_{ij} x_j \right|}\over{\sum_j |x_j|}}[/tex].
I'm pretty sure we could switch the summation order in the numerator too. But I don't know how to proceed from here.
(2) Here we have
[tex]||Ax||_2^2 = max_x {{x^* A^* A x}\over{x^* x}}[/tex] using [tex]||x||_2^2 =x^* x [/tex]
Now, I read somewhere that the x that maximizes this is an eigenvector of [tex]A^* A[/tex]. So we get
[tex]||Ax||_2^2 = \lambda_{max}[/tex], the largest eigenvalue.
My only problem with this is: How do we know that an eigenvector will maximize it?
(3) Here we have
[tex]||Ax||_\infty = max_x {{max_i \left( \left| \sum_j a_{ij} x_j \right| \right)}\over{max_j (|x_j|)}}[/tex]
using
[tex]||x||_\infty = max_j (x_j)[/tex].
Again, I don't know how to proceed from here. It seems that in order to arrive at the answer (which I like to pretend that I don't know),
I would need to choose x to be a vector where all the components are the same. Then the x_j's would cancel and i get the right answer.
But I don't see why such an x is guaranteed to maximize ||A||.
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