How to obtain orthogonal (not orthonormal) vectors from "orth" or "qr"?

Suppose there is a set of vectors v1, v2, v3, v4 with V=[v1, v2, v3, v4]. The vectors have very large number of data points ~ 100k. I wanted to orthogonalize them with each other. The classical Gram-Schmidt is not recommended by majority of authors for real data. The symbolic version of orth has skipnormalization. https://www.mathworks.com/help/symbolic/orth.html, but it does not work on orth(V). Error using orth Too many input arguments.

What would be the best way to generate orthogonal basis vectors rather orthonormal basis vectors using built in MATLAB functions? Thanks.

[Chunru]

Are "eig" and "svd" good for your application?

[FW]

I just wanted an orthogonal basis set of v1, v2, v3, and v4 but not orthonormal. V is not a square matrix for "eig" and as per the orth page, orth is obtained by SVD.

[Chunru]

svd is for rectangle matrix. orthonormal is a special case or orthogonal.

[Paul]

What is the objection to the basis being orthonormal?

Just specifying that the basis vectors are orthogonal isn't sufficient to find a unique solution. So there needs to be adiditonal constraints, or can generate the orthonormal basis and multiply the basis vector by constants.

[FW]

Paul, this orthogonalization process has applications in some chemical measurements (infrared spectroscopy).

[Torsten]

I don't understand why orth(V) does not work for you.

orth is also a numerical function and does not only work on symbolic matrices.

[Chunru]

orth is similar to svd (the undline algorithm is svd)

[Paul]

@FW

Are you looking for code that implements a specific method for finding an orthogonal basis? Or is the desire to find a specific basis regardless of the method used to find it? For example:

V = rand(10,4);
Q = orth(V);
Q = Q.*rand(1,4);

The columns of Q are orthogonal to each other, do not have unit norm, and are a basis for the range space of V. If this Q doesn't meet your requirements, please explain why not.

[FW]

@Paul, No, I am not looking for a specific method. I was just exploring built in MATLAB options. I have tried Gram Schmidt, modified Gram-Schmidt on the same data. They work fine.

As per MATLAB documentation, Q = orth(A) returns an orthonormal basis for the range of A. The columns of Q are vectors, which span the range of A. The number of columns in Q is equal to the rank of A. https://www.mathworks.com/help/matlab/ref/orth.html

The output of orth is normalized, i.e., Q*Q = I. I just confirmed.

[Paul]

@FW

it is still unclear, at least to me, what exactly you're looking for. Let's try again:

rng(100)
V = rand(10,4);
Q = orth(V);
format short e
Q.'*Q

ans = 4×4

1.0000e+00 -2.2519e-16 3.0936e-17 -1.4817e-17 -2.2519e-16 1.0000e+00 6.5204e-16 1.0469e-17 3.0936e-17 6.5204e-16 1.0000e+00 2.3952e-17 -1.4817e-17 1.0469e-17 2.3952e-17 1.0000e+00

As shown, Q is an orthogonal basis for the range space spanned by V. Unless I don't understand what you mean by "orthogonal basis." Do you agree that the columns of Q are an orthogonal basis?

Now, the columns of Q also have unit norm, so they are are an orthonormal basis, which is also an orthognal basis.

vecnorm(Q)

ans = 1×4

1.0000e+00 1.0000e+00 1.0000e+00 1.0000e+00

But for reasons yet to be explained, the desire is to find an orthogonal basis that is not orthonormal.

Here is one option for doing so

Q1 = Q.*[2 4 6 8];
Q1.'*Q1

ans = 4×4

4.0000e+00 -1.8015e-15 7.6554e-16 -2.3707e-16 -1.8015e-15 1.6000e+01 1.4509e-14 3.3502e-16 7.6554e-16 1.4509e-14 3.6000e+01 5.1859e-16 -2.3707e-16 3.3502e-16 5.1859e-16 6.4000e+01

vecnorm(Q1)

ans = 1×4

2.0000e+00 4.0000e+00 6.0000e+00 8.0000e+00

Does Q1 meet the problem requirements? If not, why not?

[Paul]

I'm still not getting it, and I'm perhaps now even more confused.

We have a vector m of dimension 2000 x 1.

We have three vectors b1, b2, b3. Are these also of dimension 2000 x 1?

We can compute an orthonormal basis for the range space of B = [b1 b2 b3]. Those basis vectors are the colums of Q = orth(B). Q = [q1 q2 q3]. I assume that's what is meant by "orthonomalize b1, b2, and b3."

Now, we bring a new aspect into the discussion: " "orthogonalize" m with respect to the three orthonormal b's." Does this mean you want to find the component of m that is orhogonal to b1, b2, and b3? In other words, do you want to find the vector w such that m can be expressed as

m = w + a1*b1 + a2*b2 + a3*b3 where dot(b1,w) = 0, dot(b2,w) = 0, and dot(b3,w) = 0, and a1, a2, a3 are scalars?

[FW]

You are right until "We can compute an orthonormal basis for the range space of B = [b1 b2 b3]. Those basis vectors are the colums of Q = orth(B). Q = [q1 q2 q3]. I assume that's what is meant by "orthonomalize b1, b2, and b3."

Now, take m, and orthogonalize m, with respect to q1, q2, q3 to obtain another vector u. This implies dot(u, q1) = dot(u,q2)=dot(u,q3) =0. However, u is not normalized. The length of u is taken as intensity.

Answer 1

[Paul]

I think this at least illustrates what you're trying to accomplish. It might not be (probably isn't) the best way

% generate some sample data
rng(100);
m = rand(1e5,1);
B = rand(1e5,3);
% verify that vectors bi are linearly independent
rank(B) == 3

ans = logical
   1

% compute the desired result
Q = orth(B);
a = sum(m.*Q);
u = m - sum(a.*Q,2); % u is the component of m that is orthogonal to the bi and the qi
% verify that u is orthogonal to the bi
u.'*B

ans = 1×3

1.0e+-11 * -0.3567 -0.6558 -0.8104

%s show that u is also orthogonal to the qi
u.'*Q

ans = 1×3

1.0e+-13 * 0.3722 0.1427 0.2941

[FW]

Thanks, I don't get the step u=m-sum(a.*Q). Note that m does not have to be orthogonal to B, it needs to be orthogonal only to Q. The dot product is relatively large as compared to classical GS. I usually see 10^-18 or 10^19.

[Paul]

The elements of a are the dot products of m with each qi.

Each column of a.*Q is the component of m in the direction of qi.

So u = m - sum(a.*Q,2) subtracts the sum of all of those components from m. What's left over, u, must therefore be the component of that is orthogonal to all the qi.

m is not orthogonal to the bi or the qi (did I say or imply that it is?). However, u is orthogonal to the bi (and the qi, of course).

Yes, I agree the dot product is relatively large, which is why I said this approach probably isn't the best way to solve the problem. I was just trying to illustrate a solution.

[FW]

Yes, I agree this probably isn't the best approach bc of large dot product. The classical method is working pretty well. As I said earlier, I was exploring built-in MATLAB functions. It seems there is none. Thanks for your effort.