# The need for the Gram–Schmidt process

As far as I understood Gram–Schmidt orthogonalization starts with a set of linearly independent vectors and produces a set of mutually orthonormal vectors that spans the same space that starting vectors did.

I have no problem understanding the algorithm, but here is a thing I fail to get. Why do I need to do all these calculations? For example, instead of doing the calculations provided in that wiki page in example section, why can't I just grab two basis vectors $w_1 = (1, 0)'$ and $w_2 = (0, 1)'$? They are clearly orthonormal and span the same subspace as the original vectors $v_1 = (3, 1)'$, $v_2 = (2, 2)'$.

It is clear that I'm missing something important, but I can't see what exactly.

• @Bye_World this is a really nice example, which in my opinion explains my question. The examples of Grand-Schmidt orthogonalization that I have seen so far deal with the problems, where I was able to find the orthonormal vectors in a matter of seconds, this is why I was curious why would one use something more complicated. It would be nice if you can post this as an answer Oct 10, 2016 at 4:18
• I do think that maybe the issue is that there are certain problems that are easy to express in an no-ON basis, but manipulation of that problem is definitely easier in an ON basis. If you can choose a priori a basis that's a thing, but certain problems are probably not easy to express in an ON basis. Oct 10, 2016 at 15:01
• I think someone should just edit that Wikipedia page to give an example where the span isn't just trivially $\mathbb{R}^n$. Two linearly independent vectors from $\mathbb{R}^3$ would be more motivating. Oct 10, 2016 at 17:33
• @asmeurer Why not be that someone? Oct 10, 2016 at 18:14
• The main reason "why you need to do all these calculations" is because people set unimaginative homework questions. The only point of the calculations is to show that you understand the method! In real-life numerical linear algebra, the idea behind G-S orthogonalization is very important, but there are much better numerical methods to calculate orthogonal basis vectors than the G-S algorithm - and you would never do such calculations by hand, in any case. Oct 10, 2016 at 21:25

Orthonormal bases are nice because several formulas are much simpler when vectors are given wrt an ON basis.

Example: Let $\mathcal E = \{e_1, \dots, e_n\}$ be an ON basis. Then the Fourier expansion of any vector $v\in\operatorname{span}(\mathcal E)$ is just $$v = (v\cdot e_1)e_1 + (v\cdot e_2)e_2 + \cdots + (v\cdot e_n)e_n$$

Notice that there are no normalization factors and we don't need to construct a dual basis -- it's just a really simple formula.

In your example, of course $\{(1,0),(0,1)\}$ spans the same space as $\{(3,2),(2,2)\}$. But let me provide an example of my own: what about $\{(1.1,1.2,0.9,2.1,4),(3,-2,6,14,2),(6,6,6,3.4,11.1)\}$? There's certainly no subset of the standard basis vectors that spans the same space as these linearly independent vectors. But this is a pretty poor choice of basis because they're not orthonormal. It'd sure be nice if we had some algorithm that could produce as ON basis from them...

• I think OP is asking why we can't use ${(1,0,0,0,0),(0,1,0,0,0),...,(0,0,0,0,1)}$ as the basis. Perhaps that could be addressed in your answer? Oct 10, 2016 at 14:06
• @user1717828 Based on the comment OP left me below his question, I think my answer suffices.
– user137731
Oct 10, 2016 at 15:26
• @user1717828 - you failed to notice that Bye_World's example is a 3-dimensional subspace of $\Bbb R^5$ which does not contain any of the standard basis elements. Oct 10, 2016 at 17:51
• @PaulSinclair, I did not fail to notice it; this is the answer I think the original question was asking about and should be addressed explicitly in the text (although Bye_World points out OP is satisfied without it). Oct 10, 2016 at 17:59

If the entire vector space we wanted a basis for all of $\mathbb{R}^n$ or a basis for, say, the $xy$-plane in $\mathbb{R}^3$, then we could certainly do this. The problem arises when we're looking for an orthonormal basis for subspaces that are more complicated.

For example, let take our subspace to be the plane $x+y+z=0$ inside of $\mathbb{R}^3$. This subspace can be written as

$$\mathrm{span} \left\{ \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} , \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \right\}$$

Running Gram-Schmidt on this gives the basis

$$\left\{ \begin{bmatrix} \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \end{bmatrix} ~,~ \begin{bmatrix} \frac{1}{\sqrt{6}} \\\frac{1}{\sqrt{6}}\\ -\sqrt{\frac{2}{3}} \end{bmatrix} \right\}$$ which generates the same subspace of $\mathbb{R}^3$. However, if we just tried to take the basis $\left\{ \vec{e}_1 = [1~ 0~ 0]^T, ~ \vec{e}_2 = [ 0 ~ 1 ~ 0]^T \right\}$, we would certainly have an ortho-normal basis, but of a different subspace.

TL;DR: Gram-Schmidt is designed to turn a basis into an ortho-normal basis without altering the subspace that it spans.

Edit Gram-Schmidt is also important in that it preserves the orientation of given basis (roughly speaking, the order in which the basis elements are introduced). Somewhat higher level and beyond the scope of an introductory course to linear algebra, but worth tucking away for later.

