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

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?

• Gram-Schmidt is a process that works in any inner product space. The cross product is only defined on $\mathbb{R}^3$ (and $\mathbb{R}^2$ if you embed it into $\mathbb{R}^3$). What you propose has, at best, an extremely narrow window of applicability. In addition, the Gram-Schmidt process has additional properties that your method does not have: when you orthogonalize the vectors $v_1,\ldots,v_n$ into the vectors $w_1,\ldots,w_n$, the vector $v_k-w_k$ is the orthogonal projection of $v_k$ onto the span of $v_1,\ldots,v_{k-1}$, important on its own; and the process works for subspaces too. Jan 26 at 18:57
• Here is a MathJax tutorial. Please use Mathjax to format the mathematics in your post. Jan 26 at 18:58

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.
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?