3
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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

  • $v_j^T A^TA v_i=v_j^T D v_i=d_{ii}v_j^Tv_i=0$
$\endgroup$
  • $\begingroup$ if $A^TA$ is not diagonal, this does not hold in general? $\endgroup$ – user519338 Apr 6 '18 at 15:34
  • $\begingroup$ @J.D Yes it doesn't work in general. $\endgroup$ – gimusi Apr 6 '18 at 15:37
  • $\begingroup$ but it is always satisfied if the eigenevctors of $A^TA$ are pairwise distinct? $\endgroup$ – user519338 Apr 6 '18 at 15:38
  • $\begingroup$ @J.D in this case it works for any orthogonal basis since $A^TA$ is diagonal since $v_j^T A^TA v_i=v_j^T D v_i=d_{ii}v_j^Tv_i=0$ $\endgroup$ – gimusi Apr 6 '18 at 15:39
  • $\begingroup$ @J.D Not necessarly $A^TA$ is always symmetric but in general it is not diagonal. $\endgroup$ – gimusi Apr 6 '18 at 15:42
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Another perspective comes from considering the singular value decomposition $A = U \Sigma V^T$ of $A$. $\endgroup$ – Omnomnomnom Apr 6 '18 at 15:31
  • $\begingroup$ if we consider the SVD, then the entries of $\Sigma$ are the eigenvalues of $A^TA$ and I think they are allways distinct. So we can say this will work always unless we have one or more singular values which are equal to zero? Is this thought correct? $\endgroup$ – user519338 Apr 6 '18 at 15:42
  • $\begingroup$ The eigenvalues of $A^TA$ are not always distinct. However, $A^TA$ is always orthogonally diagonalizable by the spectral theorem since it is symmetric $\endgroup$ – Omnomnomnom Apr 6 '18 at 15:43
  • $\begingroup$ @J.D I don't understand your question $\endgroup$ – Omnomnomnom Apr 6 '18 at 15:48
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The usual definition of an orthogonal matrix $A$ is that $A^TA=I$, not simply that this product is diagonal. $\endgroup$ – amd Apr 6 '18 at 20:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy