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I have to minimize $x^Tx$ under condition $a^Tx = 1$. Whats the geometric meanign of this task?

I think that the meaning is to find minimal vector $x$ thats is perpendicular to given vector $a$, but I am not sure.

Whats the difference if we maximize instead of minimization? Thanks for any ideas (I need the teorethical part more than solving this minimization itself - I just dont have imagination for n-dimensional space).

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1 Answer 1

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$a^Tx=1$ is a hyperplane that does not pass through the origin. Without this restriction if you make $x^Tx$ smaller and smaller you just get to $0$. However, with this restriction that doesn't happen. Instead you get the closest point on the hyperplane to the origin (which is the intersection of the line along the normal vector with the plane that passes through the origin). If you maximize you just go off in whatever direction along the hyperplane and your values get bigger and bigger (the plane is unrestricted in any directions within the plane).

EDIT1: $x^Tx=||x||_2^2$ is proportional to $||x||_2$. So you are looking for the smallest vector where the bottom tip is at $0$ and the top tip is inside the hyperplane defined by $a^Tx$ - which is obviously well defined and unique. If you maximize you are looking how big you can make your vector so that the bottom tip is at $0$ and the top tip is in the hyperplane again. This obviously has no solution since you just make your vector bigger and bigger...

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