The proof of the properties of dot product of vectors

Here is the proof of the statemengt that the dot product of n-vectors satisfies the following properties:

<A,B>=<B,A> ;
<(A+B),C=<A,C>+<B,C> ;
<(kA),B=<A,(kB)>=<k,(A,B)> ;
<A,A> is greater than or equal to 0. <A,A> is 0 if and only if A=0.

Notice that the dot product of vectors can be viewed as the product of matrices:

<A,B>=AB^T

that is the dot product of the row vector A and a row vector B is equal to the matrix product of A and the column vector B^T. This observation helps to prove our properties.

1. Notice that the dot product <A,B>=AB^T is a scalar, that is a 1 by 1 matrix. Thus <A,B> is a symmetric matrix (every 1 by 1 matrix is obviously symmetric). So

<A,B>=(AB^T)=(AB^T)^T=(B^T)^T A^T=BA^T=<B,A>

Here we used several times the theorem about transposes. Of course, this statement could be easily proved without the theorem about transposes (just use the definition of the dot product of n-vectors) but I wanted to demonstrate the relationship between the dot product and the matrix product.

2. Again we use the properties of the matrix product:

(A+B)*C=(A+B)C^T=AC^T + BC^T= <A,C>+<B,C>.

We leave property 3 as an exercise.

In order to prove property 4, we need to use the definition of the dot product: Let A=(a₁,...,a_n). Then

<A,A>=a₁a₁+...+a_na_n Thus <A,A> is the sum of squares of the coordinates of A. It is clear now that <A,A> is always non-negative and <A,A>=0 if and only if all coordinates are equal to 0.

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