• Jul 23rd 2006, 12:26 PM
luckyc1423
Let W be the subspace of R^4 spanned by :
[ 2
0
-1
3]
and
[ -6
1
5
-8]
Find a basis for W^(upside down T)

dont know the correct word for the symbol that looks like an Upside down T
• Jul 24th 2006, 02:27 AM
CaptainBlack
Quote:

Originally Posted by luckyc1423
Let W be the subspace of R^4 spanned by :
[ 2, 0, -1, 3]' and [-6, 1, 5, -8]'

Find a basis for W^(upside down T)

dont know the correct word for the symbol that looks like an Upside down T

$\{[1,\ 0,\ 0,\ 0]',\ [0,\ 1,\ 0,\ 0]',$ $\ [ 2,\ 0,\ -1,\ 3]',\ [-6,\ 1,\ 5,\ -8]'\}=\{e_1,e_2,w_1,w_2\}$

is a basis of $\mathbb{R}^4$ (proof left as an excercise for the reader).

Now $w_1,\ w_2$ is a basis for $W$, so $[w_1,w'_2]$,
where:

$
w'_2=w_2-\frac{\langle w_1,w_2\rangle w_1}{\|w_1\|^2}
$

is an orthogonal basis for $W$.

Now let:

$
e'_1=e_1-\frac{\langle e_1,w_1\rangle w_1}{\|w_1\|^2}-\frac{\langle e_1,w'_2\rangle w'_2}{\|w'_2\|^2}
$

and:

$
e'_2=e_2-\frac{\langle e_2,w_1\rangle w_1}{\|w_1\|^2}-\frac{\langle e_2,w'_2\rangle w'_2}{\|w'_2\|^2}}-\frac{\langle e_2,e'_1\rangle e'_1}{\|e'_1\|^2}
$
.

Now if I have done this right we should have $\{e'_1,\ e'_2,\ w_1,\ w'_2\}$
is an orthogonal basis of $\mathbb{R}^4$, with $\{w_1,\ w'_2\}$ an orthogonal basis
of $W$, and $\{e'_1,\ e'_2\}$ and orthogonal basis of $W^{\bot}$.

RonL