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?
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?
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.