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How can I complete a matrix to a orthogonal matrix?

Consider for example A given by

$ \begin{pmatrix} 1/\sqrt 6 & -1/\sqrt 2 \\ 1/\sqrt 6 & 1/\sqrt 2 \\ 2/\sqrt 6 & 0 \end{pmatrix} $

Then there is written that one can complete this matrix with Gram Schmidt to get a orthogonal matrix given by

$ \begin{pmatrix} 1/\sqrt 6 & -1/\sqrt 2 & -1/\sqrt 3 \\ 1/\sqrt 6 & 1/\sqrt 2 & -1/\sqrt 3\\ 2/\sqrt 6 & 0 & 1/\sqrt 3 \end{pmatrix} $

Could someone explain how to complete this matrix?

2 Answers2

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The Gram-Schmidt procedure is for the orthonormalization of a basis of a vector space. I never hear about "orthogonalize a matrix". If you "orthogonalize" a matrix as you are trying to do you will had a completely different matrix.

In any case not all matrix are "orthogonalizable", you need previously that the column or row vectors of the matrix define a basis for some vector subspace, otherwise your matrix will be not "orthogonalizable". In this case just extend this basis to a basis of the whole vector space and after apply the Gram-Schmidt procedure to this extended basis. Then the matrix defined by this basis, as columns or rows, will be orthogonal.

However observe that the orthogonalization is not unique because each orthonormal vector that you add can have at least two distinct directions.

Masacroso
  • 30,417
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You have already two columns: $v_1=\begin{pmatrix}\frac{1}{\sqrt{6}}\\\frac{1}{\sqrt{6}}\\\frac{2}{\sqrt{6}}\end{pmatrix}$ and $v_2=\begin{pmatrix}-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\0\end{pmatrix}$. Add a new column just to make $(v_1,v_2,u)$ linearly independent. You have a lot of freedom there, choose whatever is the easiest for the subsequent calculation. For example, set $u=\begin{pmatrix}1\\0\\0\end{pmatrix}$.

Now:

  • Calculate the new vector:

$$\overline{v_3}=u-\langle u,v_1\rangle v_1-\langle u,v_2\rangle v_2=\begin{pmatrix}1\\0\\0\end{pmatrix}-\frac{1}{\sqrt{6}}\begin{pmatrix}\frac{1}{\sqrt{6}}\\\frac{1}{\sqrt{6}}\\\frac{2}{\sqrt{6}}\end{pmatrix}+\frac{1}{\sqrt{2}}\begin{pmatrix}-\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\0\end{pmatrix}=\begin{pmatrix}\frac{1}{3}\\\frac{1}{3}\\-\frac{1}{3}\end{pmatrix}$$

  • Calculate $\|\overline{v_3}\|=\sqrt{\langle \overline{v_3},\overline{v_3}\rangle}=\frac{1}{\sqrt{3}}$
  • Set $v_3=\frac{1}{\|\overline{v_3}\|}|\overline{v_3}\|=\begin{pmatrix}\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}\\-\frac{1}{\sqrt{3}}\end{pmatrix}$. That is your third column.

You may get two different solutions (both are valid), depending on your choice of $u$ - just like in $3$-dimensional space you can have two different unit vectors orthogonal to an already given pair of orthonormal vectors. (If the two vectors are in a "horizontal" plane, the third vector may go "up" or "down".) It seems that you have one solution in your question, and I have found the other one.

In general, the Gramm-Schmidt process goes as follows: when adding a new vector to an already orthonormal set of vectors $v_1,v_2,\ldots,v_n$, you will choose $u$ to not be in the span of $v_1,v_2,\ldots,v_n$, and then you will set:

$$\overline{v_{n+1}}=u-\sum_{i=1}^n \langle u, v_i\rangle v_i$$

This will give you $\overline{v_{n+1}}$ which is already orthogonal to all of $v_1,v_2,\ldots,v_n$, but may not have the norm equal to $1$. Then you will "normalise" it:

$$v_{n+1}=\frac{1}{\|\overline{v_{n+1}}\|}\overline{v_{n+1}}=\frac{1}{\sqrt{\langle \overline{v_{n+1}},\overline{v_{n+1}}\rangle}}\overline{v_{n+1}}$$