| Line 29: | Line 29: | ||
| − | A | + | A: |
| + | |||
| + | <math>\begin{bmatrix} | ||
| + | v_{1} & v_{2} & v_{3} & v_{k} \\ | ||
| + | v_{1} & v_{2} & v_{3} & v_{k} \\ | ||
| + | v_{1} & v_{2} & v_{3} & v_{k} \\ | ||
| + | .... & .... & .... & .... \\ | ||
| + | .... & .... & .... & .... \\ | ||
| + | .... & .... & .... & .... \\ | ||
| + | v_{1} & v_{2} & v_{3} & v_{k} \\ | ||
| + | |||
| + | \end{bmatrix}</math> | ||
Revision as of 13:25, 8 December 2010
Statement: I am going to show that if V is a subspace of Rn. then dim(V)+dim(Vorth)=n
Because of the lack of the orthagonal symbol in the wikipedia formatting page, I will be type 'orth' in a superscript to symbolize that.
Analysis:
First, let us say we have the following:
V which is a subspace of Rn, and {v1,v2,v3,..,vk} are a basis for V. (The entries in the braces are vectors)
To refresh, a basis means those entries span V, AND are also linearly independent.
So, therefore, then dim(V)=k (k is the number of vectors in our basis, which obviously is a non-finite amount, so I use k to denote that fact.)
Now that we have those assumptions and definitions out of the way, let me construct a matrix for you.
We will call this matrix A (seems to the most common letter in the linear algebra world...but i digress)
A:
$ \begin{bmatrix} v_{1} & v_{2} & v_{3} & v_{k} \\ v_{1} & v_{2} & v_{3} & v_{k} \\ v_{1} & v_{2} & v_{3} & v_{k} \\ .... & .... & .... & .... \\ .... & .... & .... & .... \\ .... & .... & .... & .... \\ v_{1} & v_{2} & v_{3} & v_{k} \\ \end{bmatrix} $
