Revision as of 14:22, 8 December 2010

Statement: I am going to show that if V is a subspace of Rⁿ. then dim(V)+dim(V^orth)=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,..,v_k} 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=

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-[[Category:2010 Fall MA 265 Momin]]
+<br>
-=dimVplusdimVorthisn=
+= Statement: I am going to show that if V is a subspace of R<sup>n</sup>. then dim(V)+dim(V<sup>orth</sup>)=n<br> =
+Because of the lack of the orthagonal symbol in the wikipedia formatting page, I will be type 'orth' in a superscript to symbolize that.&nbsp;
-Put your content here . . .
+Analysis:
+First, let us say we have the following:
+V which is a subspace of R<sup>n</sup>, and {v<sub>1</sub>,v<sub>2</sub>,v<sub>3</sub>,..,v<sub>k</sub>} are a basis for V. (The entries in the braces are vectors)
-[[ 2010 Fall MA 265 Momin|Back to 2010 Fall MA 265 Momin]]
+To refresh, a basis means those entries span V, AND are also linearly independent.&nbsp;
+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=