You can also get into function spaces where it's not clear what the basis you can just grab from is. The Legendre polynomials can be constructed by starting with the functions $1$ and $x$ on the interval $x \in [-1,1]$, and using Gram-Schmidt orthogonalization to construct the higher order ones. The second order polynomial is constructed by removing the component of $x^2$ that points in the direction of $1$, for example.

• I don't understand this. The basis set for polynomials is just $\{1, x, x^2, \ldots\}$. The Legendre polynomials are generated by applying Gram-Schmidt on this basis with respect to a certain inner product. Oct 10, 2016 at 17:38
• Right, and that basis set is not orthonormal, making it not an "obvious basis you can just grab from". Oct 11, 2016 at 2:42
• I would also note that there is no such thing as "the" basis set. Any non-trivial vector space has infinitely many bases, each as valid as the others. There are canonical bases that are chosen out of convenience, and we have things like the standard basis for $\mathbb{R}^n$, but these are far from the only choices. Oct 11, 2016 at 17:34

Your choice of $w_1 = (1,0)$ and $w_2 = (0,1)$ fails one of the basic purposes of the Gram-Schmidt process: the result of the algorithm would not only have $\mathrm{Span}(v_1,v_2) = \mathrm{Span}(w_1,w_2)$ but also $\mathrm{Span}(v_1) = \mathrm{Span}(w_1).$

There are a few things I should mention:

(1) There are finite-dimensional vector spaces for which it is not easy to guess an orthonormal basis (or indeed any basis). Modular forms with their Petersson scalar product are an intimidating example of this.

(2) The Gram-Schmidt process is smooth in an appropriate sense, which makes it possible to use the Gram-Schmidt process to orthogonalize sections of a Euclidean bundle (a vector bundle with scalar product) and in particular to define things like the orthogonal complement of subbundles. This turns out to be important.

(3) There are also concrete and relatively elementary situations such as QR-factorization for why Gram-Schmidt orthogonalization is useful. Maybe you are already familiar with this.

• Based on the description of the algorithm, I see that not only $span(v_1, ... v_n)$ is the same as $span(w_1, ..., w_n)$, but also spans of vectors after each step are the same. But no one in the description of the algorithm it is written that this is the purpose of it: everywhere I have seen so far just tells that the span of original vectors should be the same as the span of resulting vectors Oct 10, 2016 at 4:27
• @SalvadorDali Whether this is important depends on what you want to get out of the Gram-Schmidt process. Some results like the Schur decomposition and QR-factorization depend crucially on the fact that $\mathrm{Span}(v_1,...,v_k) = \mathrm{Span}(w_1,...,w_k)$ for all $k$. Oct 10, 2016 at 4:29

To give a somewhat practical example, let’s consider the following iterative sequence:

$$x_{i+1} = \frac{A x_i}{\left |A x_i \right|},$$

Where the $x_i$ are vectors and $A$ is a quadratic matrix with a matching dimension. The first vector $x_1$ shall be random¹.

When $i$ is increased, $x_i$ will more and more be aligned to an eigenvector corresponding to the largest eigenvalue of $A$ and quickly become such an eigenvector for all practical purposes. So we can use this sequence to get a numerical approximation for the largest eigenvalue of $A$ and a corresponding eigenvector (of length 1). This approach is actually used for this purpose.

Now, if we want to obtain the second largest eigenvalue of $A$ (and the corresponding eigenvector) in a similar way, we have to remove the influence of the first eigenvalue, as it would inevitably dominate our sequence otherwise. Hence we have to work in a subspace that is orthogonal to the eigenvector for the largest eigenvalue. Gram–Schmidt orthonormalisation provides a way to do this: We start with two random vectors $x_0$ and $y_0$ and then we do a Gram–Schmidt after every step:

\begin{align} x_{i+1} &= \frac{A x_i}{\left |A x_i \right|},\\ y_{i+1} &= \frac{A y_i - \left \langle A y_i, x_{i+1} \right \rangle}{\left | A y_i - \left \langle A y_i, x_{i+1} \right \rangle \right |} \end{align}

Due to this, $x_i$ will align with the first eigenvector and $y_i$ will align with the second one. We can extend this scheme to further eigenvectors if we like.

The crucial feature of the Gram–Schmidt process that we exploit here is that the first $k$ vectors of its result span the same subspace as the first $k$ vectors of its input for any $k$. A consequence of this is that the $k$th output vector is orthogonal to all previous output vectors. Obviously, this would not work with any basis.

Now in many cases, you can determine the first eigenvectors more easily, but there are analogous problems, where you need to go the way described above. For example, it is used for numerical calculation of Lyapunov exponents, where $A$, $x$, and $y$ are subject to a complex temporal evolution.

¹ and thus we can assume that it is not orthogonal to any eigenvector of $A$

If you want to represent a subspace, the standard basis will no do.

For example, assume a plane defined by two arbitrary vectors in 3D: G-S will define an orthonormal base that spans the plane.