A few better solution than Paulo's is to use modular arithmetic. This is because Paulo's solution fails for inputs larger than 18. Yes, you could simply repeat that process, but it would get time consuming if the element was 34343454356. In effect, you are performing division by repeated subtraction, a VERY slow operation for large inputs.
Instead, reduce your vector modulo 9, replacing any elements which got mapped to zero, with a 9.
vector = [1 9 11 3 7 8 14 197 90];
newvector = mod(vector,9);
newvector(newvector == 0) = 9;
In the event that any element in the original vector was already a 0, you may choose not to convert that 0 element to a 9. This would be a simple modification to the above code, so perhaps you would have done it as...
vector = [1 9 11 3 7 8 14 197 90];
newvector = mod(vector,9);
newvector((newvector == 0) & (vector ~= 0)) = 9;
