[SOLVED] Affine subset

**Galbraith005** · February 25th 2010, 07:47 AM

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.

**Opalg** · February 25th 2010, 11:29 AM

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.

**Galbraith005** · February 25th 2010, 11:38 AM

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

**Opalg** · February 25th 2010, 12:04 PM

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?

**Galbraith005** · February 25th 2010, 12:52 PM

Thanks, you've been a great help. I think I've got it now.

**Kazzy2906** · February 26th 2010, 05:38 AM

Thanks for the help

**patpat** · February 27th 2010, 02:05 PM

We're talking about two different sets of equations here. There's the affine set M, which is the set of all vectors

such that

and

. Then there is the subspace S, which is the set of all vectors

such that

and

.

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

. Then the point is that if

then

, where

. Conversely, if

, then

for some

.

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.

**Opalg** · February 28th 2010, 01:00 AM

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.

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