# Partitioned Orthogonal Matrix

1. Dec 29, 2009

### Dafe

"Partitioned Orthogonal Matrix"

Hi,
I was reading the following theorem in the Matrix Computations book by Golub and Van Loan:

If $$V_1 \in R^{n\times r}$$ has orthonormal columns, then there exists $$V_2 \in R^{n\times (n-r)}$$ such that,
$$V = [V_1V_2]$$ is orthogonal.
Note that $$ran(V_1)^{\bot}=ran(V_2)$$

It also says that the proof is a standard result from introductory linear algebra.

So I picked up my copy of Introduction to linear algebra by Strang and did not find this.
I then looked in the Matrix Analysis book by Carl D. Meyer, and here he mentiones this under the name "partitioned orthogonal matrix". I did not find a proof though.

Is there a proper name for this "decomposition"?

Thanks.

2. Dec 29, 2009

### rochfor1

Re: "Partitioned Orthogonal Matrix"

This may be easier to see if you rephrase the problem. The columns of $$V_1$$ form an orthonormal basis for an r-dimensional subspace of $$\matbb{R}^n$$, and it is a standard result from the theory of Hilbert spaces that you may extend an orthonormal basis for a subspace to the whole space. The functional analyst in me would use Zorn's lemma to show that every orthonormal set (the columns of $$V_1$$) is contained in an orthonormal basis of the whole space, but this is overkill in the finite-dimensional case. In this case, I'd simply find a basis of $$\mathbb{R}^n$$ that contains the columns of $$V_1$$ (using your favorite argument) and then use the Gram-Schmidt process. Not that if you let the columns of $$V_1$$ be the first r vectors in the Gram-Schmidt process, they will remain unchanged (because they are already orthogonal).