Suppose there are 2 orthogonal vectors $v_1$ and $v_2$ so $v_1^Tv_2=0$. Suppose we take linear transformations of the two vectors. Let $A$ be the transformation matrix. Let $u_1=Av_1; u_2=Av_2;$ Are $u_1$ are $u_2$ orthogonal? i.e. is $u_1^Tu_2=0$? How to prove it. I tried the following way, but I'm stuck.


I'm stuck here, unable to proceed further.

If the above equation is not true for any random vectors $v_1$ and $v_2$, is it true for any specific vectors? What if $v_1$ and $v_2$ are in $Row$ $space$ $of$ $A$?


It is not true in general, but it will be true if $A^TA=\lambda\operatorname{Id}$, for some scalar $\lambda$. In particular, it is true if $A$ is an orthogonal matrix ($\iff A^TA=\operatorname{Id}$).

  • $\begingroup$ What if $v_1$ and $v_2$ are in row space of A? Is it true then? $\endgroup$ – Nagabhushan S N May 19 '18 at 8:17
  • $\begingroup$ @NagabhushanSN No, not even then. $\endgroup$ – José Carlos Santos May 19 '18 at 8:23
  • $\begingroup$ thank you. I was watching Video Lecture by Gilbert Strang on Singular Value Decomposition. In that he takes two orthonormal vectors $v_1$, $v_2$ in row-space of A. Let $\sigma_1$*$u_1$=A*$v_1$ and $\sigma_2$*$u_2$=A*$v_2$. He takes $u_1$ is orthogonal to $u_2$. How? $\endgroup$ – Nagabhushan S N May 19 '18 at 8:26
  • $\begingroup$ @NagabhushanSN Post this as another question, with a link to the video. $\endgroup$ – José Carlos Santos May 19 '18 at 8:29
  • $\begingroup$ here it is math.stackexchange.com/q/2777511/516816 $\endgroup$ – Nagabhushan S N May 19 '18 at 8:36

Not it is not true in general, just consider $e_1=(1,0)$, $e_2=(0,1)$ and


It is true when $A^TA=k\,I$, indeed

$$\langle Ae_i,Ae_j \rangle=e_i^TA^TAe_j=ke_i^Te_j=\langle e_i,e_j \rangle=0$$


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