5

Let $ W = \left \{\mathbf{x} = \begin{pmatrix} x_1 \\x_2 \\x_3 \end{pmatrix} : x_1 + x_2 + x_3 = 0 \right\}$ and find a basis for $W$

I don't really know how to do it by guess work so I tried this method:

Solve $x_1 + x_2 + x_3 = 0$ to row echelon form (which it already is in) and so we get the solution $ \begin{pmatrix} x_1 \\x_2 \\x_3 \end{pmatrix} = \begin{pmatrix} -x_2 -x_3 \\x_2 \\x_3 \end{pmatrix}$

then use a simple method to find the matrix, let $x_2 = 1$ and $x_3 = 0$ which gives us $\begin{pmatrix} -1 \\1 \\0 \end{pmatrix}$ and let $x_3 = 1$ and $x_2 = 0$ giving $\begin{pmatrix} -1 \\ 0 \\1 \end{pmatrix}$ so the basis is $\left \{\begin{pmatrix} -1 \\1 \\0 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\1 \end{pmatrix} \right\}$

Is this a valid method as I really don't like guess work (which my teacher said for us to do). I have tested and it is a basis for the vector space $W$

thanks

user113494
  • 225

3 Answers3

2

What you have done, is that you have defined the map $$\begin{eqnarray}\Bbb R^2 &\rightarrow& \Bbb W \\ \begin{pmatrix} x_1 \\x_2 \end{pmatrix} &\rightarrow& \begin{pmatrix} -x_1-x_2 \\ x_1 \\ x_2 \end{pmatrix}\end{eqnarray}$$ (Or whatever base field you are working with).

It is easy to show that this is an isomorphism. So it sends a base of $\Bbb R^2$ to a base of W. If you are familiar with isomorphism I think this may be easier to prove than checking directly that something is a basis of $W$.

benh
  • 6,605
1

Here is a hint for you to see if you are doing the right thing.

You start with a 3 dimensional space. Every independent constraint (equation) will reduce the dimension by 1. So in your problem, the space is dimension 2. Since you have two linearly independent vectors and they satisfy the constraint, you know you have the answer.

In general, this method usually works. If you know the dimension drops by $k$, pick any $k$ variables and call them the dependent variables and remaining ones the independent variables. Now go through the independent variables and set one of them zero and the rest zero. Now solve for the dependent variables. Repeat by selecting each independent variable one at a time.

If you can't solve for independent variable because of linear dependence, then pick another set of dependent variables. If you can reduce the constraints to row-echelon form, then the variables that occur in exactly one equation, become the dependent variable. Make sure you pick one variable from each equation after discarding equations that is identically zero.

Hope this long explanation helps

user44197
  • 9,730
0

Similarly to what you did...Let $x_1=-x_2-x_3$. This ensures that $x_1+x_2+x_3=0$. It is clear that $(-x_2-x_3)+x_2+x_3 = 0$. Now $(x_1,\space x_2,\space x_3) = (-x_2-x_3,\space x_2,\space x_3)$. After this simply match each variable with its respective coefficient. Notice that there is no $x_1$, so we can already see that the basis will have $2$ vectors - the coefficients of $x_2$ and $x_3$. $$(x_1,\space x_2,\space x_3)= (-x_2-x_3,\space x_2,\space x_3)= x_2(-1,\space 1,\space 0) + x_3(-1,\space 0,\space 1)$$

Therefore a basis for $W$ is $$\left \{\begin{pmatrix} -1 \\1 \\0 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\1 \end{pmatrix} \right\}$$

Zhoe
  • 2,415