
Re: Making a square matrix from two vector
Posted:
Feb 13, 2013 1:51 AM


"Steven_Lord" <slord@mathworks.com> wrote in message <kfdrs7$rn$1@newscl01ah.mathworks.com>... > > > "james bejon" <jamesbejon@yahoo.co.uk> wrote in message > news:kfdpbk$kdo$1@newscl01ah.mathworks.com... > > BrunoI really like this bsxops stuff, and have started using it in some > > of my projects at work. In the process, I've realised something which (at > > least in my view) is rather odd (though is nothing to do with your > > functions), namely the result of: > > > > (1:10).' + 1:10 > > > > % the right way: (1:10).' + (1:10) > > This is intentional behavior. > > http://www.mathworks.com/help/matlab/matlab_prog/operators.html#f038155 > > Parentheses are at level 1 in the precedence order. Transpose is at level 2. > Addition is at level 5. The colon operator is at level 6. Therefore this: > > (1:10).' + 1:10 > > computes (1:10) first (level 1 and level 6 inside the parentheses), then > transposes the result (level 2), then adds 1 to that result (level 5), then > uses that vector as the first input to the colon operator (level 6.) That > just takes the first element as documented on the reference page for COLON. > > http://www.mathworks.com/help/matlab/ref/colon.html > > "If you specify nonscalar arrays, MATLAB interprets j:i:k as > j(1):i(1):k(1)." > > This also explains why your "the right way" resolves the problem; both sets > of parentheses at level 1 are processed before the addition at level 5. > > > This is also why 8^(1/3) gives 2 instead of (1+sqrt(3)*1i) as people > expect and sometimes post about here; exponentiation [8^(1/3)] is at level 2 > while unary minus [ (8^(1/3))] is at level 3. (8)^(1/3) gives the complex > answer. > >  > Steve Lord > slord@mathworks.com > To contact Technical Support use the Contact Us link on > http://www.mathworks.com
Thanks for the explanation.

