# [SOLVED] Affine subset

• Feb 25th 2010, 08:47 AM
Galbraith005
[SOLVED] Affine subset
I've just been handed a project for my Linear Algebra course, so far I can't even begin to grasp the concept of affine subsets. Anyway I was wondering if anyone could help with the first question in which I am meant to illustrate the concept:

Show that:

M = {x=(x1,...x4) exists in R^4 : x1 - 2*x2 + x3 = 3 and x1 + 5*x3 - 2*x4=1}

is an affine subset of R^4.

Sorry about the notation here. x1 means x subscript 1.
If anyone can help I would be really grateful.
• Feb 25th 2010, 12:29 PM
Opalg
Quote:

Originally Posted by Galbraith005
I've just been handed a project for my Linear Algebra course, so far I can't even begin to grasp the concept of affine subsets. Anyway I was wondering if anyone could help with the first question in which I am meant to illustrate the concept:

Show that:

M = {x=(x1,...x4) exists in R^4 : x1 - 2*x2 + x3 = 3 and x1 + 5*x3 - 2*x4=1}

is an affine subset of R^4.

Geometrically, an affine subset is something that looks like a subspace except that it doesn't contain the origin. Algebraically, if S is a subspace and v is a vector, then the set v+S (in other words, the set of all vectors of the form v plus something in S) is an affine set, and every affine set is of this form.

In the above example, start by finding a vector in M. The easiest one to find is probably $v=\bigl(0,\tfrac32,0,-\tfrac12\bigr)$. Now let $S = \{x\in\mathbb{R}^4:x_1 - 2x_2 + x_3 = x_1 + 5x_3 - 2x_4= 0\}$. Show that S is a subspace of $\mathbb{R}^4$, and M = v + S. Therefore M is an affine set.

An affine set is simply a subspace that has been translated away from the origin.
• Feb 25th 2010, 12:38 PM
Galbraith005
Thanks...
How can you set the two eqn's equal to each other when one equals 1 and the other equals 3?
• Feb 25th 2010, 01:04 PM
Opalg
Quote:

Originally Posted by Galbraith005
How can you set the two eqn's equal to each other when one equals 1 and the other equals 3?

We're talking about two different sets of equations here. There's the affine set M, which is the set of all vectors $x = (x_1,x_2,x_3,x_4)$ such that $x_1 - 2x_2 + x_3 = 3$ and $x_1 + 5x_3 - 2x_4= 1$. Then there is the subspace S, which is the set of all vectors $x = (x_1,x_2,x_3,x_4)$ such that $x_1 - 2x_2 + x_3 = 0$ and $x_1 + 5x_3 - 2x_4= 0$.

It might have made it clearer if I had used a different letter, say y instead of x, for elements of the subspace S. Suppose I had said that $S = \{y = (y_1,y_2,y_3,y_4)\in\mathbb{R}^4:y_1 - 2y_2 + y_3 = y_1 + 5y_3 - 2y_4= 0\}$. Then the point is that if $y\in S$ then $v+y\in M$, where $v=\bigl(0,\tfrac32,0,-\tfrac12\bigr)$. Conversely, if $x\in M$, then $x = v+y$ for some $y\in S$.

Does that make it any clearer?
• Feb 25th 2010, 01:52 PM
Galbraith005
Thank you.
Thanks, you've been a great help. I think I've got it now.
• Feb 26th 2010, 06:38 AM
Kazzy2906
Thanks for the help
• Feb 27th 2010, 03:05 PM
patpat
Let $y=(1,1,1,3) \in S)$ then using your definition of $v$ we have $v+y=(1, \frac{5}{2}, 1 , \frac{5}{2}) \not\in M$.

P.S. Guys try to do this project on your own. There is no point in begging for answers on math forums.
• Feb 28th 2010, 02:00 AM
Opalg
Quote:

Originally Posted by patpat
Let $y=(1,1,1,3) \in S$ then using your definition of $v$ we have $v+y=(1, \frac{5}{2}, 1 , \frac{5}{2}) \not\in M$.

Careless typo on my part. The vector v should have been $(0,-\tfrac32,0,-\tfrac12)$. Then $v+y = (1,-\tfrac12,1,\tfrac52)$, which is in M of course.