I'm having trouble maximizing the norm of | | A x | | such that B x = b. I set up a lagrange multiplier so that L ( x , λ ) = x T ( A T A ) x − ( B x − b ) T λ. (Say that B is not invertible but b is in the colspace so that the problem isn't trivial) ∇ x L ( x , λ ) = 0 ⇒ 2 ( A T A ) x = B T λand ∇ λ L ( x , λ ) = 0 ⇒ B x = b.If I assume that A has full rank, then x = 1 2 ( A T A ) − 1 B T λ.If I plug it into the constraint, then B ( A T A ) − 1 B T λ = 2 b. How do I proceed solving for λ here?

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I&#039;m having trouble maximizing the norm of       |        |    A  x      |        |   such that   B  x  =  b. I set up a lagrange multiplier so that   L  (  x  ,  λ  )  =      x    T    (      A    T    A  )  x  −  (  B  x  −  b      )    T    λ. (Say that   B is not invertible but   b is in the colspace so that the problem isn&#039;t trivial)      ∇    x    L  (  x  ,  λ  )  =  0  ⇒  2  (      A    T    A  )  x  =      B    T    λand       ∇    λ    L  (  x  ,  λ  )  =  0  ⇒  B  x  =  b.If I assume that   A has full rank, then   x  =      1    2    (      A    T    A      )          −      1            B    T    λ.If I plug it into the constraint, then   B  (      A    T    A      )          −      1            B    T    λ  =  2  b. How do I proceed solving for   λ here?

Aryanna Caldwell · Accepted Answer

Short answer: There is no maximum, i.e. you can select   x such that   ‖  A  x  ‖ is arbitrarily large.Because if   B  x  =  b for some   x, then   B  (  x  +  y  )  =  b for all   y  ∈  Ker  ⁡  B and you can select   y arbitrarily large (assuming   A  y  ≠  0).If   B has full column rank then there is no optimization, because the solution is unique.

Lena Bell · Accepted Answer

If   B  (      A    T    A      )          −      1            B    T   is full rank then       λ    ∗    =  2  (  B  (      A    T    A      )          −      1            B    T        )          −      1        b and      x    ∗    =      1    2    (      A    T    A      )          −      1            B    T        λ    ∗    =  (      A    T    A      )          −      1            B    T    (  B  (      A    T    A      )          −      1            B    T        )          −      1        bEDIT: see the comments below. This would be a correct answer for the minimization problem, but is a wrong answer to the maximization problem here.

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