# Simple Linear algebra question but I have to have a brain freeze smh

1. Feb 10, 2012

### trap101

So I'm reading an example from my text on how to construct an isomorphism:

Let V be in R^3 be defined by the single linear equation: x1 - x2 + x3 = 0 (those are suppose to be subscripts after the x's) and let W be in R^3 be the subspace defined by the single linear equation 2x1 + x2 - x3 = 0. Since dim(V) = dim(W) = 2, V and W are isomorphic. To construc an isomorphism we choose bases for V and W. A basis for v1 = (1,1,0) and v2 = (0,1,1). A basis for w1 = (1, -1, 1) and w2 = (-1/2, 1, 0)

Question: How do I find the dim of V and W? Should be straight forward but I don't see it...smh, and how do they come up with those bases for V and W in the isomorphism?

Thanks

2. Feb 10, 2012

### tiny-tim

hi trap101!
every linear equation reduces the degrees of freedom by 1

(ie, in n dimensions, one equation gives you an n-1 space, two simultaneous equations give you an n-2 space, etc)
just choose numbers as simple as possible that satisfy the equation (and are independent) …

use 1s and 0s if possible, if not then try 2s …

3. Feb 10, 2012

### trap101

Shouldn't I be able to form a linear combination from the standard basis vectors and solve for the dim in a matrix as well?......That was the route I was taking, but it might be useful to remember that fact from now on.

4. Feb 10, 2012

### tiny-tim

not following you

5. Feb 10, 2012

### trap101

this equation: x1 - x2 + x3 = 0 from the first part of my question, My intention was to write this out as a linear combination using the standard basis vectors in R^3, and then I wanted to find the basis from the reduced-echelon form of the matrix.

Another question: How do those chosen bases relate to linear equations?

6. Feb 10, 2012

### SammyS

Staff Emeritus
I'm surprised tiny-tim didn't say "(try using the X2 button just above the Reply box )"

In terms of analytic geometry, the equations, x1 - x2 + x3 = 0 and 2x1 + x2 - x3 = 0 are equations of planes, so you can think of them as each having two dimensions.

The vectors v1 = (1,1,0) and v2 = (0,1,1) each lie in the plane, V. Similarly, w1 = (1, -1, 1) and w2 = (-1/2, 1, 0) each lie in plane W.

Last edited: Feb 10, 2012
7. Feb 10, 2012

### Dansuer

Another way you can look at it is this.
The subspace you are talking about is this

\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} with $2x_1 + x_2 - x_3 = 0$
plugging in $x_3 = 2x_1 + x_2$

$\begin{pmatrix} x_1 \\ x_2 \\ 2x_1 + x_2 \end{pmatrix} = x_1\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + x_2\begin{pmatrix} 0 \\1 \\1 \end{pmatrix}$

Those are a base for the subspace and so the dimension is 2.