A is a nxn matrix and u is a nx1 vector.. then prove the following ||Au||

Question

A is a nxn matrix and u is a nx1 vector.. then prove the following  ||Au||<=||A|| ||u||

turingtest · Answer

what is the magnitude of a matrix? its determinant?

zzr0ck3r · Answer

is this for in alg or a calc class?

zzr0ck3r · Answer

lin alg*

zzr0ck3r · Answer

what class is this for?

anonymous · Answer

in linear algebra

zzr0ck3r · Answer

is this the whole question? is the matrix orthoginal?

anonymous · Answer

how to see the answers

anonymous · Answer

||A|| is whats called the norm of the matrix A. There are many types of norms, so the poster might want to specify which we are using. The one that pops up a lot in Linear Algebra is:$$||A||=\max_{|x|=1}|Ax| $$which basically is the max length of the vector Ax when x is on the unit sphere. If this is the norm we are talking about (called the Operator Norm, http://en.wikipedia.org/wiki/Operator_norm), then we can prove the statement as follows. If x = 0, then of course the statement is true. Now let x be different from 0, and create the unit vector:$$y=\frac{x}{|x|}$$Then we have:$$||A||\ge |Ay|\Longrightarrow ||A||\ge |A\frac{x}{|x|}|=\frac{1}{|x|}|Ax|$$$$\Longrightarrow |x|\cdot ||A||\ge |Ax|$$