# [SOLVED] Gram-Schmidt

• Apr 18th 2010, 02:42 AM
Noxide
[SOLVED] Gram-Schmidt
the vectors {a1, a2, a3} are orthonormal
the vector a4 is a combination x1a1 + x2a2 + x3a3

after applying gram schmidt to the set of vectors {a1, a2, a3, a4} the resulting set of vectors will be:

{a1, a2, a3, zero vector}
is this correct?
• Apr 18th 2010, 05:41 AM
HallsofIvy
Yes, if you do it in that order- that is, normalize \$\displaystyle a_1\$ to unit length, then "Gram-Schmidt" \$\displaystyle a_2\$, etc. Since the first three are already orthonormal, you would just get \$\displaystyle a_1\$, \$\displaystyle a_2\$, \$\displaystyle a_3\$ again. And since \$\displaystyle a_3\$ is a linear combination of them, it has no "part" that is perpendicular to the space spanned by them- Gram-Schmidt would give the 0 vector.

But if you started by normalizing \$\displaystyle a_3\$, then projecting, say, \$\displaystyle a_1\$ onto that to find a vector perpendicular to \$\displaystyle a_3\$ you would get 3 completely different vectors.