Is this a simpler way to orthogonalize bases than Gram-Schmidt?

Question

This is my first post here. I have a question. In my linear algebra course we are learning the Gram-Schmidt process. But it appears to me in a much more intuitive way to do the cross product successively until obtaining orthogonal vectors.

Example 1: I have a plane in $\mathbb{R}^3$ and two vectors $v_1$ and $v_2$ that are linearly independent, but not orthogonal.

If I do $v_3= v_1 \times v_2$ and then $v_4 = v_1 \times v_3$ I will obtain that the vector $v_1$ is perpendicular to the vector $v_4$ and there I have an orthogonal base of the plane.

Example 2: I have 3 vectors $v_1,v_2,v_3$ in $\mathbb{R}^3$ that are linearly independent but not orthogonal.

If I do $v_4 = v_1 \times v_2$ , then $v_5 = v_1 \times v_4$. I get that $v_1,v_4,v_5$ are orthogonal to each other.

Is this a simpler way to orthogonalize bases than Gram-Schmidt?

I know that the cross product is defined only in $\mathbb{R}^3$, but isn't there an equivalent way to do this cross-product in $\mathbb{R}^n$ that is easier to do than the Gram-Schmidt process?

And finally, does this method of performing the cross product successively require less amount of computational operations?

Answer 1

The cross product is only available in $\mathbb{R}^3$.

In every situation you mentioned, you are only looking for an orthogonal basis of $\mathbb{R}^3$. There is a much easier solution in this case: just use the standard basis vectors.

Gram-Schmidt is more useful in cases such as:

1. Finding an orthogonal basis of a subspace of a higher dimensional space. For example, can you find an orthogonal basis of the subspace of $\mathbb{R}^5$ defined by $x_1 +3x_3+2x_4 = 0$ and $x_1 + x_2 + x_3 + 5x_5 = 0$?
2. Finding an orthogonal basis of a more abstract inner product space. For instance, define a vector space of polynomials of degree at most $n$ and give the inner product $\langle p, q\rangle = \int_{-1}^{1} p(x)q(x) \textrm{d} x$. Can you find an orthogonal basis of this space?