21
$\begingroup$

I've been given the following as a homework problem:

Find a basis for the following subspace of $F^5$: $$W = \{(a, b, c, d, e) \in F^5 \mid a - c - d = 0\}$$

At the moment, I've been just guessing at potential solutions. There must be a better method than guess and check.

How do I solve this and similar problems?

$\endgroup$
1
  • 1
    $\begingroup$ Note: Since this is a homework problem, I'm not looking for the answer. Just the strategy to solve this and similar problems. $\endgroup$ Jan 27, 2012 at 1:09

1 Answer 1

43
$\begingroup$

Let's look at the following example:

$$W = \{ (a,b,c,d)\in\mathbb{R}^4 \mid a+3b-2c = 0\}.$$

The vector space $W$ consists of all solutions $(x,y,z,w)$ to the equation $$x + 3y - 2z = 0.$$

How do we write all solutions? Well, first of all, $w$ can be anything and it doesn't affect any other variable. Then, if we let $y$ and $z$ be anything we want, then that will force $x$ and give a solution. So we have three degrees of freedom: a free choice of $w$, a free choice of $z$, and a free choice of $y$. Then $x$ will be forced. This suggests dimension $3$.

How does the choice of $w$ affect $x$, $y$, and $z$? In absolutely no way. Since choosing $w$ does not affect $x$, $y$, or $z$, this gives the vector $(0,0,0,1)$: the choice of $w$ (the $1$) does not affect the others.

How does the choice of $z$ affect $x$, $y$, and $w$? It doesn't affect $y$ and $w$. But if $z=1$, then $x$ needs to be $2$: that is, we need to get two $x$s for every $z$. This gives the vector $(2,0,1,0)$.

Finally, now does the choice of $y$ affect $x$, $z$, and $w$? It doesn't affect $z$ and $w$ (they are free), but for every $y$, we need to have $-3$ $x$s. That gives the vector $(-3,1,0,0)$.

So a basis for my $W$ consists of $(-3,1,0,0)$, $(2,0,1,0)$, and $(0,0,0,1)$. You can verify that all of them lie in $W$, and that every vector in $W$ can be written as a linear combination of these three in a unique way.

$\endgroup$
2
  • 1
    $\begingroup$ Great explanation! Looks like you just popped over the 100k reputation mark as well, congratulations. :) $\endgroup$ Jan 27, 2012 at 5:02
  • 2
    $\begingroup$ Incredibly helpful and did exactly what I asked for: gave me a strategy for solving similar problems. Thanks a bunch! $\endgroup$ Jan 27, 2012 at 17:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .