5. Show ||x||* = ||Ax||v is a vector norm.
This problem is mainly playing with ways to rename things to be more clear, i.e., dealing with a vector b s.t. Ax = b instead of Ax.
Note since A is nonsingular Ax = b will have a unique solution.
(i) WTS
||x||* ≥0
for x ≠ 0.
Note since Ax has a unique solution, we can call it b. || ||v is a vector norm, so ||b||v is a valid vector norm and ≥ 0 by definition. Since we let Ax = b, ||Ax||v ≥ 0, and ||x||* = ||Ax||v is also greater than or equal to zero.
(ii) WTS ||x||* = 0 for x = 0.
If x = 0 then Ax = 0. Since b = 0 and, by logic similar to above, ||x||* = ||b||v = 0 => ||x||* = ||Ax||v = 0. So ||x||* = 0 if x = 0.
Let Ax = b again. If ||x||* = ||b||v = 0, then b = 0 since || ||v is a vector norm. If b = 0, then A = 0 or x = 0. But A cannot be nonsingular, and the zero matrix is singular. So x = 0 if ||x||* = 0.
Thus we have shown both directions as desired.
(iii) WTS ||ax||* = |a| ||x||*.
If we have ||ax||*,
then this is the same as ||aAx||v by our definition. Let Ax = b.
So we have ||ab||v = |a| ||b||v
since || ||v is a vector norm. Replacing b by Ax again, and ||Ax||v with ||x||*, then it is readily
apparent that we have ||ax||*
= |a| ||x||*.
(iv) WTS ||x + y||* ≤ ||x||* + ||y||*.
Note ||x + y||* means ||A(x + y)||v and NOT ||(A + B)x ||v or ||Ax + y||v as some people tried to use—you replace x with (x+y).
Let Ax = b and Ay = c. Both sets of equations have unique solutions x and y again. So ||x + y||* = ||A(x + y)||v, which when A is distributed gives ||Ax + Ay||v , or ||b + c||v as just defined. Since once again || ||v is a vector norm, ||b + c||v ≤ ||b||v + ||c||v = ||Ax||v + ||Ay||v = ||x||* + ||y||*.
Some people went with the original definition of this problem and tried to show that ||x||* is a matrix norm. Would it really make sense for this to be a definition of a matrix norm if the quantity being examined, x, is a vector? Not especially.
