determine a orthogonal basis $(v_1,v_2,v_3)$ of $\mathbb R^3$, such that $Av_1,Av_2,Av_3$ are pairwise orthogonal

Question

Let $A= \begin{bmatrix} 1& 0& -4\\ 2& 1& 1\\ 2& -1& 1 \end{bmatrix} $

$A^TA=\begin{bmatrix} 9& 0& 0\\ 0& 2& 0\\ 0& 0& 18 \end{bmatrix}$

I am asked to determine a orthogonal basis $(v_1,v_2,v_3)$ of $\mathbb R^3$, such that $Av_1,Av_2,Av_3$ are pairwise orthogonal.

If we take the columns of $A^TA$ as the basis $(v_1,v_2,v_3)$, they are obviously an orthogonal basis of $\mathbb R^3$ and we can calculate

$A\begin{bmatrix} 9\\ 0 \\ 0 \end{bmatrix}=$$\begin{bmatrix} 9\\ 18 \\ 18 \end{bmatrix}$.

$A\begin{bmatrix} 0\\ 0 \\ 18 \end{bmatrix}=$$\begin{bmatrix} -72\\ 18 \\ 18 \end{bmatrix}$

$A\begin{bmatrix} 0\\ 2 \\ 0 \end{bmatrix}=$$\begin{bmatrix} 0\\ 2 \\ -2 \end{bmatrix}$

These vectors are pairwise orthogonal again.

Could someone explain why this works?

Answer 1

Note that $Av_1,Av_2,Av_3$ are pairwise orthogonal if

• $v_j^T A^TA v_i=0$

and since $A^TA$ is diagonal the condition is fulfilled since

• $v_j^T A^TA v_i=v_j^T D v_i=d_{ii}v_j^Tv_i=0$

Answer 2

Note that if $v$ and $w$ are orthogonal eigenvectors of $A^TA$ (which is to say that $v^Tw = 0$), then we have $$ (Av)^T(Aw) = v^T(A^TAw) = \lambda \ v^Tw = \lambda \cdot 0 = 0 $$

Answer 3

The fact that A$^T$A is diagonal shows that A, by definition, is orthogonal. It also means that A$^T$A's columns are multiples of the elementary vectors, and therefore orthogonal. Orthogonality is preserved by orthogonal matrices. That is, if u is orthogonal to v, and A is orthogonal, then Au and Av are orthogonal. So when you take columns from A$^T$A and multiply them by A, you get orthogonal vectors.

Note that because A is orthogonal, when you say

I am asked to determine a orthogonal basis $(v_1,v_2,v_3)$ of $\mathbb R^3$, such that $Av_1,Av_2,Av_3$ are pairwise orthogonal.

this has a trivial solution: just take the elementary vectors. When you multiply these by A, you get the columns of A, which are orthogonal.