1. [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.

2. 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.

3. Thanks...

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

4. 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?

5. Thank you.

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

6. Thanks for the help

7. 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.

8. 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.