00:01
Hello students, to prove the given inequality we need to use the hint and the definition of the operator norm.
00:06
Let z be a vector in c to the power n and let maximum z, z of j is equal to m.
00:27
Then we have mod of jk is less than equal to max j mod of jj is equal to m.
00:42
Taking the modulus squared of both the sides we get z jk whole square is less than equal to m square.
00:54
Summing over all k from 1 to n we obtain double mod z square is equal to summation k equal to 1 to n zk whole square less than equal to n m square.
01:12
Therefore double mod is less than equal to square root n m.
01:19
Now let's apply this result to the matrix a.
01:24
Let x be a vector, unit vector in c to the power cn and let y equal to y equal to ax.
01:42
Then we have y square equal to summation j equal to 1 to n axj whole square is equal to summation j equal to 1 to n summation k equal to 1 to n ajk xk mod whole square.
02:08
Using the cauchy -savars inequality we get, cauchy -savars inequality we get, we get summation k equal to 1 to n ak ajk xk is less than equal to summation aj mod x where modulus aj is equal to max k equal to 1 to n ajk mod...