This is documentation for Mathematica 4, which was
based on an earlier version of the Wolfram Language.
Wolfram Research, Inc.

3.7.4 Operations on Scalars, Vectors and Matrices

Most mathematical functions in Mathematica are set up to apply themselves separately to each element in a list. This is true in particular of all functions that carry the attribute Listable.

A consequence is that most mathematical functions are applied element by element to matrices and vectors.

The Log applies itself separately to each element in the vector.

In[1]:= Log[ {a, b, c} ]

Out[1]=

The same is true for a matrix, or, for that matter, for any nested list.

In[2]:= Log[ {{a, b}, {c, d}} ]

Out[2]=

The differentiation function D also applies separately to each element in a list.

In[3]:= D[ {x, x^2, x^3}, x ]

Out[3]=

The sum of two vectors is carried out element by element.

In[4]:= {a, b} + {ap, bp}

Out[4]=

If you try to add two vectors with different lengths, you get an error.

In[5]:= {a, b, c} + {ap, bp}

Out[5]=

This adds the scalar 1 to each element of the vector.

In[6]:= 1 + {a, b}

Out[6]=

Any object that is not manifestly a list is treated as a scalar. Here c is treated as a scalar, and added separately to each element in the vector.

In[7]:= {a, b} + c

Out[7]=

This multiplies each element in the vector by the scalar k.

In[8]:= k {a, b}

Out[8]=

It is important to realize that Mathematica treats an object as a vector in a particular operation only if the object is explicitly a list at the time when the operation is done. If the object is not explicitly a list, Mathematica always treats it as a scalar. This means that you can get different results, depending on whether you assign a particular object to be a list before or after you do a particular operation.

The object p is treated as a scalar, and added separately to each element in the vector.

In[9]:= {a, b} + p

Out[9]=

This is what happens if you now replace p by the list {c, d}.

In[10]:= % /. p -> {c, d}

Out[10]=

You would have got a different result if you had replaced p by {c, d} before you did the first operation.

In[11]:= {a, b} + {c, d}

Out[11]=