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 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?

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


*

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

A: 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.
